How about the LM317. How do I calculate it's heat dissipation at a given output voltage and current?

Listen to me carefully Zeebit..

The Tants used in the circuit are going to cause problems...replacing them with Electrolytics will cause problems too.

Not a nice circuit...as I told you. Or is that not what you want to hear??

What do you expect me to do? Like what options do I have?

 

The power dissipation of the transistor is (input voltage - output voltage) x amps.

Heat load on the transistor is highest when output voltage is low and current is high.

 

Got it. I was reading your post to fast.

How about the dissipation of the LM317?

 

Hello,

When you look at the schematic, there you will see R1 of 33 Ohms.
This will regulate the current though the large tranisitor.
The current through the LM317 will be about the emmitor-base voltage divided by R1,
so about 0.7 / 33 = 21 mA.
The dissipation of the LM317 will be low in comparisson with the large transistor.

Bertus

 

Okay thanks all. I will have to read your replies tommorow. I need to sleep.

 

Listen to me carefully Zeebit..

The Tants used in the circuit are going to cause problems...replacing them with Electrolytics will cause problems too.

Not a nice circuit...as I told you. Or is that not what you want to hear??

You actually haven't told much at all. You've said that you built this circuit a quarter of a century ago, that you may not have done the layout correctly, and that you had problems with it. You have not given any indications of what is wrong with this circuit, so how is the OP possibily supposed to choose any other circuit that doesn't have the same problem?

 

So the current through the LM317 is around 21mA, how do I know what voltage is going through it to get the heat dissipation?

 

So the current through the LM317 is around 21mA, how do I know what voltage is going through it to get the heat dissipation?

Voltage doesn't go through things, it appears across things. With that in mind, it is very simple to find the voltage across the regulator.

You have Vin (35V in the schematic).

You have the voltage across the 33Ω resistor, which is clamped by the Vbe of the bypass transistor, so that's about 0.7V leaving you with (Vin-Vbe) at the input of the regulator and Vout at the output. Hence the voltage across the regulator is

Vreg = Vin-Vbe-Vout

For simplicity sake, you can make a slightly conservative estimation that

Vreg ≈ Vin - Vout

The current is basically Vbe/Ro where Ro=33Ω in this case. Thus

Preg ≈ (Vin-Vout)(Vbe)/Ro

Using the schematic values, the worst case power would therefore be when Vout=1.2V and would be about 0.75W.

 

Thanks! I'm a little stupid when it comes to calculations.

This is good since I can just mount it on the pcb and attach a small heatsink.

 

Preg ≈ (Vin-Vout)(Vbe)/Ro

Using the schematic values, the worst case power would therefore be when Vout=1.2V and would be about 0.75W.

Just want to ask one more thing. Does the heat dissipation of the regulator change when the supply is in CC mode?

 

The average power in the regulator is always the average value of the voltage across the regulator times the current through the regulator. If that changes when you are in CC mode, then the power dissipation changes. If it doesn't, then it doesn't.

 

I'm still a little confused. I am just learning my way through all this.

So even if the supply is in CC mode, the dissipation of the regulator is still
Preg ≈ (Vin-Vout)(Vbe)/Ro?

 

Yes.

There are ways to reduce this, but that involves adding circuitry.

 

Nah, I'm fine with the setup.

In this case I can just mount it on the PCB and stick on a small heatsink. In my case the max dissipation is just 1/2W.

Thanks again!