# Simple Shunt Regulator

Discussion in 'Homework Help' started by howartthou, Aug 14, 2009.

1. ### howartthou Thread Starter Active Member

Apr 18, 2009
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Hello All

Attached is a description of a simple shunt regulator with a zener diode.

I have read this a number of times now and understand it all except for the last sentence where its concluded that the output voltage could never exceed 0.91V.

It doesn't explain the formula for deriving the maximum output voltage of the circuit.

And I have tried the formulas that I know, but no luck so far in getting the same answer of 0.91V.

Could someone please show me the formula for determining the 0.91V

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2. ### Ron H AAC Fanatic!

Apr 14, 2005
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If you left the zener out of the circuit, the output voltage would be Eout=10V*100Ω/(1kΩ+100Ω)=0.91V. This is the standard voltage divider equation, derived from Ohm's Law.
Now, if you put the zener in the circuit, it will never turn on, because Eout is much less than the zener breakdown voltage. Eout will still be 0.91V.

3. ### howartthou Thread Starter Active Member

Apr 18, 2009
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DOH! (As Homer would say)

The old "voltage divider hidden in the simple shunt regulator circuit" trick. I know Vout = Vin*R2/(R1+R2). What I don't know is why I couldn't see it.

Anyway, thanks alot Ron your answer is spot on, and the point about the zener not turning on is essential too. Thanks again, excellent answer.

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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You have to remember that in this circuit the current that is flow through
Zener diode is equal:
Iz=Is-IL
Where
Is - current that is flow through RS
IL - Load current
For your example Rs=1KΩ; Ein=10V
So for RL=∞ then Iz=Is=(10V-5V)/1K=5mA
If we connect RL=5KΩ then
Iz=Is-(Vz/RL)=5mA-1mA=4mA
If we further decrease RL the current of a Zener diode will by decrease to.
An for RL=1KΩ Iz will by 0A
And this circuit will by non longer be a voltage regulator.
So for RL smaller then 1K this circuit is a simple voltage divider

5. ### howartthou Thread Starter Active Member

Apr 18, 2009
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Hi Jony
It took me a while but I understand what you are saying. But how you get the formula Iz=Is-(Vz/RL) confuses me. This formula says the current across Z equals the current across RS minus the current across RL. I just can't see why?

6. ### howartthou Thread Starter Active Member

Apr 18, 2009
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Another question. The attached homework question wants to know Vout. So using Vout = Vin x R2 / (R1 + R2) = 24 x 1/2 = 12v.

My questions:

Is 12v corrrect?

Why disregard Q2 base current?

Why is Vout twice the voltage across R3?

To find E3 walk from Q2 base to emitter to diode. So I walk there and get to the zenner at 7.5v. Ummm, so what? How does that give me E3

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7. ### Rick Martin Active Member

Jun 14, 2009
31
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The measured output will be 16.2V

Vout is twice the voltage of R3 because R2 is identical and you have a voltage divider with equal values so the voltage on R3 will be half.

The zener has 7.5 at the emitter of Q2 and tracing back to R3 you gain approx 0.6v from the base-emitter junction of Q2 giving 8.1v across R3 and as I said above with the voltage divider you get 16.2v across both resistors resulting in 16.2v output.

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Current that flows through Rs split to Zener current and load current.
And if you apply KCL Is=Iz+IL then
Iz=Is-IL so if Is=(Vcc-Vz)/Rs=constant
This means that if load current increase diode current must decrease.
The Kirchhof law must hold.

No.
This is no longer such a simple circuit as a Zener shunt regulator.
This circuit has a negative feedback and Q2 act like a "error amplifier", Zener diode is a reference voltage and Q1 is a emitter follower.
And of course the correct answer is a 16.2V.
From II Kirchhoff (KVL) law you should know that:
VR3=Vz+Vbe2
So IR3=8.1V/1KΩ=8.1mA
And if we disregard Q2 base current then we can say that current that is flow through R2 must by equal IR3.
In reality IR2=IR3+Ib2 and that has why disregard Q2 base current.
So if we assume that IR3=IR2 then we easily calculate the voltage on a R2.
VR2=IR3*R2=8.1V and Vout=VR1+VR2=16.2V
If for some reason output voltage rise, then current that is flow through R2 increase to. So Q2 base current will rise and Q2 will turn on more. This mean that Q2 collector current increases. So now the more R1 current will flow through Q2 instead of base Q1.
So finally Q1 gets less current which decreases the output voltage.
And circuit approaching the equilibrium because negative feedback through R2.

