Yet another LM317 Current Regulator

Thread Starter

wayneh

Joined Sep 9, 2010
17,498
Will this one work? Take a look and please comment. The goal is to be able set a stable, constant current up to ~4A thru the load.

The LM317 is doing only the job of a voltage reference. I'm boxed in by a low supply voltage, so I need a low voltage at the emitter and thus a low control voltage at the base of the 2N3055. Also, a low shunt resistance at the emitter allows a higher current at a given wattage rating.

So I'm using the LM317 to give me just a volt or two, near it's bottom end. The diode is used to drop that an additional ~0.7v, allowing the base to fall below ~0.6, to allow near zero current to flow while still "on" at the FET and under control.

The biggest question I have is whether I can find a nice pot that operates at such a low resistance range to control the LM317 output. I know I don't have one in my parts cabinet. Thanks for looking.

PS: When the MOSFET is on, the ∆V across it is no more than 0.1V. I guess I haven't seen it handle a 4A load yet, so maybe it'll be a bit higher than that. Obviously not a problem in this circuit to get higher control voltage as needed. Maybe the pot should go to 150Ω.
 

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Wendy

Joined Mar 24, 2008
23,429
What is the FET for? It doesn't contribute anything as far as I can see, especially with the gate in use. It's been a long day, so I'm going to bed after this post.

Quick side note, always put parts designators on the schematic. It allows discussing exact parts without any misunderstandings.

You have a voltage regulator in the form of the LM317, 100Ω variable, and 120Ω resistors. A LM317 must have a capacitors to prevent oscillation, it is in the datasheet.

This voltage regulator feeds the BJT transistor, 1KΩ, and diode. You are feeding a constant voltage into the BJT transistor, using the diode to drop the voltage by 0.6V and reduce the voltage feed to the BJT that much more. A BJT transistor with an emitter resistor (0.25Ω) is a perfectly fine constant current source, you can eliminate the MOSFET and get good results.
 

Thread Starter

wayneh

Joined Sep 9, 2010
17,498
What is the FET for?
It's simply an on/off switch to enable current to flow only when other requirements - not shown - are met. I suppose a much smaller FET, with its logic reversed relative to the diagram, could be used to control the base voltage of the transistor. Turning "on" that FET would ground the base pin. But I've already got the IRF540 working as shown.
 

tom66

Joined May 9, 2009
2,595
It's simply an on/off switch to enable current to flow only when other requirements - not shown - are met. I suppose a much smaller FET, with its logic reversed relative to the diagram, could be used to control the base voltage of the transistor. Turning "on" that FET would ground the base pin. But I've already got the IRF540 working as shown.
It's on the low side, so any short to ground will defeat this switch.

I think you'd honestly be better off with an op-amp. With an op-amp, you could even turn the output off completely, by turning off the Vref.
 

Thread Starter

wayneh

Joined Sep 9, 2010
17,498
I think you'd honestly be better off with an op-amp. With an op-amp, you could even turn the output off completely, by turning off the Vref.
That makes some sense, to use an op-amp instead of the LM317, however I think the base current can be quite a lot more than most op-amps can handle. The current gain of the 2N3055 could be as low as under 10, so if it's running at 4A, the base current might be a few hundred mA. The LM317 has no problem with that.
 

tom66

Joined May 9, 2009
2,595
That makes some sense, to use an op-amp instead of the LM317, however I think the base current can be quite a lot more than most op-amps can handle. The current gain of the 2N3055 could be as low as under 10, so if it's running at 4A, the base current might be a few hundred mA. The LM317 has no problem with that.
Then use a darlington pair, or a power op-amp, like the LM675.
 

Thread Starter

wayneh

Joined Sep 9, 2010
17,498
Then use a darlington pair, or a power op-amp, like the LM675.
I can see where those are viable options, but I don't see that I'd necessarily gain anything over what I've already drawn. Well, maybe I could use a more standard pot.
 

SgtWookie

Joined Jul 17, 2007
22,230
Look at the L272 or L2722. Very cheap power opamps, that can source/sink up to 1A if they are heatsinked by their traces adequately. More than enough current to drive the base of a 2N3055.
 

Thread Starter

wayneh

Joined Sep 9, 2010
17,498
Look at the L272 or L2722. Very cheap power opamps, that can source/sink up to 1A if they are heatsinked by their traces adequately. More than enough current to drive the base of a 2N3055.
Well, OK, but ... why? I don't see what I'll gain from a re-design around an op-amp versus what the LM317 is doing here.
 

