Timeline for Differential Amplifier with Global Negative Feedback
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 13, 2017 at 17:21 | vote | accept | lucenzo97 | ||
| Oct 24, 2017 at 15:30 | comment | added | lucenzo97 | @Bimpelrekkie You said this: "So if Ic_Q2 is a bit larger then Ic_Q1 will be a bit smaller.", so that means if Vbe_Q2 increases a bit more then Vbe_Q1 will decrease for the same amount? | |
| Oct 24, 2017 at 13:10 | comment | added | Bimpelrekkie | You could try that in a simulator and see how much difference it makes. My bet: it will not make a lot of difference. As long as the open-loop gain is much higher than the closed-loop gain then any imbalances in the loop are suppressed. | |
| Oct 24, 2017 at 12:56 | comment | added | lucenzo97 | @Bimpelrekkie Yes, R1 should be replaced by constant current source and R3&R4 should also be replaced with current mirrors for optimal balance. | |
| Oct 24, 2017 at 12:41 | comment | added | Bimpelrekkie | Ideally R1 should be a current source but since there's almost 50 V across it, it will be good enough. The voltage at the emitters of Q1 and Q2 does not vary much (very little compared to 50 V) so the current will be reasonably constant. | |
| Oct 24, 2017 at 12:37 | comment | added | Bimpelrekkie | Oh, I misunderstood you in my previous comment! What you mean: the collector current of Q2 is not used! That's correct but it does not matter. The sum of the collector current of Q1 and Q2 remains constant. So if Ic_Q2 is a bit larger then Ic_Q1 will be a bit smaller. So we can "throw away" Ic_Q2 because Ic_Q1 contains the same information. Why is Ic_Q1 + Ic_Q2 constant? Because of R1, it forces a nearly constant current into Q1 + Q2. | |
| Oct 24, 2017 at 12:33 | comment | added | Bimpelrekkie | But how does Q2 do anything since there is no output taken on that device? There is! Through the feedback network (basically R5 and R7+C6) this is a voltage divider which takes the output voltage, divides it (by roughly 470/10470) and feeds that to Q2. I hope you understand that at the base of Q2 there is a divided-down version of the output voltage which the amplifier tries to match (compare, amplify difference) with the input voltage which is at the base of Q1. | |
| Oct 24, 2017 at 12:23 | comment | added | lucenzo97 | @Bimpelrekkie Seems logical to me. But how does Q2 do anything since there is no output taken on that device? Or does the Q2 transistor affects the Q1 transistor? | |
| Oct 23, 2017 at 17:27 | answer | added | analogsystemsrf | timeline score: -3 | |
| Oct 23, 2017 at 15:07 | comment | added | LvW | In all cases and for each amplifier type the effect of negative feedback always reduces non-linear distortions. This was proven - independent on a particular topology - for all active blocks with negative feedback (H. Black in the 1930th). | |
| Oct 23, 2017 at 14:46 | comment | added | Bimpelrekkie | Note how the name "differential amplifier" relates to the fact that the amplifier (with a differential pair at its input) amplifies the difference (voltage) at its input. By amplifying the difference and feeding it back as negative feedback, the difference is minimized. And zero difference means no distortion. | |
| Oct 23, 2017 at 14:32 | comment | added | G36 | If for some reason the Vout is different than expected the negative feedback will correct it. If Vout is larger than expected (Vin) the Q2 will reduce his IC current. But in the same time, the Q1 Ic current will goes up. Q3 and Q4 current will also increase so the Vout will drop. | |
| Oct 23, 2017 at 14:18 | comment | added | G36 | The negative feedback reduces the distortion, not the differential pair by itself. The differential pair act as an error amplifier. He compares the input signal (not distorted one) with the output signal. So, the error signal is created to cancel the distortion. Exactly in the same way as a negative feedback "brings back" the output signal to the equilibrium state (V"+" = V"-"). The negative feedback does not distinguish between types of disturbances. | |
| Oct 23, 2017 at 13:43 | comment | added | lucenzo97 | @OlinLathrop Maybe just a part of my question refers to nfb. I was mostly interested of how Q1 and Q2 prevent distortions coming from the output. | |
| Oct 23, 2017 at 13:31 | review | Close votes | |||
| Nov 7, 2017 at 3:01 | |||||
| Oct 23, 2017 at 13:08 | comment | added | Olin Lathrop | Possible duplicate of Illustrating op amp feedback without control theory. Your question is really about what negative feedback is and how it works. See my detailed answer to the above question. | |
| Oct 23, 2017 at 13:03 | history | asked | lucenzo97 | CC BY-SA 3.0 |