Linear Lab PSU kit overheating

Thread Starter

Paul_Alexa

Joined Jul 16, 2022
4
Hello all, and thank you for taking the time to read my post. I have been experimenting with a cheap psu kit and I have observed that Q4 (2SD1047) is overheating even with a heatsink on it. I have used a 1 ohm 5W resistor as a load, at a voltage of 2 volts and 2 amps, (2 amps is what I aim for, despite the PSU saying 3 in its documentation).

I am aware that Q4 has to dissipate 26 Volts X 2 Amps = 52W of heat (28 volts being my transformer's secondary voltage), but with a thermocouple between the transistor's body and the heatsink, the reading reached 148 celsius rather quickly (sub 30 seconds) which is not good, in my opinion. I even had a fan blowing over it , which managed to keep it at a sort of constant 145 celsius.

Since the kit is some cheap Chinese stuff I wondered if the 2SD1047 could be fake, and I replaced it with a 2N6307 (8 amps collector current) to no avail, it overheating just like before. Now, I have to admit that this is my first time working with a linear PSU. I have devised two scenarios:

1. It's my lack of experience and the lack of a heatsink with good thermal mass and a fan. I believe that the temperature should not be more than 60-80 on the heatsink.
2. I should substitute the transistor with two similar power transistors, in parallel, with ballast resistors to prevent thermal runaway, which would make it easier for the PSU to handle the load.

Now, the questions that I have are:
1. Should I substitute Q4 with two transistors in parallel, a beefier single one or should I just keep everything like it is and get a proper heatsink and fan?
2. What would be the normal operating case temperature for the transistor ? ( The datasheet says 150 degrees Celsius maximum temperature)

I have attached the PSU's documentation, with the schematic being on the last page.
Thank you.
 

Attachments

LowQCab

Joined Nov 6, 2012
2,502
Heat-Sinks are Big and expensive,
both are things that the makers of these Kits want to avoid at any cost.

Just install a small Fan on the Heat-Sink,
any Computer repair-shop or parts-supplier will have them.

To get decent Life-Expectancy from your Output-Transistor, keep the Temps below ~80C.
.
.
.
 

KeithWalker

Joined Jul 10, 2017
2,518
I'm not surprised that Q4 gets hot when it is dissipating 52 watts! My soldering iron is rated at 40 watts and is regulated to keep the temperature down to 360 deg.C.
Increase the size of the heat sink if you are able and make sure you use a good thermal compound between the transistor and the sink. A fan will make a big difference.
When you use the power supply, keep in mind its limitations. If you really need 2 Amps at 2 Volts, you need a different supply.
 

crutschow

Joined Mar 14, 2008
30,774
Note that, for a 28Vac input (the power supply spec says it should be 24Vac) the rectified DC is about 32V, so the dissipation is more like 30V x 2A = 60W.
That heat sink likely has a thermal resistance of about 2°C/W.

Changing the transistor and adding another will not change the amount of heat being dissipated.
A fan and/or a larger heat-sink are the only way to reduce the temperature.

Also note that the transistor rating of 150°C is it's junction temperature, which is above its case temperature by the junction-case thermal resistance (1.25 °C/W for the 2SD1047) times the power being dissipated.
You are lucky you didn't melt the junction if you had a case temperature of 145°C, which implies a junction temperature of at least 220°C.

But when will you practically need to power a 2A, 2V load?
That doesn't often happen in normal circuit testing.
 
Last edited:

MrChips

Joined Oct 2, 2009
27,114
Input voltage to the pass transistor is +32.5V.
If the transistor has to pass 3A then it has to dissipate 100W.
You need a bigger heat sink and more active heat dissipation.
 

Jon Chandler

Joined Jun 12, 2008
389
Facts about linear regulators:

PD = I × (Vin – Vout)

where

PD = power dissipated by the regulator
I = current in amps
Vin – Vout = drop across regulator

The "regulator" can be a single chip, several chips or chips and pass transistors and may include series diodes and resistors. Spreading the voltage drop across more components does not change the total power dissipation. Changing to a different linear regulator will not change the PD markedly.

Delivering 2 volts at 2 amps is almost the worst case for your power supply – the large delta V results in large PD.
 

Thread Starter

Paul_Alexa

Joined Jul 16, 2022
4
Thank you all for taking your time to reply to me, I have learned a lot by reading your answers and will keep everything in mind.
 

crutschow

Joined Mar 14, 2008
30,774
Without that heat-sink and no fan, you want to keep the dissipation below about 25W so, for the 28Vac supply, this means that the output voltage and current should be limited so it doesn't exceed (32-Vout) * Iout = 25W.

One way to reduce dissipation is to use a lower AC voltage when you need a lower output voltage.
For example you could use a dual 12V-24Vac transformer, and switch in the 12V winding when you don't need more than about 12Vdc output.
That will cut the dissipation in half for a given output voltage and current.
 

dl324

Joined Mar 30, 2015
15,098
Welcome to AAC!
I am aware that Q4 has to dissipate 26 Volts X 2 Amps = 52W of heat (28 volts being my transformer's secondary voltage)
Transformer secondaries voltages are expressed in RMS volts. A 28V secondary will give you about 38V after the bridge rectifier.
Since the kit is some cheap Chinese stuff I wondered if the 2SD1047 could be fake, and I replaced it with a 2N6307 (8 amps collector current) to no avail, it overheating just like before. Now, I have to admit that this is my first time working with a linear PSU.
What is your intended application for this power supply?