P.S. From which book you have those "homework"

Last edited: Aug 18, 2009
9. ### howartthou Thread Starter Active Member

Apr 18, 2009
96
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Hi Rick
Thanks for your answer, it explains it well and I assume a voltage divider has resistors in parallel thats why the voltage across both is double that of one?

Hi Jony!
You make circuit analysis seem so easy and you always remind me of KVL and KCL. Unfortunately I still can't read complex circuits. I start from Vcc+ now but I think author wants me to read from the minus side sometimes as my books are supposedly based on electron flow and NOT conventional current flow.

I know it doesnt matter does it? I can read it either way regardless of what the author intends?

I have to read your last response a few more times before I get it, but so far I have been able to keep up. I just can't seem to quite get there on my own yet. Once its explained, it makes sense, when left for me to do the analysis I seem to get stuck most of the time.

Anyway, I will respond to your last feedback soon, when my head is clear.

My homework is from the Cleveland Institute of Electronics and I am studying from home (Melbourne, Australia). Unfortunately I don't get any help from a tutor so I rely on the books and help from this forum - which I really appreciate.

I also have another related question:

If the "error amplifier" transistor is replaced with an Op Amp what would Vout be?

I don't think an Op Amp adds 0.6v so my guess is 15V.

Is this correct?

10. ### Rick Martin Active Member

Jun 14, 2009
31
2
The two resistors are in series mate, and when you have simple two in series you can use a voltage divider formula which is:

VR1 = (R1/(R1+R2)) * Vin

But in this case you know what the voltage across R2 is by the 7.3v + 0.6 = 8.1 so you rearrange it to look like this to find Vout:

Vout = ((R2 +R1)/R2) * VR2.

What you need to remember is that the voltage over the resistor needs to be the one above the total resitance, ie if you are trying to find R2 it is:

VR2 = (R2/(R1+R2)) * Vin

Hope this helps mate.

By the way I have a short hand version of what happens if RL increases or decreases changing Vout and how it regulates it to the original value if it will help mate.

11. ### howartthou Thread Starter Active Member

Apr 18, 2009
96
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Hi Rick
Yes the resistors are in series but they are in parallel with RL and the rest of the circuit so thats where I got confused. But I can't argue with the voltage divider formula.

Your tip about the "voltage over the resistor..." is a reallly good tip. Thanks.

And yes I am interested in your short hand version...care to explain? I save all my threads as reference material so any tips I get are appreciated.

12. ### Rick Martin Active Member

Jun 14, 2009
31
2
Sorry, yes RL is in parallel with R1 & R2. So in that case work out the voltage across each and there for both R2 & R3 and then as you know that total voltage is the voltage across RL.

So if RL say increases then the current through RL will decrease and the VRL will increase. This is the short hand way I use to explain circuit operations like this:

RL inc, VRL inc, VR3 inc, Q2B inc, Q2 IC inc, VR1 inc, Q1VB dec, Q1IC dec, Q1VCE inc so VRL dec.

Okay, the long version and this is a simple way is RL increase so obviously the voltage across RL needs to increase. If this happens the voltage arcoss R2 and R3 increases (parallel voltages) and the voltage across R3 which bias' Q2 increases. This causes more current through R1 resulting in a larger voltage drop leaving less voltage at the base of Q1. This reduces Q1's collector current resulting in a greater resistance, increasing its voltage drop, taking voltage for RL therefore reducing RL's voltage to the desired level.

Okay maybe its not shorter but this is the best way I have found to understand its operation.

Hope it helps mate. I know what its like to try get your head around this.

13. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,991
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Yes, it's doesn't matter.

When you replace Q2 with Opamp, now you can look at this circuit as a non-inverting amplifier with additional current booster BJT Q1.
The gain is equal Au=1+(R2/R3)=2[V/V] and input voltage is zener diode voltage.
So the output voltage is equal Vout=Au*Vin=2*7.5V=15V

14. ### Ron H AAC Fanatic!

Apr 14, 2005
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657
That circuit was designed before good IC op amps were available. A 5.6V zener would typically yield the best temperature stability, because the tempco of a 5.6V zener is ≈+2mV/°C, which cancels the transistor's Vbe tempco of ≈-2mV/°C, yielding a 6.2V (Vz+Vbe) reference voltage with a low temperature coefficient.

15. ### howartthou Thread Starter Active Member

Apr 18, 2009
96
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Thanks Rick.
Hmmmm, one thing your approach misses is KVL in that the Vin+ must equal the Vin-. I found myself asking why the voltage needs to increase if RL increases? I believe its because the circuit needs to balance out to zero where all the voltage rises and falls add together to equal zero (KVL).

I am imaging Jony saying "of course!" and he would be correct. I get lost in the why's and it ends up being KVL or KCL that has the answers to the questions I keeping asking that I feel I should know by now.

I am also still coming to terms with Q's. Like base vs emitter vs collector. I know collector is the sum of base and emitter currents but keep forgetting whether the circuit depends on the emitter or the collector as its output and guess thats depends on whether its NPN or PNP?

But your tips help. Thanks.

Jony
Thanks, you always amaze me with your answers, I can only aspire to think like that when I do circuit analysis...

Ron
Thanks for the tip although probably a bit beyond where I am at now. Not sure how Tempco's of the zener relate to my questions but will think about it....

16. ### Rick Martin Active Member

Jun 14, 2009
31
2
Yeah in the end it must all balance and although it sounds like a process it is an instant action.

What you should get out of it is the basic rules and roles each component has to make sure that the output voltage remains constant even with varying load currents. Transistors are a tough nut to crack but you will find if you just keep working at them and using the basic rules you will get it soon enough mate.

17. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,991
1,116
No, emittter is the sum of base and collector currents.
And emitter as the name suggests emits electrons or holes. Collector collects them.

And it's all depends on a base current.
Ic=β*Ib
Ie=(β+1)*Ib≈Ic

An when you think on a KVL. Think in a terms of a sum. The sum of all voltage drops must be equal Vcc or whatever your source is.
And thinks connect in parallel have the same voltage across them.

What is the voltage drop on R4 and R5?

Last edited: Aug 20, 2009
18. ### howartthou Thread Starter Active Member

Apr 18, 2009
96
0
Jony
"What is the voltage drop on R4 and R5?" Huh? I understand the diagrams (thanks!) but the question doesnt make sense. Which R4 and R5 for which diagram? And no voltages are shown so I assume you mean from a KVL perspective.

Thanks again for the diagrams.

Rick
Thanks again. Yes I always get the general idea, just stuck on details, especially current flow/circuit analysis. As you can see Jony is a great help and I know that my course material is lacking in KVL and KCL training so I am hunting down some material for me to practice on...if I have the time to read it with deadlines approaching...

19. ### Rick Martin Active Member

Jun 14, 2009
31
2
Kirchhoffs laws are very simple mate and as long as you understand simple series and parallel current and voltage characteristics you should be alright from that side of things.

I dont have alot of diagrams and find it hard to write alot of detail as you can easily get lost in it. Maybe if you need some finer details smoothed get someone in person who understand this to smooth the edges for you. I have found if I do the grunt work and get most of it in my head, a couple of simple questions with someone fills in the gaps nicely.

20. ### howartthou Thread Starter Active Member

Apr 18, 2009
96
0
Hi Rick
Sounds good, yes I need to do some more grunt work first, almost there. Problem is I don't have a tutor, so here I am.

Jony
In the Q diagrams the current is coming out of the collector on the second pic. So, umm, if E is the sum of B and C how is the current being "emitted" from C