OP AMP Application Part 1 (Electronics 2) — Transcript

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1. 0:00hi hello um this video is
2. 0:03um about op amp or operational
3. 0:05amplifiers
4. 0:07practical applications okay so to start
5. 0:10with
6. 0:11um we have here the negative uh sorry
7. 0:14the inverting amplifier or what we call
8. 0:16the negative scaler
9. 0:17okay so um inverting amplifier so your
10. 0:21operational amplifier can work as an
11. 0:23inverting amplifier or a negative scaler
12. 0:26so this is a closed loop gain of an in
13. 0:30of the inverting amplifier depends on
14. 0:32the ratio
15. 0:33of the two external resistors which are
16. 0:36your
17. 0:36r1 and your rf okay and it is
18. 0:39independent of the amplifier parameters
19. 0:42okay so that means um uh it doesn't
20. 0:44matter if your amplifier or your op amp
21. 0:47is 741 or other type of op amps okay
22. 0:50but when we connect it in an inverting
23. 0:53amplifier mode
24. 0:54so the gain will only be determined by
25. 0:57your r1 and your
26. 0:59rf okay so it means it scales the input
27. 1:03or it multiplies an input so it's like a
28. 1:06multiplier okay so it scales the input
29. 1:09it multiplies the input by a minus
30. 1:12constant factor
31. 1:14of k okay so this is our
32. 1:18um this is our representation for an
33. 1:21inverting amplifier so this is your
34. 1:22feedback resistance and this is your r1
35. 1:25or the internal resistance so for
36. 1:27inverting amplifier our input
37. 1:30is being placed or being um uh
38. 1:34being applied at the inverting terminal
39. 1:36so for
40. 1:37741 our inverting terminal is your pin
41. 1:40number
42. 1:41two okay so your non-inverting input
43. 1:44is grounded so our formula
44. 1:48for the voltage gain is uh the output
45. 1:50voltage divided by your
46. 1:52input voltage
47. 1:55okay so um that is equivalent to
48. 1:59negative rf all over one so from our
49. 2:01previous video we have already derived
50. 2:04why it became um negative rf all over r1
51. 2:08so your rf all over r1 can be um
52. 2:12represented by a constant okay so let us
53. 2:15let us call that k
54. 2:16so that means your input voltage can be
55. 2:20multiplied
56. 2:21k times okay so k times but we have a
57. 2:24negative sign here
58. 2:26showing that there is a 180 degrees out
59. 2:29of phase
60. 2:30between your input and your output
61. 2:32signals
62. 2:33okay so uh
63. 2:37another application for our operational
64. 2:40amplifier
65. 2:41is as a non-inverting amplifier or what
66. 2:44we call the positive scalar
67. 2:46okay so this is used when there is a
68. 2:49need for an output
69. 2:50which is equal to the input multiplied
70. 2:53by a positive
71. 2:54constant okay so positive scalar circuit
72. 2:58which uses negative feedback but
73. 3:01provides an
74. 3:01output equals the input multiplied by
75. 3:04opacity of
76. 3:05constant so there is uh no 180 degrees
77. 3:09phase reversal or your output is in
78. 3:11phase
79. 3:12with your input okay so this is our
80. 3:14representation for a non-inverting
81. 3:16amplifier
82. 3:17so we apply your input voltage as at the
83. 3:21non-inverting terminal so if we are
84. 3:23using the 741 operational amplifier
85. 3:26this is your pin number three okay so
86. 3:29this is
87. 3:29still your feedback resistance which is
88. 3:32your rf and this
89. 3:33is still your r1 however our
90. 3:37inverting terminal here is
91. 3:40grounded okay so our voltage gain
92. 3:43still equal to v output divided by v
93. 3:47input and our um value for our
94. 3:51gain is equal to r1 plus rf divided by
95. 3:55r1 or we can simplify that as
96. 3:57one plus rf all over r1
97. 4:00so this is our gain for a non-inverting
98. 4:03amplifier okay now um we have this
99. 4:07cast that when your input is placed at
100. 4:10the
101. 4:10at the non-inverting terminal okay
102. 4:14so our inverting terminal
103. 4:19uh if our input is placed at our
104. 4:21inverting terminal our non-inverting
105. 4:23terminal has no resistance or it is
106. 4:26simply grounded
107. 4:27or if our input was placed in the
108. 4:30non-inverting terminal
109. 4:31the the input resist the input voltage
110. 4:34is directly
111. 4:35coupled to your operational amplifier
112. 4:38however in practice we have this what we
113. 4:41call bias compensating resistance so if
114. 4:44you can still remember
115. 4:45from our previous um video okay so there
116. 4:49are such things
117. 4:50as uh our input uh bias currents
118. 4:54and our input offset currents that are
119. 4:56already within your
120. 4:58integrated uh within your operational
121. 5:01amplifier
122. 5:01so what does that mean so it means that
123. 5:04um even though we do not have
124. 5:05any input yet we can already read
125. 5:08something at the output
126. 5:10okay so there is already an output
127. 5:12voltage even though we don't have any
128. 5:13input signal
129. 5:14and we don't like that okay so um
130. 5:18but we cannot totally remove those
131. 5:20because those are already
132. 5:22inside okay or within your operational
133. 5:25amplifier
134. 5:26so what do we with what can we do is by
135. 5:29applying a potentiometer okay so that is
136. 5:32our nulling resistance
137. 5:34and another thing is by applying what we
138. 5:37call the bias compensating
139. 5:40resistance okay or our rc
140. 5:43okay so our rc represents
141. 5:46all or a portion of the resource
142. 5:48resistance
143. 5:49so the exact value of rc is not critical
144. 5:52because we do not usually make use of
145. 5:54your rc in computing for your gain
146. 5:57okay and the closest standard value is
147. 6:00usually sufficient
148. 6:02okay so for example this is now our
149. 6:05inverting um
150. 6:06amplifier circuit previously this is
151. 6:09just grounded
152. 6:10but in uh in actual okay in practical
153. 6:13applications
154. 6:14we place here an rc so what is the value
155. 6:17of our rc in our design
156. 6:19so your gain will still be um will still
157. 6:22be dictated by the value of the rf and
158. 6:25our
159. 6:25r1 here but your rc here is just the
160. 6:28parallel combination of your rf and your
161. 6:32rl so in case there is no exact standard
162. 6:35value
163. 6:37for this parallel combination so it is
164. 6:39okay we just have to choose the closest
165. 6:42standard value okay so this is now the
166. 6:45new
167. 6:45um representation or the new schematic
168. 6:48diagram for our
169. 6:50non-inverting amplifier so um previously
170. 6:54we don't have any resistor here it's
