Design a + 10V regulated supply for load currents from 0 to 100 mA using a zener and an emitter follower; the input voltage is +20 to +25 V.
Allow at least 10 mA zener current under all (worst-case) conditions. What power rating must the zener have?
EDIT:
What are the differences (pros and cons) between the circuit with the emitter follower and the one with zener and resistor only (attached) ?
Calculate worst-case dissipation in transistor and zener

I implemented the solution with resistor and zener diode (see attached photo),
But it is not very clear to me how to accomplish it with a follower emitter:


I understood that it has a high input impedance and this allows reducing any loss from a generic Thevenin circuit at the input ... and I also understood that the output on the emitter follows the input on the base with a voltage gain of 1.

What I do not understand is:
1) how come the current gain is high? is it mathematically proven from the small signal model ?
2) Some help on how to solve the exercise

Here are some of my thoughts on the exercise:
(a) while in the exercise with diode and resistor the 10mA of the Zener guaranteed 10V on it (and thus on Vo because Vo=Vz), I guess it is the same here. With Vz=10V the output is Vz-Vbe=9.3V --> I have to guarantee Vz=10.7V
b) the bjt is in the active zone (?) since it is not a switching application

 

The question asks for the power rating of the zener, nothing else.
Hence 10 mA x 10 V = 100 mW.
If you were to allow for 20 mA zener current, then the power dissipation is 200 mW.
A 500 mW zener would be suitable.

 

how come the current gain is high? is it mathematically proven from the small signal model ?

No.
The small-signal model is a linear model for (small) AC signals.
The current-gain is determined the large-signal DC current gain (Beta or hFE) of the transistor which is given it its datasheet.

I have to guarantee Vz=10.7V

Then you need a higher voltage Zener and possibly a voltage divider from the Zener to get the desired voltage.
For a real circuit, you may have to add a pot in the divider to allow for the voltage tolerance of the Zener.

the bjt is in the active zone (?) since it is not a switching application

Yes.

 

The question asks for the power rating of the zener, nothing else.
Hence 10 mA x 10 V = 100 mW.
If you were to allow for 20 mA zener current, then the power dissipation is 200 mW.
A 500 mW zener would be suitable.

It will be higher than that. The zener current is 10mA when the transistor base current is at its maximum. So the worst case zener current is 10mA+Ic/Hfe and happens when there is no load. Also, the zener has to be more than 10V to get 10V output.

 

Design a + 10V regulated supply for load currents from 0 to 100 mA; the input voltage is +20 to +25 V.
Allow at least 10 mA zener current under all (worst-case) conditions. What power rating must the zener have?

I implemented the solution with resistor and zener diode (see attached photo),
But it is not very clear to me how to accomplish it with a follower emitter:

View attachment 330509

I understood that it has a high input impedance and this allows reducing any loss from a generic Thevenin circuit at the input ... and I also understood that the output on the emitter follows the input on the base with a voltage gain of 1.

What I do not understand is:
1) how come the current gain is high? is it mathematically proven from the small signal model ?
2) Some help on how to solve the exercise

Here are some of my thoughts on the exercise:
(a) while in the exercise with diode and resistor the 10mA of the Zener guaranteed 10V on it (and thus on Vo because Vo=Vz), I guess it is the same here. With Vz=10V the output is Vz-Vbe=9.3V --> I have to guarantee Vz=10.7V
b) the bjt is in the active zone (?) since it is not a switching application

I think the question does not envisage the use of a transistor current booster, when the Question is to find the Power Rating of the Zener.
If we use an Opamp, the Zener rating can be just 100 mW!
My calculations show that for the simple Shunt Regulator, the maximum Power the Zener will need to Dissipate is 1.4666... Watts.
This will be when the input is 30 Volts and the Load Current is 0.

 

I think the question does not envisage the use of a transistor current booster, when the Question is to find the Power Rating of the Zener.

The stated problem is to Design a + 10V regulated supply.
It doesn't specify a method, or how well regulated it needs to be, so it would seem a transistor current booster, or any other regulator circuit, would be allowed.

 

The stated problem is to Design a + 10V regulated supply.
It doesn't specify a method, or how well regulated it needs to be, so it would seem a transistor current booster, or any other regulator circuit, would be allowed.

You could take that to an absurd level and use an LM317, and put a 5.6V zener on the output in series with 430Ω. It says that there must be a zener but not that it has to be part of the regulation circuit.

