Discrete Voltage Regulator

Thread Starter

remow7

Joined Oct 28, 2017
4
Hello.
Can anyone tell me what is wrong with my circuit . Vout seems to be the same value as Vin .
I need Vout to be between 5- 10V and Vin 18 - 20V.
I need to limit the transistor temperature at 100C and 0,4A current as a protection of overcharge.
 

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crutschow

Joined Mar 14, 2008
34,062
There should be a dot indicating a connection between Q4-C, Q6-C, and Q5-B.
Is that connection made on your circuit?

The transistor dissipation could be up to 6W so Q1 will need to be on a heatsink with a thermal resistance of ≤12.5°/W for a heatsink air ambient of 25°C.
 

AlbertHall

Joined Jun 4, 2014
12,329
The circuit also has the base and collector of Q3 shorted together - they shouldn't be.
There is a box labelled U1 with the legend 1N4148 next to it. I guess that it is supposed to be a zener diode but, in any case, it won't work with a standard diode in that position, either way round.
 

AlbertHall

Joined Jun 4, 2014
12,329
It's a diode connected transistor, but I'm not sure of it's purpose.
If the horizontal link between the base and collector of Q3 wasn't there, and if U1 was a zener diode then Q2 is a constant current source feeding the D1 zener providing the reference voltage for the PSU regulator and Q3 is a constant current source feeding Q4, Q5, and Q6, and then it looks like a working supply.

What have you got in the U1 position?
 

Ylli

Joined Nov 13, 2015
1,083
D1 is the voltage reference - a 6.2 volt Zener.

Q2 looks like it might be a current source feeding D1, with the 1N4148 providing some temperature compensation. . Q3 is confusing - what is it there for? I'm not sure the connection between Q2 base and Q3 base should be there. As said above, there should be a dot connection Q5 base, Q4 and Q6 collectors.

Oh, and with the 6.2 volt Zener, the low limit on the output will be about 6.8 volts. You would need to use a 4.3 volt Zener to get it down to 5.0 volts.
 

crutschow

Joined Mar 14, 2008
34,062
Here's my take on the circuit with a lower voltage reference (TL431, 2.5V) to allow adjustment to below 5V, and simplified of extraneous parts.

The LTspice simulation below shows an output range of 4.0V to 11.3V for 0% to 100% pot rotation, with a current limit of 420mA (yellow trace).

Q3 and Q4 are configured in a differential configuration to minimize the temperature effect of their base-emitter voltages.

C2 eliminate a high frequency peak in the loop response causing oscillations in the output, observed in the simulation.

upload_2017-10-28_23-23-0.png
 

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Thread Starter

remow7

Joined Oct 28, 2017
4
Here's my take on the circuit with a lower voltage reference (TL431, 2.5V) to allow adjustment to below 5V, and simplified of extraneous parts.

The LTspice simulation below shows an output range of 4.0V to 11.3V for 0% to 100% pot rotation, with a current limit of 420mA.

Q3 and Q4 are configured in a differential configuration to minimize the temperature effect of their base-emitter voltages.

C2 eliminate a high frequency peak in the loop response causing oscillations in the output, observed in the simulation.

View attachment 138265

thank you so much , your circuit explains it much better and it works . I did some adjustments to R8 and R5 for my output range of 5-10V.
Now i just have to implement thermal protection for Q2
 

MrAl

Joined Jun 17, 2014
11,276
Hello there,

Q3 in the original circuit is connected as a diode, and it is in series with a larger resistor than the preceding transistor so that could be for the purpose of creating a break point.

A break point is a diode or transistor connected in series with a fixed or variable resistance and is used to conduct when the voltage reaches a certain point so that it steers current away from the junction it is 'breaking'. The junction in this case is the transistor base.

I did not analyze the circuit yet so i can only guess that it may be for either current limit or temperature compensation. In any case it is there to reduce drive to the preceding transistor for some circuit conditions.

Whether or not this is a good idea or even correct i have no control over because i did not analyze the circuit but if someone already has this set up in a simulator all it takes it to find out how the circuit behaves is to simulate with and without the 'diode' and series resistor in the circuit, and also perhaps if the series resistor raised and lowered has some effect on the output, most likely under conditions of load, or under conditions of changing temperature. Keep in mind that when the temperature goes up, the base emitter characteristic voltage goes down and thus makes a transistor conduct harder, which may or may not be an important aspect of adding that 'diode' and series resistor.

Some circuit design parts have a more subtle effect on overall circuit performance so they may not be as easy to figure out. This could be a blatant error, or it could provide a very useful function. The answer lies in careful analysis ether by hand (difficult here) or by simulator under varying conditions.
 
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