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Need Explanation of Shunt Regulator

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wuchy143

Member
Hi All!,

I'm trying to figure out how this 5V shunt regulator works and am just confusing myself. I attached a schematic to avoid confusion.

I see that R4 and U1 are where the engergy is shunted to but as for how the PNP dumps this energy there to be dissipated is confusing. Does anyone have a simple explanation at to what's going on here? I know it's not too complicated....

Thanks!

-mike
 

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BrownOut

Banned
Pin 1 senses the overvoltage via R4 and "fires" U1. The current flowing through U1 also flows thru R6, and the resulting voltage at the connection of R6 and the transistor biases ON the transistor. Now, current is "shunted" thru the transistor, instead of towards the output.

At least that's how I think it works.

What is this for?
 
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wuchy143

Member
Thanks BrownOut!

This circuit is being used to regulate a +5V keyboard. It is from an older design and I wanted to get an understanding on it since I didn't do the design and will probably need to de-bug at some point.

I understand now that the excess voltage turns on the BJT which is where the current is "shunted" away from the output.

If you were to regulate this circuit to have +5V at the output node at all times how would you pick your reference voltage at U1 pin 1. I know it's only a voltage divider...but I guess what I'm really asking is how do I pick a the correct reference voltage so the zener turns on when the output goes above +5V. I've looked over its datasheet and don't get a clear answer.
 

Willbe

New Member
The V-I curve should show that as the input voltage tries to rise above design value the current increases to keep the output voltage at close to 5v. The Thevenin equiv. output impedance should also be decreasing as this happens.
 

BrownOut

Banned
Thanks BrownOut!

This circuit is being used to regulate a +5V keyboard. It is from an older design and I wanted to get an understanding on it since I didn't do the design and will probably need to de-bug at some point.

I understand now that the excess voltage turns on the BJT which is where the current is "shunted" away from the output.

If you were to regulate this circuit to have +5V at the output node at all times how would you pick your reference voltage at U1 pin 1. I know it's only a voltage divider...but I guess what I'm really asking is how do I pick a the correct reference voltage so the zener turns on when the output goes above +5V. I've looked over its datasheet and don't get a clear answer.

Give us a link to the datasheet.
 

im_in_asia_now

New Member
I think the way it works is as a zener reference regulator. The zener gives a fixed voltage to the base. The emitter voltage adjusts to 5 V to make the emitter-base voltage constant. The datasheet for tip32 says the VBE is 1.8 V, so I'd assume since it's a PNP, then the zener would be set at 3.2 V. The reference voltage across the zener is adjustable by changing the values of R3 and R4. Anybody correct me if I'm wrong. I'd assume c5 is acting as a filter to keep the voltage at the emitter more stable. This is all assuming that the voltage coming across D1 is pulsing DC.
 

MikeMl

Well-Known Member
Most Helpful Member
Hi All!,

I'm trying to figure out how this 5V shunt regulator works and am just confusing myself. I attached a schematic to avoid confusion.

I see that R4 and U1 are where the engergy is shunted to but as for how the PNP dumps this energy there to be dissipated is confusing. Does anyone have a simple explanation at to what's going on here? I know it's not too complicated....

Thanks!

-mike
Look at the attached simplified equivalent circuit taken from the LM431 data sheet. Note that if Ref (+ input of the opamp) is a mV or so HIGHER than the 2.5V reference, the gain of the opamp drives the built-in NPN into conduction, pulling the Cathode voltage down toward the Anode.

If Ref is a mV or so LOWER than 2.5V, then the NPN is off, meaning that the external circuit can pull the Cathode up as high as it can (no current flows through the collector of the NPN).

Note that in your circuit, the Ref input is driven by a voltage divider consisting of two equal value resistors (1% no less), so that sets the regulation voltage at twice the internal 2.5V reference, i. e. 5V.

Also notice that if the output of the regulator is slightly higher than 5V, that makes the Ref input slightly more than 2.5V, which causes the LM431 to sink current out of the base of the power PNP TIP34 transistor, which gets amplified by the β of the transistor, which causes a much larger current to flow from the output node to ground via the collector of the transistor.

The transistor current adds to the current flowing in R1 (1Ω resistor), which increases the drop across R1 in an amount such that the voltage at the output stays at 5V.