 

Maybe I explained myself wrongly, sorry. I added an EDIT in the main post adding a more specific question
My calculations:



@MrChips @crutschow @Ian0 @Ramussons

 

It will be higher than that. The zener current is 10mA when the transistor base current is at its maximum. So the worst case zener current is 10mA+Ic/Hfe and happens when there is no load. Also, the zener has to be more than 10V to get 10V output.

Let's say Hfe is a modest 100.
Ic/Hfe = 1 mA

I said Iz = 20 mA

If you want to get 10.7 V base voltage, put a forward biased 1N4001 diode in series with the 10 V zener.

Edit: Why does it matter if there is no load? That is what a voltage regulator is suppose to do: keep the voltage constant regardless of the load.

 

You could take that to an absurd level and use an LM317, and put a 5.6V zener on the output in series with 430Ω. It says that there must be a zener but not that it has to be part of the regulation circuit.

Okay.
But I see no point in going to an "absurd level".

 

Okay.
But I see no point in going to an "absurd level".

Just pointing out how badly the question is worded. . .

 

Maybe I explained myself wrongly, sorry. I added an EDIT in the main post adding a more specific question
My calculations:
View attachment 330522


@MrChips @crutschow @Ian0 @Ramussons

Hi,

What is it that you are questioning here?

It looks like you got the input resistor right, or close with 850 Ohms, but closer is 845 Ohms.
That is of course with worst case 20v input and 10ma zener current and 1ma base current, and a 'perfect' 10.7v zener voltage.

Worst case power in the zener will be with high line and no output current.

 

Hi,

What is it that you are questioning here?

It looks like you got the input resistor right, or close with 850 Ohms, but closer is 845 Ohms.
That is of course with worst case 20v input and 10ma zener current and 1ma base current, and a 'perfect' 10.7v zener voltage.

Worst case power in the zener will be with high line and no output current.

What I ask now is:
1) a check of the calculations I did (in post #8) .. and you did it, thanks!

2) why is R "worst case" calculated with Vi=20V and not with Vi=25V? By "worst case" I can think of the case when Ib is max, that is, when Vi=25V (and not 20V)
Ib = (Vi - Vz) / Rb

3) If Pbjt=Vce(max)*Ic(max) = Vce(max)*100mA .. is this calculation correct?
Vce(max) = Vi(max) - Rc*I(max) - Vo
= 25V - 68*100mA - 10V
= 8.2V
Pbjt = 8.2V * 100mA = 0.82W

 

What I ask now is:
1) a check of the calculations I did (in post #8) .. and you did it, thanks!

2) why is R "worst case" calculated with Vi=20V and not with Vi=25V? By "worst case" I can think of the case when Ib is max, that is, when Vi=25V (and not 20V)
Ib = (Vi - Vz) / Rb

3) If Pbjt=Vce(max)*Ic(max) = Vce(max)*100mA .. is this calculation correct?
Vce(max) = Vi(max) - Rc*I(max) - Vo
= 25V - 68*100mA - 10V
= 8.2V
Pbjt = 8.2V * 100mA = 0.82W

Hi again,

#3 looks right.

For #2, there are two "worst case" solutions:
one is for the maximum power in the zener (maybe other devices in the circuit too),
and the other is for the minimum current in the zener.
If either of these is not right (at least) then the circuit is not yet designed correctly.
This means you have to consider cases when Vin=25v and also when Vin=20v because they both lead to different "worst cases".

 

a reference voltage is often carefully set but has little current.
A voltage follower can sample the reference voltage without significantly dropping the reference voltage.

I had a 5V 5W zener it was lug mounted and it got very hot. That old technology was aweful.
Now I use an adjustable regulator with heat sink and the voltage references now are very precise
embedded band gap zener. We moved away from sagging voltage and hot zeners.
An example of carefully setting a 6,2V 1/2 W zener choose a resistor value so that the current is about 7mA
and measure the voltage. Adjust the current controlling resistance until you measure 6.200 V as you apply a load
the zener voltage drops but what happens when you use a voltage follower?

 

a reference voltage is often carefully set but has little current.
A voltage follower can sample the reference voltage without significantly dropping the reference voltage.

I had a 5V 5W zener it was lug mounted and it got very hot. That old technology was aweful.
Now I use an adjustable regulator with heat sink and the voltage references now are very precise
embedded band gap zener. We moved away from sagging voltage and hot zeners.
An example of carefully setting a 6,2V 1/2 W zener choose a resistor value so that the current is about 7mA
and measure the voltage. Adjust the current controlling resistance until you measure 6.200 V as you apply a load
the zener voltage drops but what happens when you use a voltage follower?

A voltage follower allows a lower output impedance so you can connect higher power loads and maintain relatively decent regulation.