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I am working on a multi-range current source using APEX MP38 OpAmp to generate and force-in either 100mA or 2A into a mainly resistive load. I came across Howland Current Pump configuration and I hope that its going to work, but the preliminary simulation using LTSpice fails miserably.

The resistive load that I am dealing with changes its resistance greatly due to environment temperature and the temperature induced by the current. Using discrete current sources I have observed that at room temp (25°), the resistance of this load is 20Ω and at 180° its 60Ω.

Now I want to have a current source that can output 100mA or 2A using a switch. Something like:

enter image description here

The simulation with LTSpice is not promising: enter image description here

So my question is, can you please let me know if MP38 is a good candidate for this task? What kind of opamp I should use as the voltage follower (U2). Any ideas about input/feedback resistor values and their wattage? let me know if there is some information missing!

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  • \$\begingroup\$ First, do you need bipolar current? If not, the Howland is unnecessarily complex. Second, at 180 degrees you'll need at least 120 volts to get 2 amps. \$\endgroup\$ – WhatRoughBeast Aug 1 '17 at 11:30
  • \$\begingroup\$ @WhatRoughBeast Well I only need positive current. Do you have any suggestion for a better/simpler circuit please? \$\endgroup\$ – Sean87 Aug 1 '17 at 12:00
  • \$\begingroup\$ Is the load grounded (one side connected to GND) or floating (ie, you can connect it as you wish)? Both require different circuits, and it's gonna be a lot simpler and cheaper than your expensive opamp... \$\endgroup\$ – peufeu Aug 1 '17 at 12:03
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If you don't mind a certain inelegance (and power dissipation), you can work with a standard unipolar current source. This will look like

schematic

simulate this circuit – Schematic created using CircuitLab

This is pretty simple, but it has its drawbacks, most notably efficiency.

With the current sense resistor shown, a VSET of .25 volts will give 100 mA, and 5.0 volts will give 2 amps. But.

It would be really good idea to vary the VS depending on desired load. At 100 mA, you only need a bit more than 6 volts for VS, and 12 will work very nicely. For 12 volts, worst-case power dissipation in the MOSFET is (12 - .25 - (60 x 0.1)) x 0.1 or about 1/2 watt, so that's no problem.

At 2 amps, you're in an entirely new ballgame. You'll need at least 120 volts VS (to drive 2 amps through 60 ohms), so let's say you use a 130 volt supply. With a 20 ohm load, the load will drop 40 volts, the sense resistor will drop 5 volts, so the total voltage across the FET will be 85 volts, for a MOSFET power dissipation of 170 watts. You're unlikely to get this with any reasonable FET/heatsink combination, so you'll need to go to multiple FETs in parallel, and a very beefy heatsink (multiple pounds and forced air cooling). You would also need to be very careful about running your power ground in order to avoid parasitic voltage drops causing inaccuracy or (much worse) oscillation.

Oh yes, and the current sense resistor for the value shown will dissipate 10 watts, so you'll need something on the order of a 20 watt resistor just to play it safe.

But I'm sure you'll agree that the circuit looks simple.

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  • \$\begingroup\$ Thanks for the simple circuit...I will do some experiment and report back. But for now, isn't the feedback on the positive input going to cause instability (from what I learned in school!) \$\endgroup\$ – Sean87 Aug 2 '17 at 8:16
  • \$\begingroup\$ @Sean87 - Crap. Of course it does. Fixed. Sorry. \$\endgroup\$ – WhatRoughBeast Aug 2 '17 at 12:49
  • \$\begingroup\$ Just a question, is R2 the current sense resistor? if so then what is the role of R1? \$\endgroup\$ – Sean87 Aug 6 '17 at 10:12
  • \$\begingroup\$ @Sean87 - Yes, R2 is the current sense resistor. 2 amps at 2.5 ohms gives 5 volts. R1 prevents the op amp from seeing a pure capacitive load, as some op amps have stability problems when driving a capacitor. You don't want it too high, since then it will slow down the output response, and the resulting phase shift can cause loop instability. \$\endgroup\$ – WhatRoughBeast Aug 6 '17 at 10:22
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20 ohms and 2 amps requires a voltage of 40 volts so that is the smallest voltage power supply you need at one end of the scale. 60 ohms and 2 amps requires a supply voltage of at least 120 volts and, given a few volts lost in the pass transistor and/or op-amp, the minimum supply voltage should now be regarded as 125 volts.

The problem I see is that with a 125 volt supply and a 20 ohm load that only needs 40 volts across it to conduct 2 amps, is the volt drop across the "silicon" i.e. 85 volts at 2 amps - this is a power of 170 watts. The maximum power dissipation of the APEX amplifier is 125 watts.

You need a substantially different approach and I would recommend using PWM to produce a current into your load - should you not be able to live with the ripple voltage then an inductor and capacitor will also be needed BUT, this solution will not self-heat with 170 watts.

Something like this may be of some interest topologically: -

enter image description here

It produces a constant current up to 1.5 amps into a string of LEDs and, of course, the LEDs could be replaced by your load. Or maybe something that works from a lower supply voltage and steps up to produce a bigger voltage (again this is limited to only 1 amp): -

enter image description here

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  • \$\begingroup\$ Yeah, dissipation will be enormous, OP needs to tell more, notably if the load can handle the noise from PWM/switching current source... \$\endgroup\$ – peufeu Aug 1 '17 at 13:02
  • \$\begingroup\$ Thanks for the detailed answer. I have something to say...What if I add a beefy 5watt load resistor at the ground path of my load so that I can achieve 2A with less voltage? does it make sense? \$\endgroup\$ – Sean87 Aug 1 '17 at 14:13
  • \$\begingroup\$ Your load resistance can be as high as 60 ohm and if you need 2 amps then you need a supply of about 125 volts. Adding a resistor isn't going to make it any better from what I can tell. \$\endgroup\$ – Andy aka Aug 1 '17 at 15:12
  • \$\begingroup\$ This question is still "open" so have you moved away from this concept given your latest question because, from what I can tell you still have the same problems with the new idea. \$\endgroup\$ – Andy aka Aug 7 '17 at 11:34

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