Electronic Devices - Voltage Regulators - Discussion

Please explain this answer.

Using KCL,

IR2+IR3=0, so Vo-1. 7/47+0-1. 7/12=0.
Vo/47=1. 7/47+1. 7/12,
Vo/47=0. 036+0. 141,
Vo=8. 36.

According to me the answer is 20.16 as because the zener voltage will act as non inverting voltage to opamp and due to -ve feedback the voltage below R2 is 4.1 and so the current comes to be 0.34 ma.

Thus the voltage drop across R2 is 16.05 plus the voltage just below R2. Thus the answer is 20.16 one more thing to notice there is no collector resistance hence what will be vce.

The answer is letter 20.16.

Vo = (1+R2/R3)(Voltage reference) = Answer.

Please explain me what does this mean Zener diode has rating of 1.7 V. If the voltage across zener is less than 1.7 V, it will not clamp, or will it clamp to 1.7 V.

If second is the case then, why 4.1 V is written.

As per what I believe, is the first case, and as we give 1.7 V to the zener. It will regulate and clamp the voltage to 4.1 V. In that case answer should be 20.16. Op-amp is just acting as error amplifier here, in open loop configuration.

Yeah, I also got 20. 16. So which one is true?

That's because Zener diode always has a constant Voltage (it's rating) while operating in the zener region.

However you increase the voltage (so long as it's less than PIV, 50V ), it will not change the voltage across zener, instead the current will change. so we will use 1.7V instead of 4.1V

Vo = (1+(R1/R2))Vz.
Vo = (1+(47k/12k))(1.7),
Vo = 8.36V.

From Nodal Analysis.

Vout = [(47k)(1.7k)/12k] + 1.7k.

Here you will get a 8.36V answer because the Zener diode rating is 1.7V as stated in the problem. BUT when you based your answer to the schematic, you will get 20.16V answer, because as you can see to the schematic the given more zen diode rating is 4.1V.