Constant current load for testing power supplies (guidelines)

[Dannyx]

Re: Constant current load for testing power supplies (guidelines)

The power handling and the resistance are not related, you can have 5 Ohm resistor that can only handling 1/8W (I.E. SMD resistor) of power dissipation and you can have 5 Ohms resistor that can handle 10W of power, same resistance but the physical size is not the same.

Yes, I believe that's what I said

[eccerr0r]

Re: Constant current load for testing power supplies (guidelines)

Then the other question, how accurate do you need to be...
If you don't care about 1% accuracy and just need to be in the ballpark, don't worry too much about offset voltage (most op amps are within 15mV, a lot are better than this) and just use around 100-200mV to keep most dissipation in the transistor. Just need to make sure you select op amps that can take inputs near the zero volt mark, so you don't need to bias.

[Dannyx]

Re: Constant current load for testing power supplies (guidelines)

Then the other question, how accurate do you need to be...
If you don't care about 1% accuracy and just need to be in the ballpark, don't worry too much about offset voltage (most op amps are within 15mV, a lot are better than this) and just use around 100-200mV to keep most dissipation in the transistor. Just need to make sure you select op amps that can take inputs near the zero volt mark, so you don't need to bias.

Accuracy ? Well not laboratory-grade, that's for sure...I wouldn't be able to obtain such levels of precision with junk-box parts for one thing and secondly its usefulness would outweigh the effort. To draw 10a at 100mV would require a 0.01 resistor because 0.1v/0.01ohm=10a, correct ? (bare with me, I'm just getting the hang of things). The power dissipation would then be 0.1v*10a= 1w ? Can I really get as low as 100mV with an op-amp and a control pot? Ok, it definitely has to be a multiturn pot...if I were really ambitious, I'd try and to it PWM but let's get the basics down for now

[eccerr0r]

Re: Constant current load for testing power supplies (guidelines)

I'd say screw it, if I designed for 10A and it draws either 8A or 12A, it's close enough for a load tester...

But yes, though 100mV is cutting it close for a lot of op amps out there but not in the realm of impossible. I might go and try it and see what a few op amps will do...

Now I don't have a MOSFET I can burn (pun?) on this so I'll have to go with a BJT...

[momaka]

Re: Constant current load for testing power supplies (guidelines)

Can I really get as low as 100mV with an op-amp and a control pot? Ok, it definitely has to be a multiturn pot...

Yes, you can, and *without* a multi-turn pot. The trick: 100 to 270 Ohm resistor connected in series with a diode (to the diode's anode). Connect one side of the 100-270 Ohm resistor to your 3.3V/5V/12V rail and the diode's cathode to ground. The mid-point between the resistor and the diode will be between 0.6 to 0.8V, depending on the forward voltage of the diode. Now, even with a crappy pot (I suggest 1K or 10K), you will be able to dial anywhere between 0 to ~0.7V with much better precision. Need more precision? Add a second voltage divider connected between the diode and your 100-270 Ohm resistor. Two 470 Ohm resistor will drop that 0.6-0.8V forward voltage to 0.3-0.4V. If you do that, then use a 10K pot to get any range between 0 to ~0.4V with even more precision.

[Dannyx]

Re: Constant current load for testing power supplies (guidelines)

Yes, you can, and *without* a multi-turn pot. The trick: 100 to 270 Ohm resistor connected in series with a diode (to the diode's anode). Connect one side of the 100-270 Ohm resistor to your 3.3V/5V/12V rail and the diode's cathode to ground. The mid-point between the resistor and the diode will be between 0.6 to 0.8V, depending on the forward voltage of the diode. Now, even with a crappy pot (I suggest 1K or 10K), you will be able to dial anywhere between 0 to ~0.7V with much better precision. Need more precision? Add a second voltage divider connected between the diode and your 100-270 Ohm resistor. Two 470 Ohm resistor will drop that 0.6-0.8V forward voltage to 0.3-0.4V. If you do that, then use a 10K pot to get any range between 0 to ~0.4V with even more precision.