171. 6:56just directly
172. 6:57the input is just directly connected to
173. 6:59the non-inverting terminal of your op
174. 7:02amp but uh in practical use we place a
175. 7:05biasing compensating resistance
176. 7:07and that is still equivalent to the um
177. 7:11parallel combination of your rf and your
178. 7:14ri
179. 7:14okay so we have here some design
180. 7:18considerations okay
181. 7:20so a summary of some of the major dis
182. 7:22considerations of the simplified
183. 7:25design approach to operational amplifier
184. 7:28small signal amplifier circuits
185. 7:30are as follows so number one be sure
186. 7:33that the dynamic
187. 7:34range of the op amp based on power
188. 7:37supply levels with some back off
189. 7:39will be sufficient for the desired
190. 7:42output signal
191. 7:43so that this uh if you can still
192. 7:45remember for a 741
193. 7:48you can um you can make use of
194. 7:51a power supply having a value of
195. 7:54positive or negative 12 volts so since
196. 7:57your operational amplifier
197. 7:59is dual um dual supply so you have a
198. 8:02positive and a negative voltage
199. 8:04so it can be positive 12 okay plus or
200. 8:07minus 12 it can be plus or minus 9 volts
201. 8:10it can be plus or minus 15 plus or minus
202. 8:1318 okay so in your design you should be
203. 8:17um you should be aware okay
204. 8:20on the actual power supply level so for
205. 8:23example
206. 8:23you are you need in your design to have
207. 8:26an output equivalent to
208. 8:2812 volts okay so if
209. 8:31your output must be equivalent to 12
210. 8:34volts
211. 8:34you will not be using any value which is
212. 8:37lower than 12 volts or even
213. 8:39exact 12 volts why because uh
214. 8:42we we discussed um on our previous uh
215. 8:46video that there are some back off
216. 8:50okay for that specific voltage so if we
217. 8:53have a
218. 8:54for example 12 maybe we can
219. 8:57have a maximum of output of just 10
220. 9:00volts okay so if you will be needing 12
221. 9:02volts
222. 9:03it will not be sufficient so if we will
223. 9:05be needing an output which is equal to
224. 9:0712 volts
225. 9:08better we make use of a supply which is
226. 9:11equivalent to plus or minus 15 or plus
227. 9:14or minus
228. 9:1418 and so on okay so your design must be
229. 9:18practical so your design must be um
230. 9:23what you call this you have to use um
231. 9:27the proper supply power supply levels in
232. 9:30your desired
233. 9:31output signal okay so whenever number
234. 9:34two whenever possible choose resistance
235. 9:36values in the range of
236. 9:38one kilo ohms to 100 kilo ohms
237. 9:41okay so our our
238. 9:46ideal value from our previous video
239. 9:50was 10 kilo ohms okay so
240. 9:53except avoid resistances that load the
241. 9:57output in the lower portion of the range
242. 9:59that means values of 10 kilo ohm
243. 10:03and greater are most widely used for
244. 10:05many applications because if we make use
245. 10:07of a lower value
246. 10:09um our
247. 10:12back off in our power supply will also
248. 10:17become lower
249. 10:18okay so if if our 15 volts for example
250. 10:22uh will give us an output of positive
251. 10:25plus or minus uh 13 volts um if we make
252. 10:29use of our resistance which is lower
253. 10:31than kilo ohms
254. 10:32we might get a lower supply valve
255. 10:35voltage okay
256. 10:36so number three you have to limit the
257. 10:38magnitudes
258. 10:39of closed loop gain to values much
259. 10:42smaller than the available open loop
260. 10:44gain so that's the main purpose of our
261. 10:46feedback resistance
262. 10:48okay so that is the the reason why we
263. 10:51place a
264. 10:52feedback resistance so that we can lower
265. 10:55down
266. 10:56our open loop gain which is equivalent
267. 10:59to 100
268. 11:00000 for an operational amplifier so
269. 11:03typically closed loop gain magnitudes
270. 11:06are kept under 100
271. 11:08okay and finally we have to check the
272. 11:11maximum output current levels under
273. 11:13worst case condition
274. 11:15okay so each operational amplifier has
275. 11:18its maximum um
276. 11:21output current so four seven four one so
277. 11:24it is 25 milli amperes so
278. 11:27we might uh we might um
279. 11:31over uh we have we might uh use
280. 11:34okay a maximum current greater than
281. 11:38its uh level its maximum level
282. 11:42okay which can um destroy or ruin your
283. 11:46operational amplifiers okay so for the
284. 11:49resistor values
285. 11:50um this are the most widely used so
286. 11:54this are you you must at least be
287. 11:56familiar
288. 11:57with this values okay so this is the
289. 11:59most widely used resistors okay so
290. 12:02they are low powered actual lead carbon
291. 12:05composition and film resistors with
292. 12:07tolerance of 5
293. 12:09and ten percent okay all the values
294. 12:12listed are available
295. 12:13in five percent tolerance but only the
296. 12:15bold face values are available in
297. 12:1810 tolerances okay so if you can still
298. 12:22remember your
299. 12:23um resistor color coding okay so what do
300. 12:26we mean by this so when we say
301. 12:2810 so this is available in one kilo ohm
302. 12:3210 kilo ohms 100 kilo ohms okay so 10k
303. 12:36um one mega okay so those are in those
304. 12:39following range so you have
305. 12:401.5 k okay so you have
306. 12:44um 15k okay so 150 kilo ohms like that
307. 12:50okay so
308. 12:52for other applications practical
309. 12:55applications such as instrumentations
310. 12:57okay we're in
311. 12:58we are very particular with our design
312. 13:01and we do not want any variations or
313. 13:05high variations in our design we also
314. 13:08have
315. 13:08resistor values in one percent
316. 13:11tolerances okay so
317. 13:13if we're going to make use of this
318. 13:14resistor values so they have a tolerance
319. 13:17of one percent only so therefore
320. 13:19you have of uh the the possibility of
321. 13:23getting an accurate circuit is higher
322. 13:27because your tolerance is very very
323. 13:29small okay so usually these are being
324. 13:31used for instrumentation
325. 13:33purposes so the design problem actually
326. 13:37design problems are not very very
327. 13:39difficult because
328. 13:40when you say inverting amplifier you
329. 13:42have already the design for an
330. 13:44inverting amplifier you know already
331. 13:47what does an inverting amplifier looks
332. 13:49like
333. 13:49okay so when we say design actually
334. 13:53you're just um placing your value
335. 13:57of rf and r1 in such a way that you are
