Look at this voltage regulator:
The current through Rs is I=Iz+Ib.
If Vin increases by some amount, the current through Rs will increase. But which current will this increase, Iz or Ib? Here's what I think: Since the voltage at the emitter is held constant by the zener diode, the current through RL will be the same as before. This means that Ie will be the same and Ib will not increase. Thus, from this equation: I=Iz+Ib, only the current through the zener will increase.
Is that correct?

 

You can think of it that way - as the input voltage changes, the output voltage remains constant.
However, the zener voltage is not absolutely constant as the current varies. Here is a section of a datasheet:


Looking at the line for the BZX85C3V3 (a 3.3V zener), The first two columns give the range of voltages to be expected at the current in the third column (Iz). The next column, Zz, gives the impedance of the zener when the current through it is Iz. For this zener that is 20Ω. This means that the zener behaves as if it had a 20Ω resistor in series with it so increasing the current by 1mA will increase the zener voltage by 20mV, only a small amount but nevertheless not constant.

 

Yes, you're correct if the zener is an ideal one with an infinitely sharp knee. In practice, the zener voltage will increase slightly as the current through it increases.

 

Okay, I understand. Thank you!

 

Is that correct?

The voltage regulator here is really that zener + Rs: as long as the zener is reversely conducting, the voltage acrosss it will be (almost) constant, ie. the voltage is "regulated".

But such a device has high output impedance so it is unable to drive a load. That's where the transistor comes in: the npn is a current amplifier - it lowers the output impedance of the zener regulator.

So what you have is two circuits combined.

 

Thus, from this equation: I=Iz+Ib, only the current through the zener will increase.

A real zener would increase it's output voltage in response to an increased input voltage, which would increase the output voltage to the load, which would increase the load current which would increase the base-emitter voltage drop, ...

It isn't a very accurate regulator until you add a feedback loop to take the base-emitter voltage into consideration.

 

Thanks!