Your suggestion eliminates the FET and op-amp entirely then ?

[momaka]

Re: Constant current load for testing power supplies (guidelines)

^ No, that is just for the V1 control voltage for the op-amp / comparator on the schematic you posted on the first page. Though I see now that they showed a LM741 op-amp used as a comparator - and that is a BAD idea, because LM741 is probably the lousiest op-amp you can use (believe me, I actually tried this circuit a long time ago, and until I switched it out for something better - in my case, an LM358 op-amp - the circuit would not work properly).

Basically the way this circuit works is the op-amp / comparator drives the Gate of the MOSFET based on the voltages it receives from its positive (+) and negative (-) input terminals. If the voltage on the (-) input is higher than the (+) input, the op-amp/comparator output goes low (i.e. whatever you used for Vss - the negative supply to the op-amp - in your case of that circuit, it is ground.) Conversely, if the voltage on the (-) input is lower than the (+) input, then the op-amp / comparator output swings high (whatever you use to feed Vdd to it).

So if you wonder how that circuit manages to work like that, think of it this way: the op-amp / comparator always wants to have the voltage between it's negative (-) and positive (+) inputs the same. Thus, if you set V1 to output 0.357V, then that will also be the voltage present across resistor R1. Why? Because if you set V1 to some positive voltage and there is nothing on the (-) input of the op-amp, it will output a high signal, which goes to the MOSFET gate. This turns the MOSFET ON and current starts going through R1. As the current through R1 rises, so does the voltage across R1. When the voltage across R1 becomes higher than that of V1, the op-amp switches its output back to low and the MOSFET turns OFF again. Then the voltage across R1 decreases again below that of V1, and the op-amp / comparator switches back the MOSFET ON. So in essense, this whole cycle repeats over and over.

However, it is worth noting here that this is NOT a switching circuit and the op-amp / comparator output never actually goes fully ON of OFF. Instead, after some time (called steady state), the op-amp / comparator outputs a fairly steady voltage that turns ON the MOSFET, but just only a little bit to drive enough current through resistor R1 to keep its (-) and (+) input terminals at balance. Thus, this circuit is actually LINEAR in nature, just like an LM317 regulator circuit. As such, at higher input voltages to the MOSFET, most of the power will be dissipated by the MOSFET rather than the resistor. So that FET will need a large heatsink. And even with a very oversized heatsink, you still need to make sure that you are not running your MOSFET past its safe operating area (look for an SOA graph in its datasheet).

[eccerr0r]

Re: Constant current load for testing power supplies (guidelines)

I don't think the LM741 won't work at all however. The key thing is that it requires the usual bipolar supply in order for it to work in this situation. You'll need the V- of the LM741 to go to at least -3V relative to GND, and the V+ supply probably +12V or so relative to GND. Yes the supplies are not equal to each other, but who's checking? Op amp won't care...

---

If one can get away with just using a LM317, it's a lot easier to just implement it this way as everything but the sense resistor is inside the LM317 (op amp and dissipation transistor). Using the LM317 means you have to get all the way to 1.25V, so even in this prepackaged solution it's a fairly high sense voltage.

About the largest problem with the LM317 is that it's only 1.5A max. A LM350 will get you to 3A and LM338 to 5A, though each of these would require beefier and beefier sense resistors. A 10W resistor should handle all cases (5A * 1.25V = 6.25W, which is within 10W).

[Dannyx]

Re: Constant current load for testing power supplies (guidelines)

Thanks for the input...my workplace is being reorganized at the moment (actually we're on the verge of disbanding, but nvm), so I won't have time/space to mess around with little projects for a while now, since work is where I spend most of my time and do stuff like this. I'm much too beat after a day at work to work on anything at home, so I only do the "design and research" here

So another important factor appears to be the choice of op-amp.... even a split-supply would not be too much of an issue, since I do have some center-tapped transformers lying around which could use a good project

[momaka]

Re: Constant current load for testing power supplies (guidelines)

I don't think the LM741 won't work at all however. The key thing is that it requires the usual bipolar supply in order for it to work in this situation. You'll need the V- of the LM741 to go to at least -3V relative to GND, and the V+ supply probably +12V or so relative to GND. Yes the supplies are not equal to each other, but who's checking? Op amp won't care...