336. 14:01um
337. 14:02you are asked for
338. 14:05okay so what what must be the value of
339. 14:08rf and r1 that is being required in the
340. 14:12design okay
341. 14:13so we have here an example using a five
342. 14:17percent
343. 14:18resistance values okay design and
344. 14:20inverting so vcvs is a voltage
345. 14:23controlled voltage
346. 14:24source um amplifier okay
347. 14:27to provide a voltage gain of negative 10
348. 14:30so we want a voltage gain of negative 10
349. 14:33and what are some other considerations
350. 14:36so it says here that the input impedance
351. 14:39must be no less than
352. 14:4010 kilo ohms okay so therefore our
353. 14:43minimum requirement is 10 kilo ohms
354. 14:46to avoid loading the given signal source
355. 14:49okay so the signal level will be
356. 14:52sufficiently high
357. 14:53to the thermal resistance noise and the
358. 14:56bias current effects are
359. 14:58not felt to be a critical problem
360. 15:00provided that the resistance
361. 15:02values less than 500 kilo ohms are used
362. 15:05so therefore we are given a very broad
363. 15:08range of our resistances that we can
364. 15:11actually
365. 15:12make use of okay so it just says here
366. 15:15that
367. 15:15it must not be less than 10k okay
368. 15:18or it it must not be greater than 500k
369. 15:22okay so but as long as it is within the
370. 15:24range your design is
371. 15:27is correct okay so you don't have to
372. 15:30worry
373. 15:30as long as you can comply with the
374. 15:33requirements okay
375. 15:35so this is an example this is just one
376. 15:37example
377. 15:38of the design for that specific problem
378. 15:42because our um
379. 15:46because our um requirement there
380. 15:49is uh you have to have a gain
381. 15:52equivalent to negative 10 so since there
382. 15:55is a negative
383. 15:56already in our requirement so we already
384. 16:00know that you are going to make use of
385. 16:02an inverting amplifier
386. 16:03because if you have an inverting
387. 16:05amplifier automatically you get the
388. 16:07negative
389. 16:08gain there okay so the only problem is
390. 16:10what combination of rf and ri
391. 16:12are you going to make use of so that you
392. 16:15can have a value which is equivalent to
393. 16:1710 okay so our um
394. 16:21our formula for the voltage gain for
395. 16:24your
396. 16:24inverting amplifier is equivalent to
397. 16:27negative rf all over r1
398. 16:30so you must must use a value of rf up
399. 16:35and r1 in such a way that if we divide
400. 16:38them
401. 16:38your answer will be equivalent to 10 so
402. 16:41in this case
403. 16:42it make use of 220 kilo ohms
404. 16:46all over um 22
405. 16:49kilo ohms so automatically we have the
406. 16:52negative sign there so 220 divided by
407. 16:55um 22 kilo ohms so that will give us
408. 16:58again which is equivalent to
409. 17:00negative 10 which is your required
410. 17:04gain in the in the problem in your
411. 17:07design problem
412. 17:08okay so the 20 kilo ohms there okay
413. 17:11where did it came from
414. 17:12so the 20 kilo ohms there is your um
415. 17:15compensating resistance
416. 17:17and your compensating resistance is just
417. 17:19the
418. 17:20parallel combination of your rf which is
419. 17:23220 kilo ohms
420. 17:25parallel your 22 kilo ohms here
421. 17:28okay so if you solve for that so that is
422. 17:31220 k
423. 17:33times 22 k divided by
424. 17:36220 k plus 22 k
425. 17:41so that will give you an exact value of
426. 17:4420
427. 17:45kilo ohms so luckily um these are all
428. 17:48available
429. 17:49so if we're going to look back
430. 17:54okay so if we're going to look back at
431. 17:57our resistor values where is that
432. 18:00so this is your value 20 220 kilo ohms
433. 18:04and make use of
434. 18:0522 kilo ohms actually we can make use of
435. 18:09where from from this 22 up to 51 here
436. 18:14okay so we cannot uh we can uh
437. 18:17we can also use this one so we can use
438. 18:19100 kilo ohms divided by 10k
439. 18:22okay so this is included in our
440. 18:27uh in in uh this is included in our um
441. 18:30option okay so you can also make
442. 18:32yourself 150 kilo ohms
443. 18:34and 15 kilo ohms for your r1 so that can
444. 18:37still give you a gain equivalent to 10
445. 18:40so you can make use of um 470 kilo ohms
446. 18:43and 47 kilo ohms
447. 18:45that will still give you a gain which is
448. 18:47equivalent to negative 10
449. 18:49okay but can we use the 56 here so if we
450. 18:52make use of the 56 here
451. 18:54you can have um 560 kilo ohms for your
452. 18:59rf and 56 kilo ohms for your
453. 19:01r1 however it says in the problem that
454. 19:04we must not exceed 500 kilo ohms
455. 19:08okay so therefore we are only permitted
456. 19:10to use
457. 19:11up to 47 470 kilo ohms and 47 ohms okay
458. 19:16so
459. 19:16so in a design type of a problem class
460. 19:20you can have several answers okay so you
461. 19:23can have several
462. 19:24answers as long as you follow all of the
463. 19:27conditions
464. 19:28okay that is asked in the design problem
465. 19:31so no problem so
466. 19:33there is no such thing as a wrong design
467. 19:35as long as you follow
468. 19:37all the specific conditions okay
469. 19:41so we have here um using the standard
470. 19:44one percent resistance value
471. 19:46design of vcvs non-inverting okay so
472. 19:50what do we mean by non-inverting
473. 19:51amplifier so in this case
474. 19:53we have to place your input at the
475. 19:56non-inverting terminal okay
476. 19:58so to provide a voltage gain of 18 so
477. 20:01remember
478. 20:02that for uh for a
479. 20:05non-inverting amplifier your gain
480. 20:08okay so your voltage gain is equal to rf
481. 20:11all over r1 plus one so we need to get
482. 20:16a combination okay so you have to have a
483. 20:19combination
484. 20:21where in your rf all over r1
485. 20:25must be equivalent to 17 okay so that if
486. 20:29we add them
487. 20:30your resulting gain is 18 and the
488. 20:33positive answer there
489. 20:35or the positive gain there will be um
490. 20:38will be
491. 20:39uh what you call this uh
492. 20:44will be possible if we make use of a
493. 20:46non-inverting because the gain for a
494. 20:48non-inverting amplifier is
495. 20:50positive okay so in this case we have an
496. 20:53example
497. 20:54so we have here um two kilo ohms we have
498. 20:5734 kilo ohms and we have here
499. 20:59uh 1870 ohms okay
500. 21:02so if we divide 34 kilo ohms by
501. 21:062 so we will be having 17 uh
502. 21:09for that okay then 17.