Ron H

Joined Apr 14, 2005
7,063
Well, OK, but ... why? I don't see what I'll gain from a re-design around an op-amp versus what the LM317 is doing here.
Your circuit's output current will be very unstable as the transistor temperature changes, and/or as ambient temperature changes.
You can take an op amp, a low-cost reference, and a Darlington transistor or MOSFET, and make a very solid, variable current sink (or source).
You will need some serious heatsinking for the transistor, no matter how you do it.
 

Thread Starter

wayneh

Joined Sep 9, 2010
17,498
Your circuit's output current will be very unstable as the transistor temperature changes, and/or as ambient temperature changes.
You can take an op amp, a low-cost reference, and a Darlington transistor or MOSFET, and make a very solid, variable current sink (or source).
You will need some serious heatsinking for the transistor, no matter how you do it.
Understood. Thanks for the education. The flaw in my thinking was that the ∆V (base - emitter) would be constant. That's sort of true, but as you note it can walk around and respond to temperature, causing havoc in my circuit. Especially if the temp effect leads to positive feedback.

I'll re-design. Actually I think I'll just retreat to using the LM317 below 1.5A, or gang two together if I need more juice. I just happened to have the 2N3055 on hand, but it's going back in the drawer for now.

Yikes, I just became a senior!! :eek:
 

Ron H

Joined Apr 14, 2005
7,063
Understood. Thanks for the education. The flaw in my thinking was that the ∆V (base - emitter) would be constant. That's sort of true, but as you note it can walk around and respond to temperature, causing havoc in my circuit. Especially if the temp effect leads to positive feedback.

I'll re-design. Actually I think I'll just retreat to using the LM317 below 1.5A, or gang two together if I need more juice. I just happened to have the 2N3055 on hand, but it's going back in the drawer for now.

Yikes, I just became a senior!! :eek:
ΔVbe is the main culprit, because it will be the transistor that takes the big temperature hit. However, the diode ΔV works in the same direction, but is mainly affected only by ambient.
 

Thread Starter

wayneh

Joined Sep 9, 2010
17,498
OK, I took the good advice offered here and collected some data, and now have an improved design. Take a look.

The FET I was using for switching the big current on and off has been replaced by just the comparator that used to control the FET. I reversed the logic so the comparator pulls the transistor's base to ground when it's "off", instead of pulling up the FET's gate when "on".

I looked at the 2N3055 under expected conditions and found a current gain of 50X; 40mA at 2A. My previous measurement was made with a lower Vce and the base current was much higher when Vce dropped. Using a low ohm shunt keeps the emitter voltage low and thereby maintains Vce.

So I could have used a medium current op-amp to control the 2N3055 directly, but since I had an un-used op-amp onboard and plenty of spare transistors, I went with the Darlington arrangement.

Without adding more components I can't figure out how to light up an indicator LED when it all turns "on". This was easy when the FET was controlling below the power transistor. But that's about the only thing I'm unhappy about. Otherwise, I think this is a much better and more versatile design than I first posted. Comments?
 

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Ron H

Joined Apr 14, 2005
7,063
Your comparator needs to be connected to the +input (wiper of the pot). Grounding the -input will send the output to the positive rail.
Connecting to the +input will still allow a little current flow, equal to the comparator saturation voltage divided by 0.25Ω. Add a resistor between the wiper and the +input to minimize it.
 
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Thread Starter

wayneh

Joined Sep 9, 2010
17,498
Your comparator needs to be connected to the +input (wiper of the pot). Grounding the -input will send the output to the positive rail.
Oops, of course that's right. My hand sketch was right but I botched the drawing. So much for working late.
Connecting to the +input will still allow a little current flow, equal to the comparator saturation voltage divided by 0.25Ω. Add a resistor between the wiper and the +input to minimize it.
OK, I can see that since the comparator isn't ideal and doesn't really pull all the way to ground, a small current could still be allowed to flow through the shunt.
 

Thread Starter

wayneh

Joined Sep 9, 2010
17,498
Here's a way that avoids the small OFF current.
Brilliant! And this also gives me back a way to light up an LED when the circuit is turned "on" by releasing the pull-up of the -input.
v.3 attached

[UPDATE] I breadboarded this and it works fine. Only time will tell how stable it is over several hours of operation. I added a 470Ω resistor between the base of the smaller transistor and the driving op-amp, to protect the op-amp output. If the voltage drops too much across the load, the Vce collapses around the darlington pair, and the base current goes too high for the op-amp to keep up. I didn't fry it, but thought it better to add the protection. Thanks to all for helping me arrive at a very nice solution.
 

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