Simple linear power supply design

[wakoko79]

Hi, I'm doing a project now and it includes making a regulated supply. The supply isn't really the focus of my project so I didn't bother doing an SMPS.

The circuit I use gets its power from a transformer (20Vrms). I'm using an lm317 regulator with 2 pass transistors (pnp). Each transistor is calibrated so that the voltage across the 0.27 ohm resistor will redirect the flow of current back to the lm317(which has internal current limit) so that each pass transistor will have a maximum current of about 3A flowing through them.

Please comment. Help me improve this. =) thanks..


View attachment 62885

 

[MrAl]

Hi,

Just for starters with the amount of current it looks like you will be drawing you probably want more than 3300uf for the main filter cap.

 

[wakoko79]

well, I think we have some 4.7mF.. I think I should have about 33mF but that is just too big and I dont think I can buy them locally.

 

[MrAl]

Hi,

Well in wall warts they put at least 2200uf per amp i think. We can do a little calculation if you tell us your input voltage and required max output voltage and max output current. We can then tell if you'll see any ripple on the output if the caps are too low in value.

 

[wakoko79]

Umm.. The nominal output voltage is 24Vdc and max output current is 6Adc. And as I've said earlier, the input is off-line. a transformer (maybe a centertap, but I'll opt for a full wave rectification anyway.) I hope I can buy a 20Vrms (28.28Vpeak) transformer with a rating of 6A somewhere, I think a 24Vrms transformer will generate a lot of wasted heat.

I was also thinking of connecting several capacitors in parallel. I'll take your advice. =)

 

[k7elp60]

Just as a sidenote when using brute force filtering in power supplys, donot exceed maximum continous current of rms(.566) or the transformer will heat up as the peak charging current will exceed the ratings of the transformer.

 

[ronsimpson]

Q4 emitter and collector are shorted together.
Q2, Q4 do not function!

 

[wakoko79]

Just as a sidenote when using brute force filtering in power supplys, donot exceed maximum continous current of rms(.566) or the transformer will heat up as the peak charging current will exceed the ratings of the transformer.

Huh? Please explain. I didn't read up anything like that. Thank you for that info sir! And where did .566 came from?

Q4 emitter and collector are shorted together.
Q2, Q4 do not function!

Oh right! But that's weird. Simetrix did the simulation pretty well without errors so I thought it had no errors.

So will this correct it?
View attachment 62889

I think this circuit still is wrong though. I included 2 pass transistors that have current limiters so that thermal runaway won't be much of a problem. But with this config, if one transistor undergoes thermal runaway, current will be piped to LM317. And if there is overcurrent, LM 317 will halt functioning, right? Help me here please >.<

 

[alec_t]

Not clear what you're trying to do here. The purpose of pass transistors is to off-load current which would otherwise have to pass through the regulator IC; yet you seem to be trying to dump current from the pass transistors back through the regulator?

 

[MrAl]

Not clear what you're trying to do here. The purpose of pass transistors is to off-load current which would otherwise have to pass through the regulator IC; yet you seem to be trying to dump current from the pass transistors back through the regulator?

Hi,

The basic idea is to try to pump the current into the input of the 317 rather than pass it to the output (bypass the pass transistors) when the current exceeds a limit set by the 200mOhm resistor. This puts the total load back on the 317 which then goes into current limit. Might be a better way, but that's one way.

 

[ronsimpson]

Not clear what you're trying to do here. The purpose of pass transistors is to off-load current which would otherwise have to pass through the regulator IC; yet you seem to be trying to dump current from the pass transistors back through the regulator?

If the current through the LM317 is low Q1, Q3 does nothing.
If the current is enough that 0.6 volts appears across R4 (2.7 ohms) then Q1, Q3 turn on and help lift the load.
The circuit works!
The current limit is a little strange, but works.

 

[alec_t]

Perhaps something like this? I've only shown one pass transistor plus its current-limiting add-ons. Pretend the regulator U1 is a 317. Ignore resistor 'Reg' which is just for the simulation so that regulator current can be monitored.
The graphs show currents and output voltage for loads between 0.01Ω and 10Ω.
When Q1 collector current tries to rise above ~3A the voltage drop across R1 turns on Q2, raising Q3 base voltage and so reducing Q1 base current flowing through Q2 and R3, thus clamping Q1 collector current at ~3A.

 

[ronsimpson]

alec_t,

I like your current limit. I have never used Q3.

 

[wakoko79]

Hi,

The basic idea is to try to pump the current into the input of the 317 rather than pass it to the output (bypass the pass transistors) when the current exceeds a limit set by the 200mOhm resistor. This puts the total load back on the 317 which then goes into current limit. Might be a better way, but that's one way.

Yes. Pass transistors are there to get more current to the load that the lm317 won't be able to handle otherwise. But the pass transistors also have a limit. So to stay in the SOA, the pass transistor should also be current limited. Thermal runaway will be fine (I guess in this circuit) if the load current is only about 2A~3A, but if the load current is around 6A, thermal runaway will make one of the two pass transistors really hot (or even melt it).