503. 21:12so we have here um uh
504. 21:1634 kilo ohms divided by 2 kilo ohms so
505. 21:19that will give you 17 plus one so that
506. 21:22will be
507. 21:23equivalent to
508. 21:27so that will be equivalent to 18
509. 21:30okay and then for the compensating
510. 21:33resistance all we have to do is to
511. 21:35um uh get the parallel combination of
512. 21:38two and 34 okay which is eighteen
513. 21:41thousand one thousand eight hundred
514. 21:43seven ohms okay so if we're going to
515. 21:46look at those values here
516. 21:48so we have um 34 kilo ohms so where is
517. 21:50your 34 kilo ohms
518. 21:52so this is your 34
519. 21:55kilo ohms so where is your 34 kilo ohms
520. 21:59this one so 340 so this is 34
521. 22:03and this is sure 2 kilo ohms so we're
522. 22:06sure 2 kilo ohms
523. 22:07here 200 and then we have the 1
524. 22:11okay so we have one eight seventy
525. 22:16one eight seventy so we have a one eight
526. 22:20seventy so luckily we have all
527. 22:22but uh when you get the parallel
528. 22:25combination and you didn't get any exact
529. 22:27value
530. 22:28it's okay all you have to do is to
531. 22:32get the one with the nearest value so
532. 22:35you can have any other answer okay so
533. 22:37you can
534. 22:37you can uh look at this value and then
535. 22:40as long as it will give you a gain of 18
536. 22:42that means you will have a ratio which
537. 22:44is equal to 17
538. 22:46therefore your answer is correct okay
539. 22:49so our non-inverting amplifier is the
540. 22:53counter part of your common base
541. 22:55configuration
542. 22:56for our um bjt amplifiers or the
543. 23:00common um source also common gate
544. 23:03configurations for
545. 23:05for fat okay and then the inverting
546. 23:07amplifiers
547. 23:10okay is the common emitter configuration
548. 23:14for bjt
549. 23:15or the common uh common source
550. 23:19configuration for
551. 23:20four okay so we have another
552. 23:24type of application so we're in we call
553. 23:27this
554. 23:27as your voltage follower or unity
555. 23:30follower
556. 23:31okay so voltage follower or unity
557. 23:33follower provides
558. 23:34gain of unity without any phase reversal
559. 23:38so what is your output what is your
560. 23:41input that will also be your output
561. 23:43okay so um
562. 23:46this is your spool as a buffer or
563. 23:49isolation amplifier
564. 23:51because it allows input voltage to be
565. 23:53transferred as output voltage
566. 23:56while at the same time preventing load
567. 23:58resistance from loading down the input
568. 24:00source
569. 24:01okay so in your logic circuits if you
570. 24:03have this is
571. 24:04the the opposite of an inverter so an
572. 24:07inverter will give you the same output
573. 24:10amplitude
574. 24:11okay or magnitude but with a phase
575. 24:13reversal
576. 24:14okay so if our input for example is
577. 24:16negative one your output is positive one
578. 24:19so for voltage follower or unity
579. 24:21followers so it is a buffer okay so
580. 24:24when you see block diagrams having a
581. 24:26buffer so you can make use of an
582. 24:28operational amplifier for that
583. 24:31so if our input is one millivolt your
584. 24:34output is also
585. 24:35one millivolt okay so this is just being
586. 24:38used for
587. 24:39electrical isolation purposes okay
588. 24:42so how do we how do we connect
589. 24:44operational amplifiers in such a way
590. 24:47that it will be uh functioning as a
591. 24:51voltage follower or a unity follower
592. 24:53so this is your
593. 24:56this is your connection okay all we have
594. 24:59to do
595. 25:00is to place an input at the in
596. 25:02non-inverting terminal because
597. 25:04uh we have to have the same phase
598. 25:06reversal so
599. 25:07our input must be placed in the input
600. 25:09terminal and then this is our
601. 25:12output okay so you just have to place a
602. 25:14ground
603. 25:15between your your inverting terminal
604. 25:18and your output for your feedback
605. 25:22so it is just like um
606. 25:26so it is just like that um for example
607. 25:30our input here is um
608. 25:34so for example we have here your upper
609. 25:37operational amplifier so let us call
610. 25:40this your va
611. 25:41okay and then let us call this your
612. 25:44vb where in your vb this is your
613. 25:47non-inverting side
614. 25:50so this is your input voltage okay
615. 25:53um this is your
616. 25:57feedback here and we have here your
617. 26:00output
618. 26:01okay so always remember that if we if we
619. 26:04make use of this configuration
620. 26:06uh your ri okay so your r
621. 26:09i is infinite okay so we don't have any
622. 26:13input here we don't have any input
623. 26:15resistance
624. 26:16so this is infinite and our feedback
625. 26:19resistance
626. 26:19is what this is shorted so this is
627. 26:22equivalent to
628. 26:24to zero okay so in this case we can say
629. 26:27that your va
630. 26:29um is equal to your
631. 26:32v out okay so if you are going to
632. 26:35remember
633. 26:36our voltage here virtual ground so the
634. 26:39difference between va and vb here must
635. 26:41be equal to
636. 26:42zero so to do that our va
637. 26:46must be equal to
638. 26:49our va must be equal to to vb
639. 26:52okay so but our vb is
640. 26:55equal to your input voltage okay which
641. 26:58is your
642. 26:59v in but your vb is equal to your v
643. 27:02in okay so your vb is equal to your ven
644. 27:06and your vb is equal to va so that is
645. 27:09your v in is equal to
646. 27:11your v out
647. 27:15so in solving for the voltage gain okay
648. 27:17so your voltage gain is v
649. 27:20out all over v in and since they are
650. 27:22equal your voltage gain is equal to
651. 27:25one okay so if you wanted a buffer
652. 27:28circuit so
653. 27:28in our communication subject we can see
654. 27:31block diagrams having buffers circuits
655. 27:34in their blocks
656. 27:35so one option is to make use of an
657. 27:37operational
658. 27:38amplifier okay so what is its
659. 27:41corresponding counterpart for bjt
660. 27:44so this is just like your emitter
661. 27:47follower or
662. 27:48our common collector configuration