That's another reason I switched out the one in my circuit for a 358. But again, you will find that LM741 is just soooooo slow and at certain loads, you may get some pretty nasty oscillation. With 358, my circuit was much more stable and regulating much better.

If one can get away with just using a LM317, it's a lot easier to just implement it this way as everything but the sense resistor is inside the LM317 (op amp and dissipation transistor). Using the LM317 means you have to get all the way to 1.25V, so even in this prepackaged solution it's a fairly high sense voltage.

About the largest problem with the LM317 is that it's only 1.5A max. A LM350 will get you to 3A and LM338 to 5A, though each of these would require beefier and beefier sense resistors. A 10W resistor should handle all cases (5A * 1.25V = 6.25W, which is within 10W).

Actually, if you want to build a really simple constant current source, your best bet is with a BJT, preferably a Darlington (or if you don't have one, use a small BJT to drive the BJT just like Darlington does, and you will achieve the same thing). Simply wire the Emitter to one end of your sense resistor and the other end of the sense resistor to ground. Then connect the Collector to whatever it is you want to sink constant current from.
And the base? Drive it on with a simple diode biasing network and a pot to fine-tune the biasing of the Darlington BJT. For better thermal stability of the circuit, use small signal transistors instead of the diodes and thermally couple to the same heatsink as the Darlington BJT.

This is by far the simplest circuit and doesn't require any ICs. It's actually almost the same thing as what stereo/music amplifiers use at their output stages.

[Dannyx]

Re: Constant current load for testing power supplies (guidelines)

I messed around with a simulator I've used before and came up with this. It seems to work as intended with a BJT rather than a FET for some reason: with a BJT connected the way Momaka suggested, the current follows the control voltage all the way up to the calculated 10a (with a 0.1 ohm resistor) as it should, whereas with a FET, it would max out at around 1.8a while the op-amp's output goes all the way up to the preset "rail" (15v).

This is just a simulation done with a tool that's far from professional and uses rather generic components, so it could just be due to plain buginess or it could indeed be accurate and perform similarly with real life components as well. The BJT's beta was set to 100 in case you're curious, so at least it's got some parameters you can tweak for some of the components. It also means I was able to dial in a very precise control voltage down to the millivolts range, which may not be so easy to achieve...it's a good start though

[eccerr0r]

Re: Constant current load for testing power supplies (guidelines)

I wouldn't call it simple if you need a bunch of other components to bias and amplify the voltage back - a lot of components are needed. A very common current control is use the sense resistor tuned at the voltage of the base-emitter junction of another transistor, but this requires another package to pull down the base drive of the Darlington, and itself needs biasing.

Or perhaps just do away with the sense resistor and just rely on the fact BJT collector current in common emitter is completely a function of the base drive. However, the pot would vary widely depending on luck of the draw part you got - the gain of the Darlington, though high, is not consistent between parts and negative feedback is necessary.

[and as an edit response to Dannyx: yes. Darlington is necessary. The base drive is a significant portion of the power that will be dissipated in the resistor that will give you error. With the currents we're talking about here, the gain of a non-Darlington BJT will likely be significantly less than 50, which will cause error, not to mention most op amps probably can't supply enough current to allow a straight BJT to pass many amperes.]

At least the LM317 would be fairly consistent, 1.25V±40mV or so, it's got some internal temperature compensation, so just this and one external part (the resistor) is luck of the draw. Plus only two components and only 3 nodes (in, GND, and resistor to out pin on LM317) - which can even be assembled without a PCB, how's that for quick and simple?

However there is one major drawback of the LM317 - you can't easily pot tune a LM317 circuit. You'd potentially need a fairly hefty watt-rated pot that can deal with a significant amount of current to adjust the current. The Darlington and op amp solutions at least have high gain and a 1/4W pot is usable.

