Elementary Voltage Regulator

[dkace]

Hi,
I am reading the cookbook of power supplies and in linear regulators, I found this schematic ( load and by pass capacitor is missing in BJTs' emmiter).
It is clear there that the amplifier is used to correct the error voltage in the output regarding the reference voltage in the + pin. And that the resistors in the voltage dividor are elected in a way that Vref=Voltage in the - pin. My stupid question: If Vref=V- then Vout is 0, no? If Vb=0 then the transistor is closed, no? How is it possible to have a down conversion then? I am sure I miss something, so ... help me find it!

D.

 

[aProgrammer]

If Vout is not zero, this add a negligible voltage between V+ ans V-,
because the voltage gain of the opamp is very high.

So in fact, Vout can be any value when V+ equal V- because Vout/GAIN is negligible.

 

[dkace]

Oups!!!
Sorry guys and girls ! I forgot to draw the negative feedback in the opamp circuitry!
I will do it as soon as my computer memory will be freed from 4 tasks I am running...
D.

I think I solved it.
Here how it is: The opamp circuit is a negative feedback amplifier. Therefore if we didn't have the Vref the - input would be in virtual graound. What is done is that we take care with the voltage dividor this point to be in Vref voltage . So any excurtion from this will result in different( amplified) output, which moves the operation line up or down thus regulating the emmiter voltage accordingly.
Is that right now?

 

[Apollo13]

The circuit u have given is perfect. The amp has negative feedback.. And load would be connected at emitter of the BJT. It is a simple linear regulator.
I donot understand the problem. Stability could be an issue as in parallel to load one needs a output cap.