Another way to think of this is to suppose that the external load on the power supply was suddenly cut in half. The reduction in current would naturally cause the output voltage to rise. However, the TL/LM431 sees the increase in output voltage and reacts by turning on the TIP34 harder. This "shunts" current to ground exactly offsetting the reduction in load current keeping the drop across R1 (and the output voltage) constant.
 

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wuchy143

Member
http://www.fairchildsemi.com/ds/TI/TIP32.pdf

Above is the datasheet BROWNOUT requested

Some great explanations here. Thanks everyone.

im_in_asia_now I understand everything you say until the end

This is all assuming that the voltage coming across D1 is pulsing DC.
1. Why is R1 and D1 even in the circuit?
2. What's the significance of having a pulsing DC signal. I assume you mean it's just alternating between 0V and some voltage at some frequency?

MikeMl I am still looking into your response. I apologize but please stay posted because I will have a response shortly.

Thanks again

-mike
 

BrownOut

Banned
I meant the data sheet for U1, sorry. It's OK though, I looked it up. Like you, I first thought it was a triac, but it's really a programmable voltage reference that "turns on" according to the equation:

Vo = (1 + R1/R2) Vref. - Where R1 and R2 are R3 and R4 in your schematic, respectively.

From the data sheet, Vref is near 2.5V. Using the approximation, we can calulate R1/R2

R1/R2 = Vo/Vref - 1. Since Vo/Vref =2, R1/R2 = 2-1 = 1. Thus, R1 and R2 are equal value.
 

im_in_asia_now

New Member
Pulsating DC is what you get after you rectify AC. The schematics looked similar to a zener reference regulator, (though not the same because of the variable zener).

I'm not sure what's meant by the letters ISRC_R. Plus there are no other labels for what the input at the cathode of D1 is. Given the diode D1, I made the assumption that the input at the cathode of D1 was probably some form of AC, and that would make the function of D1 a half wave rectifier.

One of the uses of the transistor is as a regulator. I can't explain how it works very well in this circuit because it's PNP, and I learned mostly NPN. You could probably do some research on the internet about it if you're unfamiliar with using a transistor as a regulator.

If you could take voltage readings at points 1 and 2 of U1 that would make it a lot more easy for us to figure out how it and the circuit work.
 
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wuchy143

Member
Thanks brownout

I think I have a much better grasp now on the schematic.

I have a couple glaring questions though. One of which I'm a little embarrased I don't know the answer..but I'm going to ask anyway.

1. How did they pick this transistor. I know that it's in the datasheet as an example application so I would know to use a PNP but what about predicting the base current so that the transistor is 100% on when U1 conducts. I guess what I"m asking is the transistor is normally cutoff.(no regulation is happening) Then when the supply voltage goes above 5V the transistor is saturated letting current run through it. Do I just assume that no matter what U1 will hold Q3's base at 2.5V which will definitly make the 1.8V difference b/t the base and emitter?

2. What is the significance of R1 and D1?

Thanks

-mike
 

im_in_asia_now

New Member
from

SERIES REGULATOR

"SHUNT REGULATORS use a variable resistance placed in parallel with the load. Regulation is achieved by keeping the resistance of the load constant. Characteristically the resistance of the shunt moves in the opposite direction of the resistance of the load."

14179_230_2.jpg

For PNP just flip it so the PNP emitter is in the place of the NPN collector, and the - is the same as your ground.
 
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BrownOut

Banned
Q3 never saturates. As the voltage at it's emitter ( the one with the arrow ) attempts to rise higher, it will conduct harder and harder to maintain a constant voltage. The increase in conduction of Q3 is driven by U1, according to the schematic of U1 that MIKEMI provided. The purpose of R1 is to provide a voltage drop from the unregulated portion of the power supply to achieve 5V at the output. Without the resistor, there would be no control of the voltage, no matter how hard Q3 was driven.

As for D1, I'm not sure. It's existance is dependent on the type of power input. All I can say is that it prevents current from flowing up, away from the rest of the circuit.
 

wuchy143

Member
Got it. Thanks. I re-read what he wrote after reading your post about R1 and now it makes sense.

You guys are great. Thanks again.
 
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