663. 27:50because in our common collector
664. 27:51configuration or our ammeter
665. 27:54configuration this is that is in
666. 27:55phase and no voltage gain or your v out
667. 27:58is all
668. 27:59approximately equal to your to your v
669. 28:02in or for fat it is equivalent to your
670. 28:05common drain
671. 28:06configuration okay so that is the
672. 28:10that is the third practical application
673. 28:13of your op-amp so you just get your op
674. 28:15amp
675. 28:15you place an input okay so you put an
676. 28:18input voltage at the
677. 28:20non-inverting terminal and then you
678. 28:22place a wire
679. 28:23okay shorting your inverting terminal
680. 28:25and your output here
681. 28:27um you can have your inverter with you
682. 28:30okay so we can place some resistors here
683. 28:33so these resistors here are for
684. 28:35compensating resistors only okay but
685. 28:39anyways it will give you the same
686. 28:42um expected output okay
687. 28:48so another practical application for our
688. 28:52operational application
689. 28:54amplifier is other or summer
690. 28:57okay so what is other or summer
691. 29:01okay we have here um this provides an
692. 29:04output voltage
693. 29:05proportional to or equal to the
694. 29:07algebraic sum
695. 29:08of two or more input voltages okay
696. 29:12each multiplied by a constant gain
697. 29:14factor
698. 29:15so this is um basically similar to a
699. 29:18scalar
700. 29:19except that it has more than one input
701. 29:21okay so
702. 29:22for example this is our um circuit for
703. 29:26an adder or a summer so you will make
704. 29:28usually of a
705. 29:29com inverting amplifier but instead of
706. 29:32having just
707. 29:33one input we have several inputs okay
708. 29:36and then
709. 29:36our sum is the thus our
710. 29:40output rather is the sum of all the
711. 29:42input voltages here
712. 29:44multiplied by a certain factor
713. 29:47okay so how do we uh how do we evaluate
714. 29:51this type of
715. 29:53it
716. 29:57okay so for this type of circuit
717. 30:04okay so let's just say um we have
718. 30:08several currents that will pass through
719. 30:11your
720. 30:11operational amplifier okay so those
721. 30:14currents uh will can be added up
722. 30:18using um the total current which is your
723. 30:20i1
724. 30:22plus your i2 plus your i3 here
725. 30:25okay so we have here but what is your
726. 30:28i1 okay so our i1 here is equivalent to
727. 30:32um
728. 30:33your your input voltage v1 at that
729. 30:36terminal
730. 30:37okay divided by your certain resistance
731. 30:39r1
732. 30:40okay and then we have here your i2 but
733. 30:43what is your i2 there
734. 30:44so that is equivalent to the voltage
735. 30:46that is being applied across the
736. 30:48terminal
737. 30:50divided by your r2 and then what is your
738. 30:53i3 so the voltage across the terminal
739. 30:55which is your v3 divided by your
740. 30:58your your r3 here okay so that is our
741. 31:02answer okay so if you're going to
742. 31:04remember what is our v
743. 31:06out okay for a non for an inverting
744. 31:09amplifier because that configuration is
745. 31:11an
746. 31:11inverting amplifier because we place all
747. 31:13the inputs
748. 31:14at the n at the inverting terminal so
749. 31:17that will give us
750. 31:18minus rf all over r1
751. 31:22multiplied by our input
752. 31:25voltage so that is your output voltage
753. 31:29so if we combine them okay so if we
754. 31:31combine them
755. 31:32we can have your total output voltage
756. 31:35which is equivalent to minus okay
757. 31:40that is your rf since rf feedback
758. 31:42resistance is common
759. 31:44to all of your inputs so your rf there
760. 31:47will be so your rf there
761. 31:51is equivalent to sorry rf all over r1
762. 31:56multiplied by your input which is your
763. 31:58v1 we have here plus
764. 32:00okay rf all over r2 plus
765. 32:03multiplied by your v2 you have here plus
766. 32:07rf all over r3 multiplied by
767. 32:10v3 here okay so that
768. 32:14is depending on the value of your r1
769. 32:17okay but you have the same
770. 32:21but you have the same rf and r1
771. 32:24okay so if we wanted to simplify
772. 32:28okay so if we wanted to simplify your
773. 32:31circuit
774. 32:32where in each of the value of r are
775. 32:35equal
776. 32:36okay so let us say your r1 is equal to
777. 32:39your
778. 32:40r2 that is equal to your r3 and that is
779. 32:43equal to your
780. 32:44rf we can simplify this further by
781. 32:47having minus
782. 32:48okay r let us just call that r
783. 32:52all over r multiplied by
784. 32:56v1 okay so we have here your
785. 33:00um r all over r
786. 33:03multiplied by v2 plus let's say
787. 33:06um you have here
788. 33:10um r all over r multiplied by
789. 33:13v three okay since our r is constant
790. 33:16we can put it outside and that is r over
791. 33:19r so this is constant which is
792. 33:20equivalent to one
793. 33:22so that will give you a v1
794. 33:25plus v2 plus v3 which is our
795. 33:29output voltage okay so
796. 33:32as we can see what did we have here so
797. 33:35we have
798. 33:36a sum of all the input
799. 33:39voltages so v1 plus v2 plus v3
800. 33:42so this is your other okay so usually
801. 33:47in practical applications we have here
802. 33:50your mixers for example
803. 33:52so you you have your mixers for in for
804. 33:54example you have in up in uh
805. 33:56you you wanted to have a microphone and
806. 33:58then you wanted to have a guitar for
807. 34:01example or a piano for example and you
808. 34:03wanted to blend
809. 34:04the music or you wanted to amplify the
810. 34:08the microphone further so you can hear
811. 34:10the voice
812. 34:11compared with the with the auxiliary
813. 34:14input for example
814. 34:15so you can make use of a uh other or a
815. 34:18summing
816. 34:19amplifier okay so this is your
817. 34:22application for adding or summer
818. 34:26okay so this is if simple addition or
819. 34:29inversion is
820. 34:30desired okay so you're compensating
821. 34:32resistance here is
822. 34:34still equivalent to the
823. 34:37to the parallel combination of all the
824. 34:40resistors that you can find
825. 34:42in the circuit okay so that is the
