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I'm designing a constant-current dummy load for testing my power supplies. This load should be able to handle a maximum of \$2\mathrm{A}\$ at \$24\mathrm{V}\$. My current design, so far, is loosely based on the design found at EEVblog #102.

Original design:

schematic

simulate this circuit – Schematic created using CircuitLab

A simplified version of the original design from the website is shown above. In the schematic, \$V_1\$ and \$V_2\$ are, respectively, the control voltage, and the power supply under test. The number of amps that this load draws is equal to the number of volts on the input of the op-amp.

For example, when \$V_1=1\mathrm{V}\$, the op-amp turns on MOSFET \$\mathrm{M}_1\$ until the voltage at its inverting (minus) input is also \$1\mathrm{V}\$. This voltage is applied across \$R_1\$. Since, ideally, no charge flows into or out of the op-amp's inputs, we can use Ohm's law to determine the \$R_1\$'s current, \$1\mathrm{V} / 1\mathrm{\Omega} = 1\mathrm{A}\$. Therefore, \$1\mathrm{A}\$ is drawn from \$V_2\$ (the PSU under test) through \$\mathrm{M}_1\$ and \$R_1\$.

If the control voltage is, for example \$1.5\mathrm{V}\$, then the current draw would be \$1.5\mathrm{V} / 1\mathrm{\Omega} = 1.5\mathrm{A}\$. As you can see, the current is equal to the control voltage because of the \$1\mathrm{\Omega}\$ resistor.

The problem:

As previously stated, I intend to build a power supply what can handle \$2\mathrm{A}\$ at \$24\mathrm{V}\$. Using the previous schematic, I ran some quick estimates at max load. . .

$$ V_1 = 2\mathrm{V} $$ $$ V_2 = 24\mathrm{V} $$ $$ R_1 = 1\mathrm{\Omega} $$

For the resistor \$R_1\$: $$ P_d = 1\mathrm{V} \cdot 2\mathrm{A} = 2\mathrm{W} $$

For the MOSFET \$\mathrm{M}_1\$: $$ P_d = 23\mathrm{V} \cdot 2\mathrm{A} = 46\mathrm{W} $$

The major problem with this design is that the MOSFET has to dissipate \$46\mathrm{W}\$! This is probably not very healthy for the TO-220 device. Therefore. . .

Schematic 2:

schematic

simulate this circuit

In this schematic, the load is split across 2 sets of MOSFETs/resistors. My hope is that each one of the components would only have to dissipate half of the original heat. This means that \$R_1\$ and \$R_2\$ would have to dissipate \$1 \mathrm{W}\$ each, and \$\mathrm{M}_1\$ and \$\mathrm{M}_2\$ would have to dissipate only \$22 \mathrm{W}\$ each.

I tested this design using CircuitSim, and it works! However, practically, I can see a potential issue with this implementation.

  • Do \$\mathrm{M}_1\$ and \$\mathrm{M}_2\$ have to be matched?
  • Do \$R_1\$ and \$R_2\$ have to be matched?
  • Is thermal runaway an issue?

One of my major concerns is that, outside of the ideal, virtual world of the circuit simulator, the MOSFETs' parameters vary, and one MOSFET would hog all of the current, destroying itself. Then, the other MOSFET would quickly follow suit.

What can I do?

  1. Do nothing.
  2. Connect the MOSFETs' sources.
  3. Add another op-amp for \$\mathrm{M}_2\$.

What if I use bipolar transistors? What kind of op-amp should I use? Should I use an op-amp at all? What are the limitations of using op-amps in this application? Please feel free to explore other options or suggest something else!

Thanks for your time! Hopefully this topic can be expanded upon for the benefit of others in the future.

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Directly paralleling the transistors is a terrible idea- the threshold voltages of the MOSFETs will be different and one transistor will take much more current than the other- further you are only measuring one current so the total current may be greatly in error. It's worse than with BJTs.

Fortunately op-amps are cheap and you can just use two or more (one per MOSFET), fed from the same control voltage, each fed back from an individual source resistor. Use a single supply op-amp, not an LM741. That way each current will be precisely the desired fraction of the total, and the total will also be accurate. There is no need to match transistors or resistors (but the resistors should be accurate enough for your desired current accuracy). The currents from each transistor will simply add.

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  • \$\begingroup\$ Looks to me like the threshold voltage can vary by 2V, according to the datasheet so, in theory, one could be just about completely off. \$\endgroup\$ – Spehro Pefhany Jan 3 '15 at 4:07
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    \$\begingroup\$ Completely agree. From the MTP3055 datasheet the threshold voltage can vary from 2.0 to 4.0 volts which is far too much to make the original schematic function properly. \$\endgroup\$ – horta Jan 3 '15 at 4:08
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I would use a separate op-amp for each switch, each with its own resistor so that you have multiple independent current control loops in parallel. The general rule of thumb is it OK to use MOFSETs in parallel because as they heat up their on-resistance increases, and so they generally share current well, but this is for use as a on/off switch. When used in an analog fashion such as this, I'd imagine differences between components could result in very different currents through each transistor for a given gate voltage.

