OpAmp - Negative Feedback operation

[shaikss]

Hi Folks,

I was going through some analytical paper on Opamps and got a doubt regarding the -ve feedback operation.
I have attached the circuit.
The Q is to label which of the terminals 'A' and 'B' are +ve and -ve so that the entire circuit works in -ve feedback.

Let me explain how I analysed the circuit,
Here terminal B is fixed to 1V and the current source of 1mA is driving the collector of BJT.
For me, it looks like a current monitor circuit.
So, I feel that the terminal which is connected to collector should be inverting terminal and so terminal A should be inverting terminal to make the circuit work in negative feedback mode.

Am I correct?
If wrong, Please correct me and let me know how to analyse the circuit.

Thanks!

 

[LvW]

So, I feel that the terminal which is connected to collector should be inverting terminal and so terminal A should be inverting terminal to make the circuit work in negative feedback mode.

Hi shaikss,

no, the terminal A has to be non-inverting to establish negative feedback.
Explanation: A rising opamp output voltage causes further opening of the transistor. This leads to a falling collector potential and compensates the output increase.

---------- Post added at 13:49 ---------- Previous post was at 13:27 ----------

I suppose it could be useful (if not necessary) to increase the reference voltage to ensure proper BJT operation.

 

[varunkant2k]

...label which of the terminals 'A' and 'B' are +ve and -ve so that the entire circuit works in -ve feedback.
Am I correct?

Hi Shaikss,
Terminal "\'A' should be non inverting to make the loop in -ve f/b. Transitor always have negative polarity from BJT's , Base to collector.(check y). So 'A' should be +ve terminal.
Next, in this circuit, opamp infact used to dc bias the BJT and provide CM.

 

[shaikss]

So, I feel that the terminal which is connected to collector should be inverting terminal and so terminal A should be inverting terminal to make the circuit work in negative feedback mode.

Hi shaikss,

no, the terminal A has to be non-inverting to establish negative feedback.
Explanation: A rising opamp output voltage causes further opening of the transistor. This leads to a falling collector potential and compensates the output increase.

---------- Post added at 13:49 ---------- Previous post was at 13:27 ----------

I suppose it could be useful (if not necessary) to increase the reference voltage to ensure proper BJT operation.

Hi LvW,

Can you please explain me in detail?

---------- Post added at 05:21 ---------- Previous post was at 05:19 ----------

---------- Post added at 05:21 ---------- Previous post was at 05:21 ----------

Hi Shaikss,
Terminal "\'A' should be non inverting to make the loop in -ve f/b. Transitor always have negative polarity from BJT's , Base to collector.(check y). So 'A' should be +ve terminal.
Next, in this circuit, opamp infact used to dc bias the BJT and provide CM.

Hi Varunkant2k,

I didn't understand. Can you explain me how to analyse it?

 

[permute]

assume terminal A is at 5V. clearly higher than terminal B's 1V. if A is the inverting input, then the opamp's output should go as low as possible -- preventing the transistor from turning on at all. This is a stable state. Likewise, if A were 0V, and B were 1V, then the opamp would force as much current into the base as it could, turning the transistor on fully -- again a stable state. indeed, this would be a latch.

Now assume in the 5v/1v example that A were non-inverting. the transistor would turn on, forcing the terminal away from 5V. Likewise in the 0v/1v case, the transistor would turn off allowing the voltage to rise. This would be the correct action.

 

[shaikss]

assume terminal A is at 5V. clearly higher than terminal B's 1V. if A is the inverting input, then the opamp's output should go as low as possible -- preventing the transistor from turning on at all. This is a stable state. Likewise, if A were 0V, and B were 1V, then the opamp would force as much current into the base as it could, turning the transistor on fully -- again a stable state. indeed, this would be a latch.

Now assume in the 5v/1v example that A were non-inverting. the transistor would turn on, forcing the terminal away from 5V. Likewise in the 0v/1v case, the transistor would turn off allowing the voltage to rise. This would be the correct action.

Thanks permute. Now, I understood.
Can you suggest me some text book where i can solve this kind of problems?

 

[permute]

the 3rd edition sedra-smith book (microelectronic circuits) is good in general, though might be a bit rigorous if you are looking for something lighter. It does have a lot of questions on opamps as well as feedback. Basic opamp circuits have plentiful literature on TI/national/maxim-ic/analog's websites. Basically any company that makes opamps.

this circuit is often called the "simulated zener", or the "shunt regulator".

 

[Chipmunk]

You can find some suggestions for Op Amp literature in

Recommendations for Op Amp Books

The second paragraph contains a link to "Op Amps for Everyone" from Texas Instruments which is quite good (and freely avaialable).

I would also second permutes suggestion of Sedra / Smith, they have lots of exercises. But I think the latest edition is the 6th, not the 3rd.

 

[varunkant2k]

---------- Post added at 05:21 ---------- Previous post was at 05:19 ----------

---------- Post added at 05:21 ---------- Previous post was at 05:21 ----------



Hi Varunkant2k,

I didn't understand. Can you explain me how to analyse it?

Hi Please read transistor operation. If you get voltage response from base to collector in Common Emitor configuration..you will get it.