I know this is not the best config, especially the one I fixed. It will work for sometime, but when the temperature effects kick in, it won't be able to reach the set max current of 6A. At least that's what I think will happen in theory.

 

[ronsimpson]

Because there are separate emitter resistors on Q1, Q3 the thermal runaway should be less of a problem than if the two emitters are connected together. If Q1 is hotter and wants to draw MORE power then its emitter resistor will help turn off Q1 and shift more of the load to Q3.

I would have added base resistors to Q1, Q3 and had Q4, Q2 connect to the base of the individual transistors. This way each transistor (1,3) will have independent current limiting. This is only to watch for uneven current at and near current limit caused by thermal runaway. Also to balance between two transistors with unequal parameters.

 

[wakoko79]

Perhaps something like this? I've only shown one pass transistor plus its current-limiting add-ons. Pretend the regulator U1 is a 317. Ignore resistor 'Reg' which is just for the simulation so that regulator current can be monitored.
The graphs show currents and output voltage for loads between 0.01Ω and 10Ω.
When Q1 collector current tries to rise above ~3A the voltage drop across R1 turns on Q2, raising Q3 base voltage and so reducing Q1 base current flowing through Q2 and R3, thus clamping Q1 collector current at ~3A.

This is great! Thanks a bunch! I think with this, there wont be thermal problems. =)

 

[wakoko79]

So ok, I found out that I was actually using a 24Vrms in simulations as input. I adjusted the circuit alec_t made and it look like this:View attachment 62906

And here is the power of one of pass transistor (in blue) and the power in the LM317 (in red).
View attachment 62907
The average power disipated in each of the transistors is about 15.8W. I'm guessing that will be really hot, but how hot do you think? how big a heat sink should I use (with TO-220)?
I'm really new to this, this temperature calculations are not even taught in our school.


PS: I changed the supply from 24Vrms to 20Vrms. The LM317 then didn't regulate the voltage to 24V. I think it has something to do with the drop-out voltage of the chip. So am I stuck now with 220V to 24V tx?

I'm also concerned with the thing k7elp60 said:

Just as a sidenote when using brute force filtering in power supplys, donot exceed maximum continous current of rms(.566) or the transformer will heat up as the peak charging current will exceed the ratings of the transformer.

Is that true? Please confirm. thanks again!

And here is a simetrix file for simulation.

 

[ronsimpson]

The current rating on a transformer is for a resistive load not a full wave bridge. k7elp60 said to use the transformer at 1/2 rating. (.566) I think 3/4 rating is OK at room temperature. There are several sources that talk about de-rating by .6 to .7.

Heat sinks are rated by temperature/watts. example 10C/1watt. This heat sink with 15.8W will increase 158 degrees above room temp. (20C)
Most heat sinks will also have a graph of air flow. Even a little fan makes a big difference!
I would aim for 50C on the case of the transistor. That is at the point where you want to remove your finger in 2 seconds. 50C-20C=30C so the temp rise is 30C.
30C/15W=2C/1W You need a big heat sink. or A heat sink with air flow.

 

[k7elp60]

Hi tried to post a reply as in ronsimpson's post, but had some difficulty doing so. Here is my thoughts.
What I am sayings is if you want the power supply to supply a continuous 3A then the transformer needs to be rated at 3/.566. This figure came from two sources, 1 Triad Transformer Co. and 2 from a transformer manufacture friend of mine. As you know the peak voltage of the transformer is 1.414 times the rms. The filter capacitor will charge to near this value. Since the average load curreuld be 3A then the peak charge current is close the double that or 3/.566 so if the transformer is not rated at near 6A rms load current the winding supplying the current will heat up.
I am on a short vacation so I am away from home and don't have my notes with me so when I say the average value of a brute force filter capacitor is 3000uf per amp. I know there is a formula that has ripple voltage frequency and load current to calculate the exact value of capacitor required. I have built a great number of linear power supplies and have been real successful with the 3000uF/amp. I once built a high current linear power supply and the transformer was getting hot, so a did some research on max current.

 

[ronsimpson]

k7elp60,

I have rethought the transformer heating/current rating problem.

>The transformer is rated for a resistive load at 3 amps. That's RMS.
>When the transformer is driving through diodes into a capacitor the peak current goes up and the duty cycle goes down.
>There is a way of doing this with math but I want a simple answer. so.....
>Lets say the current in the diodes is on for 1/2 the time. The current must be 2X or 6amps for 50%. (rough numbers)
>Watts=I^2 x R watts lost in copper is current squared times resistance in the wire.
>To make the math easy R=1 ohm.
>With resistive load 3a^2=9 watts of heat.
>With a diode+cap+resistor load. Current = 6A @ 50%
(6^2)/2=19 watts.
With a diode+capacitor the current goes up as the duty cycle goes down and the heat goes up.

I did find the transformers ran 100C (or more) at full power.