You could use bipolar transistors with a similar circuit as long as the op-amp output current capability is sufficient for whatever the transistor gain requires. You have to be careful using MOSFETs for analog circuits like this because some (especially trench types) are designed for fast switching full-on or full-off type applications and fry quickly when left in-between too long with significant current. This MOSFET isn't of that type but check the "Maximum Rated Forward Biased Safe Operating Area" graph in the datasheet. Even with two in parallel you are dangerously close to the edge of the safe operating area. I'm not a transistor expert but typically find BJTs more robust for analog circuits (I've killed many more MOSFETs than BJTs).

For projects like this I think one of the biggest issues you need to think about is how you are going to dump heat. I think using multiple transistors is a great idea in addition to big heat sinks, fans, etc. Junction to ambient thermal resistance for T0-220 packages is usually around 60 degC per watt without any heat sink. So 22W would rise the junction temperature of the transistor 1,320 degC over ambient! That is definitely going to release all magic smoke! You usually need to keep the junction temperature below 150degC or so depending on the part.

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You are right to worry about that power level in an MTP3055, but it's nowhere near the limit of the package. For instance, the Vishay IRL540 is a TO220, and it's rated for 150 watts. You just need to make sure you've got a good heat sink. And by good heat sink I mean a big honker. No clip-on jobbers.

Assuming you still want to worry, your alternative approach will probably work. It's true that one of the MOSFETs will be running open loop, but I'd guess you'll be all right. Matching the transistors would be a good idea, but I doubt it's necessary. Use 1% resistors. Isolating the gates from each other with about a 100 ohm resistor for each gate is also nice.

The ultimate way to go if you insist on using wimpy MOSFETs is simply to duplicate your load circuit, with each load set to pull 1 amp. Put them in parallel and you're set. Op amps are cheap, although I'd strongly advise upgrading from 741s. At the least, use single-supply op amps and save yourself the trouble of providing dual supplies - and with 741s you really need dual supplies.

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  • \$\begingroup\$ @bt2 Your comment appears irrelevant because he says to use a heatsink. \$\endgroup\$ – horta Jan 3 '15 at 3:41
  • \$\begingroup\$ oops, deleted, though 150W even with a heat sink seems high for a TO-220. \$\endgroup\$ – bt2 Jan 3 '15 at 3:50
  • \$\begingroup\$ @bt2 Agreed, the only way you'd get 150W is if you had active cooling which hardly ever happens in the field. \$\endgroup\$ – horta Jan 3 '15 at 3:52
  • \$\begingroup\$ I think you're underestimating the part. Yes, 150 watts requires active cooling, but the OP is only asking for 50 watts. Data sheet says junction to case thermal impedance is 1 deg/W, and case to sink is ~0.5. That's a total of 1.5 deg/W, or 75 deg junction to sink. Since the chip can run at junction 175 C, the heat sink can be at 100 C. Like I said in the answer, you need a good heat sink (and I don't mean a clip-on jobber either, but a big-ass extrusion), but I don't really see how this deserves a downvote. \$\endgroup\$ – WhatRoughBeast Jan 3 '15 at 4:18
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    \$\begingroup\$ MOSFETs have the ability to switch "really" fast, so there is always the possibility, particularly with several in parallel, to get high-frequency oscillations, particularly with a non-optimal physical layout such as (no offense intended) a newbie might produce. With 100 ohm on the gates, the combination of the resistor and the gate capacitance produce a low-pass filter for the gate drive and make this unlikely. It's not common in the first place but it can happen. Just being careful or paranoid, take your pick. \$\endgroup\$ – WhatRoughBeast Jan 4 '15 at 0:22
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The datasheet for the power transistors states the threshold voltage varies @Troom from 2V up to 4V.

If you plan to sink 2A (1A per transistor) and you use 2 Ohm sense resistors, the voltage accross these would be 2V. In worst case this means one transistor will conduct more than the other.

You can do several things:

  • if you plan to use this only on 24V systems, rise the control voltage / introduce a divider at the sense line. This means you will have higher voltage on sense resistor, higher power dissipation on the sense resistor and better current sharing between the two transistors.
  • use bipolar transistors. The mismatch in Vbe is small compared to the voltage accross the sense resistors. This will solve the root of your problem, but will introduce additional power dissipation in the OPA, something you must take into account (you may have to buffer the OPA with an additional transistor).

That beeing said, when you buy semiconductors, they are usually from the same production batch and the matching of the various parameters is ussualy much better then that shown in the datasheet. So in essence you could simply buy the transistors, measure them to see how well they match, calculate back the ratio of the current sharing and if satisfied - simply use them.

This is the route I would go.

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