[eccerr0r]

Re: Constant current load for testing power supplies (guidelines)

That's another reason I switched out the one in my circuit for a 358. But again, you will find that LM741 is just soooooo slow and at certain loads, you may get some pretty nasty oscillation. With 358, my circuit was much more stable and regulating much better.

I thought the LM741 was comparable to the LM358... GBW (0.5MHz min, 1.5MHz typ) is not great, but a lot of op amps of the time period aren't that high, the LM358 is also only 1MHz.

Not saying the LM358 is bad or LM741 is good by any means, the LM741 IS horrible by today's standards. But it's not an unworkable solution, hence that's why the LM741 got so popular. It works for so many things that everyone knows what a 741 is! (Ask a random person what LF355, LM358, LM741, LM1458, NE5532, LM6142, LM324, TL062, TLV2372 are and see which ones are most recognizable.)

My problems with LM741:
- power rails that go beyond output voltage swing necessary is by far the worst problem why I don't want to design with them. LM358 only the positive rail needs to go beyond output range. RRIO op amps are better than LM358.
- not perfectly linear operation. It's not unworkably bad but there are a lot better ones out there.
- It's slow and noisy
- Eats significantly more power than LM358.
- input impedance is not as good as JFET or MOSFET. However the LM358 is about the same as the LM741, which is better than some bipolar RRIO op amps.

BTW probably a lot of the music heads will instantly recognize the NE5532 most likely...

[budm]

Re: Constant current load for testing power supplies (guidelines)

"whereas with a FET, it would max out at around 1.8a while the op-amp's output goes all the way up to the preset "rail" (15v)." May be your simulator has bad MOSFET modeling. You can just set up a another simulator and remove the OP Amp and replace the Gate drive with variable Voltage source to see what the simulator will show.
https://cdn.badcaps-static.com/pdfs/...3b3483e353.pdf

[momaka]

Re: Constant current load for testing power supplies (guidelines)

whereas with a FET, it would max out at around 1.8a while the op-amp's output goes all the way up to the preset "rail" (15v).

That might be bug in the software or something to do with the MOSFET model in the software. MOSFETs are voltage-controlled devices - i.e. the current they pass depends on their Gate voltage. So at 15V at the Gate, that MOSFET should be fully conducting.

The BJT's beta was set to 100 in case you're curious

For large power transistors passing lots of current, your beta / current gain will be much lower - probably in the range of 10-50, and in some cases, even single-digit numbers.

I wouldn't call it simple if you need a bunch of other components to bias and amplify the voltage back - a lot of components are needed.

You are looking at two small-signal BJTs (or diodes if you like), a Darlington (or build your own Darlington pair), a pot, and maybe two more resistors max. Not to mention the versatility of this circuit. Beats the LM317 constant current circuits any day.

Or perhaps just do away with the sense resistor and just rely on the fact BJT collector current in common emitter is completely a function of the base drive.

That's a bad idea.
As the BJT becomes warmer, its Vbe voltage drop will become lower. If you are using a pot or resistor to control the current going into the Base, that current will increase due to the now-lower Vbe drop. This will make the transistor pass even more current and get even warmer, further dropping the Vbe and biasing itself to death - a phenomenon commonly known as "Thermal Runaway" (if you've dabbled with discrete transistor audio amplifiers, you've probably read about this). Hence why you always need an Emitter sense resistor. The only way you can avoid it is if you have a heatsink large enough that won't allow the transistor to get significantly hotter past a certain point, so the thermal runaway will eventually tapper off. But it is still risky and you will get a lot of drift.

[and as an edit response to Dannyx: yes. Darlington is necessary. The base drive is a significant portion of the power that will be dissipated in the resistor that will give you error. With the currents we're talking about here, the gain of a non-Darlington BJT will likely be significantly less than 50, which will cause error, not to mention most op amps probably can't supply enough current to allow a straight BJT to pass many amperes.]

