Voltage Regulator: TR1 & TR2 Current Regulation

[M P]

Homework Statement

https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/66/66455-bb02bb0d3c839e914352eb0379b453f2.jpg

Homework Equations

The Attempt at a Solution

Transistors TR2 and TR1 amplify the current being regulated through the regulator. The current gain of transistors is not important since the regulator will increase/decrease the current through pass transistors to maintain output voltage.

Anyone can help if I am on the right track?

 

[berkeman]

Homework Statement

https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/66/66455-bb02bb0d3c839e914352eb0379b453f2.jpg

Homework Equations

The Attempt at a Solution

Transistors TR2 and TR1 amplify the current being regulated through the regulator. The current gain of transistors is not important since the regulator will increase/decrease the current through pass transistors to maintain output voltage.

Anyone can help if I am on the right track?

At first glance, I don't think that's what the extra transistors are doing. Look at the equivalent circuit for what's inside of a typical 3-terminal linear regulator like an LM7805. What is different about that regulator's internal main circuitry compared to the extra transistors outside the 3-terminal regulator in this problem? It's a subtle but important difference, which is part of the answer to this problem, I believe...

 

[CWatters]

TR1 is to increase the current capability but TR2 has a slightly different purpose. I had to think about it for quite awhile. Hint: What does the regulator have (or usually has) that the transistor TR1 doesn't have?

Edit: Berkeman posted his reply while I was still thinking about it.

 

[M P]

At first glance, I don't think that's what the extra transistors are doing. Look at the equivalent circuit for what's inside of a typical 3-terminal linear regulator like an LM7805. What is different about that regulator's internal main circuitry compared to the extra transistors outside the 3-terminal regulator in this problem? It's a subtle but important difference, which is part of the answer to this problem, I believe...

All I could see is input output ground

 

[berkeman]

View attachment 83390
All I could see is input output ground

I mean the equivalent circuit for what is inside the regulator...

 

[berkeman]

BTW, as I think about it more, I think CWatters is probably correct. I initially thought that the extra transistors may be to support low dropout regulation when the 3-terminal regulator does not have sufficient input-output voltage difference across it. But with the way this is put together, it's not a low-dropout add-on. Go with CWatter's hints on this one...

 

[M P]

I mean the equivalent circuit for what is inside the regulator...

let me bring this to another level

So I read post of berkeman and downloaded a spec of this beauty he advised and I ma just reading...
Starting current
Thermal protection
Ref V
Series Pas element
SOA protection
Current Gen etc...

 

[M P]

What does the regulator have (or usually has) that the transistor TR1 doesn't have

Small current flows through?

 

[M P]

Small current flows through?

All I thought is those TR 1 and 2 are transistors pass switched on by a drops from R1 and R2.

 

[berkeman]

One way to start to approach a problem like this is to just come up with some voltage numbers and resistance values, and visualize what is going to happen. Then based on that first cut, refine your voltage numbers and resistance values to get closer to what the real behavior would be.

So assume the 3-T regulator is an LM7805, so the output voltage is 5V, right? Then make Vin = 8V and have a 1V drop across R1 to make 7V into the input of the 3-T regulator. Assume you get about 0.5V across R2, and work out what happens in the circuit. Can you post your thoughts about that starting point? That's what I did to start to understand CWatter's hints versus what I initially thought the circuit did...

 

[M P]

One way to start to approach a problem like this is to just come up with some voltage numbers and resistance values, and visualize what is going to happen. Then based on that first cut, refine your voltage numbers and resistance values to get closer to what the real behavior would be.

So assume the 3-T regulator is an LM7805, so the output voltage is 5V, right? Then make Vin = 8V and have a 1V drop across R1 to make 7V into the input of the 3-T regulator. Assume you get about 0.5V across R2, and work out what happens in the circuit. Can you post your thoughts about that starting point? That's what I did to start to understand CWatter's hints versus what I initially thought the circuit did...

Thanks berkeman tha you not giving upon me
So we have 6.5V and that need to be rectified by TR2?

 

[berkeman]

Thanks berkeman tha you not giving upon me
So we have 6.5V and that need to be rectified by TR2?

No, rectifying is not what TR2 does. Can you post a copy of the circuit with the voltages that you work out for the first-cut input voltages I mentioned?

 

[M P]

That is what I think you were saying right?

 

[M P]

That is what I think you were saying right?

So the volt drop should turn those TR1 and TR2 ??

