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I'm trying to make a dummy load inspired by this one by GreatScott, and I'm using parts I have on hand. Here's the constant-current source design I'm using: Current Source  Schematic

I'm hoping to test a 12V power supply with this, and only use up to 2-3A. From the datasheet of the IRF630B it can handle up to 200V/9A (I know that number is a bit inflated, but 12V/3A seems to be within the safety range).

There are three areas where I'm not sure:

  1. I understand that MOSFETs required a V(th) to turn on, and that the IRF630B I'm using has a V(th) between 2-4V. I'm not really sure how to calculate it, or if I'm supposed to? My google-fu has yet to yield any helpful information about that. It seems like that would be pretty important for controlling the MOSFET.
  2. The ACS712 current sensor I'm using has a zero-current voltage of 2.5V. It seems like I maybe need to correct for that, or does the current design work with that anyway? I know very little about op-amps, except they strive for balance between their inputs, so it should just "correct" (raise/lower output voltage) until the current sensor outputs the same voltage as the DAC, or am I wrong?
  3. The input resistors to the op-amps. Is 1.5k ok for the ACS712? I'm planning to use 200 (not 2k as in the schematic) for the DAC I'm using to limit the current (25mA max). Should there be one between the LM358s? There was no max input current on the datasheet... so it should be ok without one, right?
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3 Answers 3

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  1. To turn fully on the IRF630B requires 10V Vgs. To start to barely turn on (250uA) it needs 2-4V. To get 10V Vgs you should have a supply voltage on the op-amp of about 12V. Somewhere in between might work too, depending on the variations of the transistor itself. Under typical conditions (which a good designer would never rely upon) the transistor needs around 5V plus whatever is dropped across that fuse (should be < 100mV at 2-3A). The output voltage of the LM358 goes to the supply voltage minus a couple of volts, depending, which implies a typical requirement of a 6-7V supply voltage, and 12V will work for sure.

  2. If the circuit you show worked, it would give zero output for DAC voltages < 2.5V and controlled current sink for higher DAC voltages. The input voltage range of the LM358 goes to the supply voltage minus a volt or two. Safer to assume two. So you're getting a bit marginal at 3.45V in with a 5V supply. It is (just) guaranteed to work at 25°C but not over a wide temperature range. Not great. It's perfectly fine with a 12V supply (which you would need anyway as explained in 1.)

  3. The resistors do nothing much of value here so there's nothing to calculate.

I think it rather likely the circuit you show will be unstable under load. If you build this I would suggest having an oscilloscope at hand. There is a lot of gain (both op-amps and the transistor add gain) and a lot of phase shift around the feedback loop.

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  • \$\begingroup\$ Do you think the second op-amp stage is even needed? I think I had a bit of a brain fart and I thought the op-amp could only output a max of 3.3V... doh. I see know the supply rails are what limit that. I have it connected to 24V right now - do you think that's overkill? \$\endgroup\$
    – wootie11
    Aug 12, 2021 at 17:40
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    \$\begingroup\$ It's not overkill, but the Vgs absolute maximum rating is 24V so you'd be exceeding it somewhat under fault or oscillation conditions. Unlikely to cause actual issues. Adding a voltage divider would cause additional issues by adding another pole. An LM7812 might be better if you want to do something. \$\endgroup\$ Aug 12, 2021 at 17:52
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The exact Vth for the MOSFET won't matter as long as the LM358 will exceed it by at least a few volts since the feedback loop will cancel any variation out so to speak. You're powering your LM358 with 24V, this won't be a problem.

I think the design assumed that you'd program the offset into whatever was driving the DAC. In my experience with these types of current sensors, there's a bit of zero offset that you'll have to adjust for anyway and being able to change it in software makes it easy.

The input resistors are likely there to equalize bias current.

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  1. Vth also stands for Vgs which is the gate to source voltage. A better thing to do is look at Id (or Rdson) vs Vgs (from the datasheet), you can see that the FET will not start to conduct until ~3V and it is temperature dependent.

enter image description here
Source: https://media.digikey.com/pdf/Data%20Sheets/Fairchild%20PDFs/IRF630B.pdf

  1. The ACS712 measures current via a hall effect sensor, they make the output voltage to be centered between the rails (half of Vcc) so the device can output both positive and negative currents.

Quiescent output voltage (VIOUT(Q)). The output of the device when the primary current is zero. For a unipolar supply voltage, itnominally remains at VCC⁄ 2. Thus, VCC = 5 V translates into VIOUT(Q) = 2.5 V. Variation in VIOUT(Q) can be attributed to the resolution of the Allegro linear IC quiescent voltage trim and thermal drift

Source: https://www.allegromicro.com/-/media/files/datasheets/acs712-datasheet.ashx

so it should just "correct" (raise/lower output voltage) until the current sensor outputs the same voltage as the DAC, or am I wrong?

You can think of it like this, for any DAC voltage above the 2.5V of the output of the hall effect sensor, (so say 2.6V) U1A will try and raise the voltage of it's output, which in turn will also raise Vgs of the mosfet (through U1B which has a gain on it) and allow it to conduct more current. The will increase until the output of the ACS712 matches that of the DAC (I assume the DAC is a voltage DAC).

  1. the 2k and 1.15k resistors are probably for current limiting if either terminal hits the rails, during normal opamp operation (if not close to the rails) the opamp has an input bias current of 150nA or lower and so the resistors will be negligible. You still need them to avoid burning out the protection diodes if the input voltage goes outside the range of GND or Vcc
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