826. 34:45definition for your compensating
827. 34:48resistance okay so what are
828. 34:52an example okay let's have an example
829. 34:57so we have here um in a certain
830. 34:59application
831. 35:00it is desired to combine two signals v1
832. 35:03and v2
833. 35:04to form a signal vo according to the
834. 35:07following combination
835. 35:08so your vo must be equivalent to
836. 35:11negative v1
837. 35:12minus 10 v2 okay so
838. 35:16the minimum the minimum um
839. 35:19input impedance for both inputs should
840. 35:22no less than 10k
841. 35:23so design a circuit to met to to meet
842. 35:27the
843. 35:29to meet the requirements okay so how are
844. 35:32we going to
845. 35:33solve for that
846. 35:38so how are you going to solve for that
847. 35:40so actually class
848. 35:42we have here your answer okay so you
849. 35:45have two inputs
850. 35:46okay so we just have to make sure
851. 35:53so we just have to make sure that
852. 36:06okay so we just have to make sure that
853. 36:09we can
854. 36:10have this output so vo
855. 36:13must be equivalent to negative v1 minus
856. 36:1610
857. 36:17v2 okay so since you share
858. 36:22the same rf okay so we have to make sure
859. 36:25that your v1
860. 36:27will have a gain which is equal to 1
861. 36:30so that is equivalent to 100
862. 36:34divided by 100 so that is equivalent to
863. 36:361
864. 36:38so this is negative okay so we have here
865. 36:41um
866. 36:43um 100 kilo ohms divided by 100 kilo
867. 36:48ohms so this will give you a gain which
868. 36:49is equivalent to one
869. 36:51times v1 and then you have to
870. 36:54uh choose a value that when you divide
871. 36:57it by 100
872. 36:58the answer will become 10. okay so that
873. 37:02is 100 kilo ohms here divided by 10k so
874. 37:06this
875. 37:06ratio here will give us a value which is
876. 37:09equivalent to 10
877. 37:11multiplied by our v2 so in this case
878. 37:16so that will give you minus v1 plus
879. 37:1910 v2
880. 37:25so that is your
881. 37:29design considerations
882. 37:32okay so if you might be wondering um can
883. 37:36i choose 200 here
884. 37:37yes but you have to use 200 here
885. 37:41okay and if you choose 200 here you have
886. 37:43to use
887. 37:4420 here okay as long as you can have
888. 37:47your uh you can have the required
889. 37:51design um
890. 37:54problem
891. 38:00okay so you have many options okay so
892. 38:03you have many options
893. 38:04now what is your
894. 38:07rc there okay so where did the
895. 38:118.33 kilo ohms came from so that is just
896. 38:14your
897. 38:16um
898. 38:19parallel 100k parallel 10 kilo ohms
899. 38:23so all the resistors that you can see
900. 38:26there in your sig
901. 38:27in your circuit that will be the
902. 38:30that will be or that will cover your
903. 38:33compensating
904. 38:34resistance okay so i have here
905. 38:38a problem and that will serve as your
906. 38:41activity okay so it says here in a
907. 38:46certain signal processing system
908. 38:48okay it is necessary to combine and
909. 38:51modify the levels of three signals v1 v2
910. 38:54and v3
911. 38:55so the desired vo is 2v1 plus
912. 38:585v2 minus 10 v3
913. 39:01okay so um what we need
914. 39:05is a non-inverting output
915. 39:09okay since this is positive but uh
916. 39:12it says here that you can make use of
917. 39:15linear combination circuits and or
918. 39:18inverting amplifiers only
919. 39:20okay so you might be wondering ah this
920. 39:23is positive
921. 39:24so i will be using a non-inverting
922. 39:26amplifier so that i can have a positive
923. 39:28answer
924. 39:30but it says in the problem that you are
925. 39:33to use
926. 39:33only inverting amplifiers okay um
927. 39:37and also so i need a positive output so
928. 39:41if that is an inverting i will be
929. 39:42placing a negative
930. 39:44input okay but it also says here that
931. 39:47you are not to use okay so do not use
932. 39:50negative voltages
933. 39:51as input so you have to use only
934. 39:53positive
935. 39:54input so you have to design okay so you
936. 39:58have to design
937. 40:00using inverting amplifiers okay so in
938. 40:03the design
939. 40:04no nothing is wrong as long as you can
940. 40:07get
941. 40:08this output voltage okay but you have to
942. 40:11take into considerations that you will
943. 40:12be using inverting amplifiers on
944. 40:15also only but you can have more more
945. 40:18than one
946. 40:19okay and you have to make use of
947. 40:21positive inputs only okay
948. 40:24so this is your uh task okay and
949. 40:27i want you to answer this uh
950. 40:31using our um using what you have learned
951. 40:35from our
952. 40:36discussion okay so this is one
953. 40:39application okay so this is one
954. 40:42application for
955. 40:43our operational amplifier okay
956. 40:46so it can be used as a positive scalar
957. 40:49it can be used as a
958. 40:50negative scalar it can be used as a um
959. 40:54unity gain okay or a buffer and then it
960. 40:57can be used as a summing amplifier
961. 40:59okay and then what will be our next
962. 41:04so our next is if it can add likewise it
963. 41:08can also
964. 41:08subtract okay so if can if it can add
965. 41:12it can also subtract okay so what are we
966. 41:15going to make use of where are we going
967. 41:18to place the input
968. 41:19so that we can
969. 41:22uh we can have your your
970. 41:25operational amplifier work as a
971. 41:28subtractor
972. 41:30okay so um a subtractor or a closed-loop
973. 41:33differential amplifier is used to
974. 41:36provide an output proportional to or
975. 41:38equal to the difference okay subtractor
976. 41:40to the difference of two input signals
977. 41:42so this time your inputs are supplied
978. 41:45both at the inverting and the
979. 41:47non-inverting terminals okay so
980. 41:49for summing amplifiers all of your
981. 41:52inputs are being placed only at the nun
982. 41:54at the inverting terminal so if we
983. 41:56wanted to make use of your op
984. 41:58amp as a subtractor therefore we have to
985. 42:01place an input
986. 42:02both to your inverting and to your
987. 42:04non-inverting terminal so how does this
988. 42:07look like