 

[berkeman]

So the volt drop should turn those TR1 and TR2 ??

Closer. Remember I said 8V in and 0.5V drop across R2. So the emitter of TR1 should be at 7.5V, not 6.5V.

So now to get closer to operation, we need less voltage drop across R2, so say it is only 0.3V. Will that turn on TR1? And if so, do you see what CWatters was hinting at, about providing extra current to the output?

And to start to see what TR2 does, increase the input voltage to 10V, and adjust R2 so that TR1 is on with 7V at the input to the 3-T still. What does changing the value of R2 do? What happens when you get about 0.7V across it?

 

[CWatters]

let me bring this to another level

So I read post of berkeman and downloaded a spec of this beauty he advised and I ma just reading...
Starting current
Thermal protection
Ref V
Series Pas element
SOA protection
Current Gen etc...

Expand on SOA. That includes the feature I was thinking of in post 3.

 

[M P]

Expand on SOA. That includes the feature I was thinking of in post 3.

Thank you for your answer CWatters!
"SOA is defined as the voltage and current conditions over which the device can be expected to operate without self-damage"
So TR2 protects the circuit by the sound of it?

 

[M P]

Closer. Remember I said 8V in and 0.5V drop across R2. So the emitter of TR1 should be at 7.5V, not 6.5V.

So now to get closer to operation, we need less voltage drop across R2, so say it is only 0.3V. Will that turn on TR1? And if so, do you see what CWatters was hinting at, about providing extra current to the output?

And to start to see what TR2 does, increase the input voltage to 10V, and adjust R2 so that TR1 is on with 7V at the input to the 3-T still. What does changing the value of R2 do? What happens when you get about 0.7V across it?

I did not understand the first bit and you are saying we need to rise input to 10V. Would not that give 9.5V at R2?
From clues of CWatters I think TR1 is a transistor pass and TR2 protects the circuit from self damage.

 

[CWatters]

I note the load isn't specified. What sort of load might cause damage to the regulator?

 

[berkeman]

I did not understand the first bit and you are saying we need to rise input to 10V. Would not that give 9.5V at R2?
From clues of CWatters I think TR1 is a transistor pass and TR2 protects the circuit from self damage.

That's pretty accurate. Can you describe how TR2 does this protection task? It's a very common circuit in power supply design...

 

[M P]

That's pretty accurate. Can you describe how TR2 does this protection task? It's a very common circuit in power supply design...

hello guys again and thanks for helping me..
I was reading stuff about LM79XX features like berkeman advised and read something might fit this situation...
TR2 perhaps reduce the difference between input and output. greater the difference more power is dissipated in the regulator so it gets hot??

 

[berkeman]

hello guys again and thanks for helping me..
I was reading stuff about LM79XX features like berkeman advised and read something might fit this situation...
TR2 perhaps reduce the difference between input and output. greater the difference more power is dissipated in the regulator so it gets hot??

No... What happens when the current through R2 and through TR1 gets large enough that you get close to a diode drop across R2?

 

[M P]

No... What happens when the current through R2 and through TR1 gets large enough that you get close to a diode drop across R2?

TR2 will kick in and regulate the current..

 

[berkeman]

TR2 will kick in and regulate the current..

Pretty much. When it stars to turn on, where does it divert input current to?

 

[M P]

Pretty much. When it stars to turn on, where does it divert input current to?

it goes towards TR1

 

[CWatters]

Try again. It's a surprising answer.

 

[M P]

Try again. It's a surprising answer.

Not sure to be honest goes to voltage regulator?

 

[M P]

Not sure to be honest goes to voltage regulator?

Unless is something to do with R1?

 

[CWatters]

Not sure to be honest goes to voltage regulator?

Exactly.

The regulator chip has a built in current limit but T1 (on it's own) doesn't. T2 adds current limiting to T1.

If the load current increases the current in R2 goes up.
When the drop across R2 exceeds Vbe T2 starts to turn on
That pulls the base of T1 up reducing/limiting the current through T1
and sending more current through the regulator which has it's own internal current limiting

 

[M P]

Exactly.

The regulator chip has a built in current limit but T1 (on it's own) doesn't. T2 adds current limiting to T1.

If the load current increases the current in R2 goes up.
When the drop across R2 exceeds Vbe T2 starts to turn on
That pulls the base of T1 up reducing/limiting the current through T1
and sending more current through the regulator which has it's own internal current limiting

Thank you guys for all your help!