Exactly!

[eccerr0r]

Re: Constant current load for testing power supplies (guidelines)

You are looking at two small-signal BJTs (or diodes if you like), a Darlington (or build your own Darlington pair), a pot, and maybe two more resistors max. Not to mention the versatility of this circuit. Beats the LM317 constant current circuits any day.

Except when you're trying to build one without a PCB There are so many things that the LM317 solution abstracts away: no power source for op amp, no external voltage reference, no biasing needed for feedback resistor, don't have to worry about gain, it's all done inside the IC as long as you don't fry it or the sense resistor...

(the one transistor HFE*Ib current sink)'s a bad idea.
As the BJT becomes warmer, its Vbe voltage drop will become lower. If you are using a pot or resistor to control the current going into the Base, that current will increase due to the now-lower Vbe drop. This will make the transistor pass even more current and get even warmer, further dropping the Vbe and biasing itself to death - a phenomenon commonly known as "Thermal Runaway" (if you've dabbled with discrete transistor audio amplifiers, you've probably read about this). Hence why you always need an Emitter sense resistor. The only way you can avoid it is if you have a heatsink large enough that won't allow the transistor to get significantly hotter past a certain point, so the thermal runaway will eventually tapper off. But it is still risky and you will get a lot of drift.

Agreed, it's not a good idea, however it still would be "minimal components" solution when it comes down to it if you can deal with the temperature nonlinearity (of course design such that the expected heat dissipation is handled by the heatsink, and the limited base current would limit conduction). However I'd think this is the only other way to build a stop gap current sink with only 2 components like the LM317 solution - but if you don't count the voltage regulator...

[momaka]

Re: Constant current load for testing power supplies (guidelines)

Except when you're trying to build one without a PCB

Meh. Even then you could get creative and use that hardened cardboard material that is used as back support on a lot of furniture (the technical name for it escapes me for the moment). Then drill holes and connect stuff with wire. Actually, that's how the entire filter board is done for the Sansui SP-7500X speakers I found recently.

Copper clad board is another alternative. Or even breadboard - heck, everyone who's considering playing with electronics more seriously should own one - it's such a useful prototyping tool!

There are so many things that the LM317 solution abstracts away: no power source for op amp, no external voltage reference, no biasing needed for feedback resistor, don't have to worry about gain, it's all done inside the IC as long as you don't fry it or the sense resistor...

Uhmm, the CC source I was talking about in the last several posts does not use any op-amps. Only 1-2 transistors, 1-2 diodes (or transistors), and 2-3 resistors. That includes the shunt resistor and power transistor. In fact, if you wanted to go very simple, you could probably skip the diodes completely. It's just the circuit will be a bit more "unstable" when trying to set the current to the desired value.

That said, the headphone amplifier I build for a university project had its output transistors biased exactly like that - no diodes/transistors. Instead, each transistor was biased separately by a network of resistors and a pot. I didn't put any Emitter feedback resistors either, so the amp could get into thermal runaway. However, I countered that with two massive Xbox 360 CPU heatsinks - one for each upper and lower transistor set (so I could also get away without using silicone pads and washers to isolate the tabs of the transistors from the heatsink - i.e. the heatsinks were "live"...). Didn't have Darlington transistors either, so I build my own pair from discrete components, lol.

[eccerr0r]

Re: Constant current load for testing power supplies (guidelines)

I was not talking about specifically the two transistor or transistor-diode solutions, but also covering the initial op amp solutions - after all, the LM317 contains an op amp. However it still stands, two components (IC regulator + resistor) is less than 4 components (two transistors and two resistors) and building a device with point to point construction with 4 components is more difficult than with 2...

In fact, most of these "2 transistor" current sinks were designed for the milliamp range where transistor gains are high and heat sinking isn't a problem. I do wonder what happens when things get scaled up into the ampere range and once again the base drive becomes significant with respect to the current sink needed, and consistency will start to falter with transistor/diode variation.