989. 42:08okay so this is your subtractor
990. 42:11okay so you have an input
991. 42:14so if you have a non-inverting amplifier
992. 42:18this v1 here is grounded if you have an
993. 42:21inverting amplifier this v2 here
994. 42:24is grounded okay if you have a summing
995. 42:27amplifier
996. 42:28um all of the inputs v1 and v2 are being
997. 42:31placed at the
998. 42:32inverting terminal if you have a buffer
999. 42:35okay
1000. 42:35our input is in this one and then we
1001. 42:37just have to place a wire between the
1002. 42:39inverting and the
1003. 42:41output terminal okay so for a subtractor
1004. 42:44we need to place an input both at the
1005. 42:46inverting terminal and the non-inverting
1006. 42:48terminal
1007. 42:49okay so how do we arrive with this
1008. 42:53uh v2 minus v1 here okay so
1009. 42:58if you're going to remember from your um
1010. 43:01from your circuit analysis okay so we
1011. 43:04have what we call the superposition
1012. 43:06theorem
1013. 43:07so what does the superposition theorem
1014. 43:11um function so how does the super
1015. 43:15positions during functions so we're just
1016. 43:17going to work on only one
1017. 43:20input terminal at a time okay so that
1018. 43:22means
1019. 43:23when we are dealing with the with your
1020. 43:25v1
1021. 43:26so we are going to uh place your v2
1022. 43:29grounded
1023. 43:30and at the same time when we're dealing
1024. 43:32with veto already
1025. 43:33we are going to ground your v1 okay so
1026. 43:36let us start first with v1
1027. 43:40so when we are working on v1 okay so
1028. 43:43that means your v2
1029. 43:44is grounded so when your v2 is grounded
1030. 43:48okay so when your v2 is grounded it's
1031. 43:50just like you're working on
1032. 43:52as as an inverting amplifier
1033. 43:56so when you are working as an inverting
1034. 43:58amplifier
1035. 43:59okay so v1 is acting alone
1036. 44:03so that is an inverting amplifier so
1037. 44:05that will give
1038. 44:06us an output so let us call that output
1039. 44:09as v01
1040. 44:10okay so your output is a non in
1041. 44:13is an inverting amplifier and that is
1042. 44:15equivalent to rf
1043. 44:17all over r1 multiplied by your
1044. 44:20v1 okay so this is just like an
1045. 44:23inverting amplifier
1046. 44:28so this is just like an inverting
1047. 44:30amplifier
1048. 44:32okay so what if v2 acts alone
1049. 44:36so that means we are going to ground
1050. 44:38your v1 so if we ground your b1 v1
1051. 44:41it is just like working as a
1052. 44:43non-inverting
1053. 44:46amplifier so when you have a
1054. 44:50non-inverting amplifier
1055. 44:51okay let us call this vo2
1056. 44:55let us call this vo2 and that will give
1057. 44:58you rf
1058. 44:58all over r1 plus 1 okay multiplied by
1059. 45:03v2 okay so for superposition through
1060. 45:07them up uh
1061. 45:08upon getting the individual answers
1062. 45:12we are going to combine them so we're
1063. 45:14going to either add them or subtract
1064. 45:16them
1065. 45:16but since this is of negative sign so if
1066. 45:18we add them
1067. 45:19we will be having a negative sign here
1068. 45:22okay so for simplification
1069. 45:24for a simplification purposes we can
1070. 45:27just assume that your rf
1071. 45:28plus r1 rf all over r1 plus 1 is
1072. 45:32approximately equal to rf all over r1
1073. 45:35okay so combining the two so we have
1074. 45:37your v out
1075. 45:38and your v out is equivalent to vo2
1076. 45:42minus vo1 okay because of the negative
1077. 45:46sign there
1078. 45:47okay or plus so vo2 is um rf
1079. 45:51all over r1 times v2 plus
1080. 45:54okay minus but that's a negative sign so
1081. 45:56we have rf
1082. 45:57all over r1 multiplied by v1
1083. 46:00okay so factoring out your rf all over
1084. 46:03r1
1085. 46:04we have here an answer which is
1086. 46:06equivalent to v2
1087. 46:08minus v1 okay so if we are going to
1088. 46:14if we are going to simplify this further
1089. 46:18and let us say rf will be equal to r1
1090. 46:21so we will be getting an output which is
1091. 46:23the difference
1092. 46:25okay so we will be getting an output
1093. 46:27which is the difference between your two
1094. 46:29input voltage
1095. 46:30which is your v2 minus v1 so we have
1096. 46:34a subtractor okay so that is
1097. 46:37how your operational amplifier can be
1098. 46:41um can be applied as
1099. 46:44an as a subtractor so as you can see
1100. 46:48i'm just changing where are you going to
1101. 46:51place your
1102. 46:52input you have several applications
1103. 46:56for your operational amplifier so that
1104. 46:58means your operational amplifier class
1105. 47:00is very versatile
1106. 47:02okay and it can add it can subtract
1107. 47:06okay so actually it was called
1108. 47:08operational amplifiers
1109. 47:09because it it was really designed it was
1110. 47:12really first designed
1111. 47:14to um to perform
1112. 47:17mathematical operations that's why it
1113. 47:19was called operational
1114. 47:21amplifiers okay so this is a very simple
1115. 47:24type of a subtractor
1116. 47:26however in practical use in practical
1117. 47:30use
1118. 47:30we have here your practical application
1119. 47:34for a
1120. 47:35subtractor okay so this is
1121. 47:38what we call this is how we uh
1122. 47:43this is how we design or how we use your
1123. 47:47um uh
1124. 47:51actual subtractor class okay so where it
1125. 47:53we have four
1126. 47:54resistors here okay so this is like a
1127. 47:58voltage divider for your v1
1128. 48:00and your v2 okay so for example um
1129. 48:04we have here your v2 alone for example
1130. 48:07uh if we ground your v1 here so it is
1131. 48:09working as an uh
1132. 48:11as a non-inverting amplifier so we can
1133. 48:14get your
1134. 48:15first output which is minus r4 which is
1135. 48:18your feedback
1136. 48:19all over r3 multiplied by our
1137. 48:24v2 here okay and if your v1
1138. 48:27is acting alone and your v2 is grounded
1139. 48:30okay so the input voltage will be
1140. 48:34uh so the input v1 here will be divided
1141. 48:37between
1142. 48:38r1 and r2 and the input is the voltage
1143. 48:41across your r2
1144. 48:42because this will serve as a uh voltage
