Help with analyze a simple analog cirucit

[eladta]

Hi all,
I have a pretty simple circuit but i'm not sure how to analyze it. Can anyone give me a hand ?


Thanks.

 

[godfreyl]

The voltage at the non-inverting input of the opamp = Vin. (The 200Ω resistor makes no difference).

The opamp drives the transistor to set the voltage at it's emitter = Vin.

The 100Ω and 20K resistors at the output form a voltahe divider, so Vout = Vin * (20000/20100) with no load, and the output impedance = 20K||100Ω.

 

[eladta]

so this is just a simple buffer ?
is there a situation where the BJT is off ?

 

[godfreyl]

If Vin = zero, then Vout = zero too, so bjt is off.
Also, if you connect a resistor between Vout and +Vcc, that can switch the bjt off, and Vout won't be able to follow Vin anymore.

Otherwise, yes, it's a simple buffer.

 

[eladta]

so why do we need the bjt for ?

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another question, how did u calculated the output impedance ?
Thank u !!

 

[godfreyl]

so why do we need the bjt for ?

I don't know, maybe to get higher output current?

another question, how did u calculated the output impedance ?

It's the parallel combination of the resistor from the emitter to Vout and the resistor from Vout to ground. Maybe I made a mistake, I thought they were 100Ω and 20K, now it looks like 200Ω and 20K.

 

[eladta]

It's the parallel combination of the resistor from the emitter to Vout and the resistor from Vout to ground. Maybe I made a mistake, I thought they were 100Ω and 20K, now it looks like 200Ω and 20K.

but they are not parallel connected, they are in series (the values are 20k and 100Ω)

 

[godfreyl]

but they are not parallel connected, they are in series

Yes, but they are both connected to Vout, so any external load will see both of them as if they were in parallel

If Vin is kept constant and a load resistor is added between Vout and ground, the output voltage will change and the current through both resistors will change. The current through the load will equal (change in current through 100Ω resistor) + (change in current through 20K resistor)

(the values are 20k and 100Ω)

You sure? The first digit looks exactly the same.

 

[eladta]

Yes, but they are both connected to Vout, so any external load will see both of them as if they were in parallel

The basic definition for parallel conection is that EACH leg of the resistor will be connected to the other. In here they are only connected through Vout. the second wire of each resistor is connected to another point.

 

[crutschow]

The basic definition for parallel conection is that EACH leg of the resistor will be connected to the other. In here they are only connected through Vout. the second wire of each resistor is connected to another point.

The assumption here is that the output impedance of the transistor emitter is essentially zero due to the negative feedback. Thus, from a Thevenin view-point, the two resistors are in parallel to ground.

 

[eladta]

The assumption here is that the output impedance of the transistor emitter is essentially zero due to the negative feedback. Thus, from a Thevenin view-point, the two resistors are in parallel to ground.

According this assumption you are right, but how can you justify that assumption ?

 

[crutschow]

According this assumption you are right, but how can you justify that assumption ?

Because the output impedance of the circuit is approximately the open loop output impedance of the emitter follower divided by the loop gain of the feedback loop. Given the high open-loop gain of a typical op amp, the calculated value for the output impedance will likely be a small fraction of an ohm.

The calculation of this is left as a exercise to the reader. ;-)

 

[eladta]

Thanks a lot for your kind help !