If you intend to use it for low voltage at high current, you'd be better off with switching regulator or a different transformer.
What would be the normal operating case temperature for the transistor ?
An allowable case temperature could be too hot for you to touch. Junction temperature is more important.

These devices specify an unusually high allowable junction temperature (200C). But at a case temperature of 200C, the allowed power dissipation is 0W.
1658162492150.png
 

BobaMosfet

Joined Jul 1, 2009
2,028
Hello all, and thank you for taking the time to read my post. I have been experimenting with a cheap psu kit and I have observed that Q4 (2SD1047) is overheating even with a heatsink on it. I have used a 1 ohm 5W resistor as a load, at a voltage of 2 volts and 2 amps, (2 amps is what I aim for, despite the PSU saying 3 in its documentation).

I am aware that Q4 has to dissipate 26 Volts X 2 Amps = 52W of heat (28 volts being my transformer's secondary voltage), but with a thermocouple between the transistor's body and the heatsink, the reading reached 148 celsius rather quickly (sub 30 seconds) which is not good, in my opinion. I even had a fan blowing over it , which managed to keep it at a sort of constant 145 celsius.

Since the kit is some cheap Chinese stuff I wondered if the 2SD1047 could be fake, and I replaced it with a 2N6307 (8 amps collector current) to no avail, it overheating just like before. Now, I have to admit that this is my first time working with a linear PSU. I have devised two scenarios:

1. It's my lack of experience and the lack of a heatsink with good thermal mass and a fan. I believe that the temperature should not be more than 60-80 on the heatsink.
2. I should substitute the transistor with two similar power transistors, in parallel, with ballast resistors to prevent thermal runaway, which would make it easier for the PSU to handle the load.

Now, the questions that I have are:
1. Should I substitute Q4 with two transistors in parallel, a beefier single one or should I just keep everything like it is and get a proper heatsink and fan?
2. What would be the normal operating case temperature for the transistor ? ( The datasheet says 150 degrees Celsius maximum temperature)

I have attached the PSU's documentation, with the schematic being on the last page.
Thank you.
Can you find some other electronics scrap (bad power-supply) anywhere to steal a heat-sink from? I always salvage electronics for this sort of stuff. Q4 gets hot because they are operating it in the linear region, rather than saturation- that's not necessarily bad design, but not great design either. They are using the transistor to control voltage while passing current for the load. If you want current and voltage of any significance, you're going to have to deal with watts- and usually that's a heatsink and/or fan to give the component the overhead to avoid thermal damage.

A power-dissipation curve can be quite handy if you work it out for your problem (they look like this):

1658240525441.png

If the load-line falls to the left of the Pd curve, the component won't fry.
 
Last edited:

crutschow

Joined Mar 14, 2008
30,774
Q4 gets hot because they are operating it in the linear region, rather than saturation- that's not necessarily bad design, but not great design either.
Not sure what you mean that it's "not great design", since it's normal for a transistor to be in the linear region for the operation of a linear regulator (how else would you do it?).

For the transistor to operate in saturation would require it to be in some type of switching regulator.
 

Thread Starter

Paul_Alexa

Joined Jul 16, 2022
4
I have upgraded the coolers on my computers a couple of years back and have found a barely used CPU heatsink with a good fan on it, and I will use that for the transistor. I have already replaced the 24 volt regulator with a 12 volt one.
Thank you very much for the additional info.
 

MrChips

Joined Oct 2, 2009
27,114
I have upgraded the coolers on my computers a couple of years back and have found a barely used CPU heatsink with a good fan on it, and I will use that for the transistor. I have already replaced the 24 volt regulator with a 12 volt one.
Thank you very much for the additional info.
If you replaced the 24V regulator with a 12V regulator then the 12V regulator will get hot.
You cannot escape the simple math and physics.
33V x 3A = 100W which has to be dissipated as heat somewhere.
You need to reduce the input voltage. One way to do that is to replace the power transformer with one that outputs a lower voltage.
Can you show us the specifications of the transformer?
 

KeithWalker

Joined Jul 10, 2017
2,518
If you replaced the 24V regulator with a 12V regulator then the 12V regulator will get hot.
You cannot escape the simple math and physics.
33V x 3A = 100W which has to be dissipated as heat somewhere.
You need to reduce the input voltage. One way to do that is to replace the power transformer with one that outputs a lower voltage.
Can you show us the specifications of the transformer?
I believe the regulator he changed was the one for the fan, which will take very little current.
 

Thread Starter

Paul_Alexa

Joined Jul 16, 2022
4
I believe the regulator he changed was the one for the fan, which will take very little current.
Yes, I was talking about the regulator for the fan, my bad. I will also look for a transformer with a lower output voltage .
Thank you all again, you have been very helpful to me!
 

BobaMosfet

Joined Jul 1, 2009
2,028
Not sure what you mean that it's "not great design", since it's normal for a transistor to be in the linear region for the operation of a linear regulator (how else would you do it?).

For the transistor to operate in saturation would require it to be in some type of switching regulator.
There are other means for regulating current without relying on thermal dissipation, other than using a single transistor. Let your considerably knowledgeable imagination soar with the possibilities :)
 
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