DMM reads voltage across MOSFET DS with zero G voltage

Discussion in 'General Electronics Chat' started by theoldwizard, Jan 29, 2016.

  1. theoldwizard

    Thread Starter Member

    Jul 17, 2005
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    I should know this the answer. I probably used to know this, but after being retired for 7 years, you forget some things ! This is automotive related but it is really a general electronics question.

    Auto tech is diagnosing a problem. He unplugs a solenoid and probes the 2 pins in the connector on the harness with a DMM. One pin is connected to Vbatt, the other (I assume) to the Drain of a MOSFET (I don't know if there is any additional back EMF protection outside of the MOSFET). There is no signal to the Gate. The DMM reads Vbatt.

    Plug a 12v test light between Vbatt and MOSFET and nothing.

    Now I know this is related to the difference in impedance of these 2 test devices (10MΩ vs about 50Ω), but I can't explain it !
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    If the gate is floating, then you have no idea what the state of the MOSFET is in. What, exactly, do you mean by "no signal" to the Gate?

    What makes you think that the off resistance of the MOSFET is 10 MΩ. Could it be 1 MΩ? If so, and if your DMM has a resistance of 10 MΩ, then you have a voltage divide in which about 90% of Vbatt is appearing across the DMM.
     
  3. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    The difference is the 9,999,950 ohms one can infer from your description.

    With the high impedance meter a small leakage can look like a full on battery driver, while the test lamp, which needs lots of real current, will not light.

    That's why some people keep an old school analog meter around for testing. (Mine is a Simpson 260.) They have a much lower input resistance (20K ohms per volt is common) which while pretty high is low enough so most leakage current look like an open.
     
  4. SLK001

    Well-Known Member

    Nov 29, 2011
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    The gate may be holding a charge. I used to test MOSFETs by FIRST applying 0 volts to the gate, then measuring VDS, which should read near infinite, then applying Vdrain to the gate and remeasuring VDS, which now should read very low.

    EDIT: I used the DIODE function on a DMM and it was the voltage from the DMM that I used to charge the gate.
     
  5. theoldwizard

    Thread Starter Member

    Jul 17, 2005
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    I understand this. The DMM "load" is only about 1μA as opposed to about 25mA for a test light.

    But WHERE is this "leakage" occurring ? Internal to the MOSFET ?

    Would you expect different results if it was a Darlington ?
     
  6. theoldwizard

    Thread Starter Member

    Jul 17, 2005
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    No access to the actual transistor. It is inside of a sealed module.
     
  7. SLK001

    Well-Known Member

    Nov 29, 2011
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    I wasn't suggesting you test it. I meerly suggested that the reason that you see VBATT is because the gate is still holding a charge. The charge will bleed off in a minute or two, but while it is there, the MOSFET will be conducting.
     
  8. crutschow

    Expert

    Mar 14, 2008
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    The charge won't necessarily "bleed off" since the leakage path for the charge could be to either source or drain.
    The leakage could actually add charge.
     
  9. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    I would have no idea since I do not know what this is connected to. My strongest hunch is there is a capacitor in there you are reading, but if the two measurements keep repeating then something else is going on.

    I don't have a guess what a transistor would do here, but whatever is there has a lot of leakage to make the meter respond, but not enough to make the lamp light. Would be interesting to see what an ammeter says.

    A charge on the gate of a MOSFET truly turns it on, and it will light the bulb.
     
  10. crutschow

    Expert

    Mar 14, 2008
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    MOSFETs have a very high input impedance as the gate looks like a capacitor, with leakage that can be due to the internal diodes that some have for gate over-voltage protection.

    Darlingtons (or any BJT) are a different story as they have a relatively low impedance (look like a forward biased diode) and thus can't build up any significant charge on the base.
     
  11. theoldwizard

    Thread Starter Member

    Jul 17, 2005
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    I don't have access to the transistor or a schematic either, but based on my experience it is likely connected directly to the output of a μprocessor (or I*IC to digital out).

    The readings are repeatable (if you call a test light NOT lighting a "reading") are repeatable back to back to back ...
     
  12. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    Yeah I call that repeatable. If there was a cap holding the voltage up then you could read it once until you tried the lamp, which would discharge the cap and then the meter would no longer have the large voltage to read.

    If you stick an ammeter there what do you see?
     
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