1145. 48:45divider
1146. 48:46okay so for vo2
1147. 48:50okay so let us call that vo2 so that is
1148. 48:53a non-inverting amplifier so we have one
1149. 48:56plus
1150. 48:56r4 all over r3 okay
1151. 49:00um by the input voltage but what is the
1152. 49:04input voltage so the input voltage is
1153. 49:06the voltage across r2
1154. 49:08so we have a voltage divider between r1
1155. 49:11and r2 multiplied by our
1156. 49:14v1 here okay so that is your
1157. 49:19that is your v2 so to get the total
1158. 49:22value so to get the total output voltage
1159. 49:26so that is a combination of your um
1160. 49:30r3 1
1161. 49:33plus r4 all over r3 multiplied by
1162. 49:37r2 all over r1 plus r2
1163. 49:40multiplied by v1 so minus
1164. 49:44you have here r4 all over r3 multiplied
1165. 49:47by
1166. 49:49by v2 okay so just what we did earlier
1167. 49:53this time we just have your voltage
1168. 49:55divider configuration here
1169. 49:57okay so for instrumentation purposes we
1170. 50:00make use of
1171. 50:01the balance closed loop differential
1172. 50:03amplifier
1173. 50:04so the balance closed loop differential
1174. 50:07amplifier is the most important case of
1175. 50:09closed loop differential amplifier
1176. 50:11so it is said to be balanced if we get
1177. 50:14if we make use of these values
1178. 50:16so your r here species specifies a
1179. 50:20certain
1180. 50:21value that are equal okay and
1181. 50:24this a here specifies again
1182. 50:28okay so uh the two gains are equal
1183. 50:31and opposite so how are we going to
1184. 50:34attain this balance
1185. 50:35close loop differential amplifier if
1186. 50:38we're going to follow this
1187. 50:40suggested values for r1 r2 r3 and
1188. 50:43r4 okay so if we're going to um
1189. 50:48if we're going to
1190. 50:51substitute this value in our equation
1191. 50:55here
1192. 50:56so if we're going to substitute those
1193. 50:57values in our equation
1194. 50:59so that will give us okay so your v out
1195. 51:02is equal to 1 plus what is supposed to
1196. 51:05be the value of
1197. 51:06r4 r4 is equivalent to a
1198. 51:09r3 or let us call that ar divided by
1199. 51:12r3 which is equivalent to r okay and
1200. 51:15this is equivalent to what is our r2
1201. 51:18r2 is equivalent to ar all over
1202. 51:21r1 is equal to r plus r2 is equivalent
1203. 51:25to
1204. 51:26ar okay multiplied by v1
1205. 51:29so this is r4 r4 is equivalent to
1206. 51:32a r
1207. 51:36ar all over r multiplied by v2
1208. 51:40okay so we have here um a
1209. 51:43r here will be cancelled out so i have
1210. 51:46here 1 plus
1211. 51:47a and then i have here a r
1212. 51:50i can factor out your r here so this is
1213. 51:53equivalent to 1
1214. 51:54plus a okay so i can cancel out your
1215. 51:58r here and i can cancel out your 1 over
1216. 52:01a
1217. 52:01i here so i have a remaining a there for
1218. 52:04my
1219. 52:04v1 minus i can cancel out your r there i
1220. 52:08have here
1221. 52:09a multiplied by your v2 okay so your a
1222. 52:13can be factored out so this is simply v1
1223. 52:16minus v2 so our v out for a balanced
1224. 52:20amplifier is simply the difference
1225. 52:24okay so that is simply the difference
1226. 52:28between v1 and v2
1227. 52:31multiplied by a certain gain but you
1228. 52:35have to make sure
1229. 52:36okay but you have to make sure that you
1230. 52:38follow
1231. 52:39this specific resistor requirement
1232. 52:43okay so this is the derivation
1233. 52:50so we have here an example a circuit is
1234. 52:52desired for an
1235. 52:53instrumentation application in which the
1236. 52:56difference between two signals p1 and v2
1237. 52:59is to be amplified by a factor of 10 so
1238. 53:02that
1239. 53:02is 10 v1 minus v2 design a circuit to
1240. 53:06perform the required function
1241. 53:08so the required function is that your a
1242. 53:16so your a is equivalent to 10
1243. 53:19okay so so we have to choose a value for
1244. 53:23r okay which are similar to every values
1245. 53:27of
1246. 53:27our four resistors that is in our
1247. 53:29circuit
1248. 53:30having a resistance which is equal to 10
1249. 53:33okay sorry again which is equal to 10 so
1250. 53:36what is our
1251. 53:37um answer for that so we can make use of
1252. 53:41this one okay
1253. 53:42so in this case your r3
1254. 53:46is equal to r okay and your r4 is equal
1255. 53:50to ar
1256. 53:51our gain is 10 okay so 10 times 10
1257. 53:55so we have here 10 100 kilo ohms so our
1258. 53:58r1 is equal to r so let's say this is 10
1259. 54:01kilo ohms
1260. 54:02and then this is equal to 10 times r
1261. 54:05which is equivalent to 100 kilos
1262. 54:07so can we make use of 20 yes we can make
1263. 54:10use of 20. so this is 20k
1264. 54:13this is 200k so this is 20k this is 200k
1265. 54:17so we can make use of 30. so this is 30k
1266. 54:20this is 300k this is
1267. 54:2210 30k this is 300 kilos so that is
1268. 54:25what uh uh balance close
1269. 54:29loop differential amplifier operates
1270. 54:32okay
1271. 54:32so don't forget the don't don't be
1272. 54:36confused with the circuit because the
1273. 54:38circuit is just the same
1274. 54:40so when you say this sign all you have
1275. 54:42to do is to change
1276. 54:44the value of your resistances
1277. 54:47okay so there are more upper
1278. 54:50mathematical operation so what else can
1279. 54:53uh can
1280. 54:54an operational amplifier do so aside
1281. 54:56from subtracting aside from adding aside
1282. 54:58from inverting aside from multiplying
1283. 55:01okay so it can also integrate it can
1284. 55:04also differentiate and it can also
1285. 55:06compare
1286. 55:07so we have an integrator differentiator
1287. 55:10and
1288. 55:10comparator and i will be discussing that
1289. 55:13on the next
1290. 55:14video okay so that's it for now
1291. 55:17okay so i will be ending our discussion
1292. 55:20here and wait for the second
1293. 55:22uh the follow-up video for the other
1294. 55:24practical applications of
1295. 55:26our operational amplifiers okay so thank
1296. 55:29you very much and don't forget to hit
1297. 55:31the like button
1298. 55:32after watching okay thank you and god
1299. 55:39bless