19.5 The noninverting buffer op-amp configuration is shown...

Name: 19.5 The noninverting buffer op-amp configuration is shown in Fig. P10.3. Assuming that the op-amp has infinite input resistance and zero output resistance, what is ?? If A=1000, what is the closed-loop voltage gain? What is the amount of feedback (in dB)? For Vs=1V, find Vo and Vi. If A decreases by 10%, what is the corresponding percentage decrease in Af? (10 points)
Uploaded: 2024-07-30T16:36:16-08:00
Duration: 5 min 4 s
Channel: Adi S
Description: 19.5 The noninverting buffer op-amp configuration is shown in Fig. P10.3. Assuming that the op-amp has infinite input resistance and zero output resistance, what is ?? If A=1000, what is the closed-loop voltage gain? What is the amount of feedback (in dB)? For Vs=1V, find Vo and Vi. If A decreases by 10%, what is the corresponding percentage decrease in Af? (10 points)

19.5 The noninverting buffer op-amp configuration is shown in Fig. P10.3. Assuming that the op-amp has infinite input resistance and zero output resistance, what is ?? If A=1000, what is the closed-loop voltage gain? What is the amount of feedback (in dB)? For Vs=1V, find Vo and Vi. If A decreases by 10%, what is the corresponding percentage decrease in Af? (10 points)

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Transcript

00:01 So in this question it is given that the output voltage that is denoted by v0.

00:10 So this is connected to the negative terminal with a feedback gain to be equal to 1.

00:28 That is we can say that here b is equals to 1.

00:34 Then we can calculate the gain in db.

00:39 That is b db will be equals to 20 log based in b so this will be equals to 20 log 10 into 1 as the value of b is equal to 1 so this will be corresponds to the value of 0 db so now we have our required formula for the output voltage and this is given by v0 is equal to vs tars minus v0 into aol.

01:24 So this will be further equals to v .s .aol minus v0aol.

01:38 This will be equal to 1 plus aol into v0 is equal to vs aol.

01:46 Aol.

01:49 That means v0 by vs here is equal to aol divided by 1 plus aol.

02:01 So this will be equals to acl.

02:07 As we have got our required ratio for v0 by vs, so here the closed loop gain again that is denoted by a .cl is equal to a .cl.

02:23 A0l plus divided by 1 plus a0l that we have derived of.

02:31 Putting down the respective values, we get 1 ,000 upon 1 plus 1 ,000 and this will be equal to 0 .99, which will be approximately equal to 1.

02:46 Now proceeding to the next step of the solution where it is given in the question that the value for a0l is decreased by 10%...