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I'm fooling around with a PMOS, a Fairchild FDD6637, and pondering whether it would be possible to use Rdson as a sort of shunt-resistance for measuring the voltage drop, to get the amount of current flowing through drain and source.

The circuit will be using 2A at most. Here I'm thinking that if there's a short circuit somewhere, and the current rushes, I want to be notified when the current goes above 3A.

With Vgs = -4.5V, Rdson is 20mohm. At 2A, the Vds-drop will be 40mV. At 3A, the drop is 60mV.

So I had this idea: feed the drop into an opamp and pre-amp it to a more feasible level. Exactly how it's supposed to be coupled is unknown at the moment (I'm just playing with the idea), but let's say I couple the opamp to convert 60mV drop to, say, 4V (Vcc is 5V). Then let's say I deliver these 4V to another opamp (comparator) where one of the reference inputs is set to a fixed 3.9V, making it trip when the 4V arrives. That way I'll get a signal telling me there's a current rush.

Question: Am I thinking correct, or am I attacking this problem from the wrong direction?

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  • \$\begingroup\$ IME a mosfet specifies only its maxiumum RDSon, never its minimum. I think you can see how this could be a problem for your plan. \$\endgroup\$ – Wouter van Ooijen Feb 26 '17 at 11:42
  • \$\begingroup\$ @WoutervanOoijen Usually these days the Rds(on) plots in a datasheet can be reasonably relied on for these purposes. At the least for power FETs. \$\endgroup\$ – Asmyldof Feb 26 '17 at 11:49
  • \$\begingroup\$ For a one of design ok with calibration but tolerances of >30% NG. Is this a cost reduction? or lazy design? \$\endgroup\$ – Sunnyskyguy EE75 Feb 26 '17 at 14:58
  • \$\begingroup\$ @TonyStewart.EEsince'75: Neither. It's a learning-as-I-go-design. \$\endgroup\$ – bos Feb 26 '17 at 22:33
  • \$\begingroup\$ then stick with a 50mV shunt R for accurate results and compare with MOSFET differences. \$\endgroup\$ – Sunnyskyguy EE75 Feb 27 '17 at 0:20
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What you are proposing is a very common practise in DC/DC converters. Many of the driver chips made for P-FET buck conversion have a 100mV current fold-back trippoint, which can be created by a FET's own resistance, or an external resistor.

There are some reasons people might choose to use a resistor, like the turn-on time of a heavy power FET, in which time the apparent resistance may be higher than at steady state, causing false trips. But if you take proper care of such occurrences it can work fine, for example if inrush currents are limited at turn on, or you ignore it for about a ms when you turn on.

You could use a standard comparator to detect 60mV or 100mV directly. There's quite a few with more than enough sensitivity and only 10mV or 20mV hysteresis built in, so that you need only one part.

If you want the current to be measured by a micro-controller as well, your approach may be the best way, and you could Google something like "Using an Op-Amp as differential amplifier", I'm sure that'll result in some interesting reading to get you started on converting the 5V and 4.94V input into a 4V output.

Make sure the input common-mode range of the Op-Amp (or Comparator) is rail-to-rail, as you will be measuring at the VCC Rail with both pins.

Be aware though, that you can't trip at 50mV and expect that to be exactly 2.5A. You need to allow for some error margins.

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Question: Am I thinking correct, or am I attacking this problem from the wrong direction?

well, conceptually sound but fundamentally wrong.

the basic concept of measuring current by measuring voltage drop over a known resistance is sound.

however, that "known resistance" in this case can vary widely / wildly based on different temperature, devices, driving voltage, .....

there is so much uncertainty there that you would have a hard time to attribute change in voltage drop to just the current, thus rendering the current sensor not as useful.

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  • \$\begingroup\$ So, how about adding a 10mohm 1206-resistor just before thr PMOS, and read from that one? \$\endgroup\$ – bos Feb 26 '17 at 12:41
  • \$\begingroup\$ it would work. watch the power dissipation and thermal coefficient of the resistor. you may also look for high side current sense -> a linear solution to your problem. and may offer more flexibility than your "digital" solution. \$\endgroup\$ – dannyf Feb 26 '17 at 14:29

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