[momaka]

Re: Constant current load for testing power supplies (guidelines)

I was not talking about specifically the two transistor or transistor-diode solutions, but also covering the initial op amp solutions - after all, the LM317 contains an op amp. However it still stands, two components (IC regulator + resistor) is less than 4 components (two transistors and two resistors) and building a device with point to point construction with 4 components is more difficult than with 2...

Okay it is... but LM317 can pass only 1.5 Amps of current, and that is only if operating it up to a certain input voltage (in other words, observe the SOA / power dissipation limits).

With transistors, it's much easier to build a high-current CC source.

In fact, most of these "2 transistor" current sinks were designed for the milliamp range where transistor gains are high and heat sinking isn't a problem. I do wonder what happens when things get scaled up into the ampere range and once again the base drive becomes significant with respect to the current sink needed, and consistency will start to falter with transistor/diode variation.

That's not how it works. See attached sample constant current circuit below.

In this circuit, you can set up the current flowing through the transistor to be pretty much independent of its current gain (beta / HFE), provided of course, that you are not pushing the transistor near its maximum specs (particularly near maximum current, as the HFE will start to drop significantly).

Basically, this circuit relies on the fact that transistors have a fairly constant Base-Emitter voltage drop, Vbe. For normal transistors it is about 0.7V. And for Darlington it is usually twice that (i.e. 1.4V, as there are two transistors inside it, hence two Vbe drops).

Therefore, if we set the voltage going to the transistor's Base to be 2V (Vb = 2v), then the Emitter will be at approximately Ve = 2 - 1.4 = 0.6V above ground (note that I am doing the calculations here for the circuit with the TIP120 Darlington transistor). This will hold true regardless of the resistance of Re, provided of course that it is not so low to cause excessive current draw - in which case, the current gain will start to become the limiting factor of how much current will flow through the circuit (and we want to avoid that). That means if Re = 1 Ohm, then the transistor will pass about 0.6 Amps (Ie = Ve / Re = 0.6V / 1 Ohms = 0.6 Amps). And if Re = 0.3 Ohms, then Ie = 2 Amps. Alternatively, we could set the voltage at the Base to be 3.4V. Then the Emitter voltage will be Ve = 3.4 - 1.4 = 2V. If we use the 1 Ohm Emitter resistor, Ib = 2 Amps again.

So essentially, you can control the current by changing Vb and/or choosing appropriate values for Re. But generally, the lower the resistance of Re, the more sensitive the circuit will become to Base voltage changes of the transistor. On that note, I have drawn R1/R2 resistor divider network connected to the same voltage source as the transistor for simplicity. But in reality, you don't want to do that, especially if V1's voltage tends to vary with the load. If you want very good current regulation, then R1 should be connected to a separate voltage source with tight voltage regulation.

Another note: I put resistor Rb there simply to separate the transistor's base voltage from the voltage divider network. But in reality, you don't need Rb at all. However, you could replace Rb with a potentiometer if you want to fine-tune the current. Though, if you want to do that, it is probably better to connect the potentiometer in series either with Ra or Rb - depending on what values you picked for Ra and Rb.

[eccerr0r]

Re: Constant current load for testing power supplies (guidelines)

I've already mentioned that as long as you don't exceed the SOA of the LM317, it will work and work very well. And you still have LM338 and LM350 if needed. Simplicity/component count is the reason for using the LM317, not flexibility. However it is very stable and only two components to worry about variations.

The two transistor topology I was talking about normally does not use a Darlington - the common 2-transistor current sink uses two standard transistors, where one of them is the feedback amplifier. The 2-transistor sink circuit uses the lower current secondary transistor's VBE as the reference voltage and as it uses standard transistors, gain is an issue. I suspect converting it to a 3-transistor (or using a Darlington for the pass transistor) then the gain won't be as big of an issue anymore.

Now here you consider a slightly different different topology, I had not considered this topology, but this topology still needs some reference whether a CVVS or a diode/zener drop. This definitely requires a Darlington, the base current would otherwise be a problem.