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I have this very simple circuit - actually there are 6 of them on a board.

schematic

simulate this circuit – Schematic created using CircuitLab

The MCU is connected to an MCU, and it is used to drive RGB led strips with PWM singals. My problem is that the MOSFET is not fully turned off. Or so it seems. When the MCU line is high and the Q1 transistor is fully turned on, then the gate of the M1 MOSFET is at 0.6V (one junction). Even though the threshold voltage of M1 should be about 2V, it is not 100% turned off at 0.6V. The result is that the LED strip glimmers a bit. It is almost nothing, but the LED strip is 20 meters long, and all of that small glimmering adds up. Basically, I cannot use this circuit as it is, because of that small glimmering. It becomes very distrubing in room that is used to watch movies in the dark.

Would it be possible to turn the FET fully off? My only idea is to use a dual power supply, and put the emitter of the Q1 transistor at -0.7V instead of 0V. However, creating a -0.7V supply just for this seems to be an overkill.

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    \$\begingroup\$ Why not replace the 2N2222 with a VN2222 logic-level MOSFET? It should control the IRFZ44 quite well and will require much less current from your MCU. (Put a weak pull-down resistor from G-S on the VN2222 to keep it off if the MCU output is tri-state). \$\endgroup\$ Commented Mar 6, 2019 at 15:17
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    \$\begingroup\$ "When the MCU line is high" the BJT should be on and not open circuit. \$\endgroup\$
    – Andy aka
    Commented Mar 6, 2019 at 15:37
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    \$\begingroup\$ Using words like "open" and "closed" to describe transistors is confusing. Do you mean open like a valve or open circuit? Also, you can't just say "the transistor" when there are two transistors in your circuit. \$\endgroup\$ Commented Mar 6, 2019 at 15:55
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    \$\begingroup\$ First, please please please edit your post. An "open circuit" is one that does not conduct electricity. An "open valve" is one that passes fluid. To avoid confusion no one in electronics talks about an "open transistor" unless they mean on that has failed as an open circuit. Go through your post, and replace "open" with "on" and "closed" with "off" -- that will un-confuse those of us who are trying to help. \$\endgroup\$
    – TimWescott
    Commented Mar 6, 2019 at 16:16
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    \$\begingroup\$ 2.2k is a very strong pull-up. 50k should work a lot better. \$\endgroup\$
    – dandavis
    Commented Mar 6, 2019 at 19:34

1 Answer 1

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If the "MCU" signal is going up to 3.3V then Q1 should be fully saturated, and it should pull its collector down to 0.2V or so. If it's only going down to 0.6V then you have a defective transistor in there, you've put it in backwards (emitter where the collector should be, collector where the emitter should be), or your MCU doesn't have enough drive to bring the "MCU" line up to its rail (it should be almost at the rail; it certainly should be within 200mV, i.e. 3.1V for a 3.3V rail, 4.8V for a 5V rail).

I suspect you just plugged Q1 in backwards. Double-check your pinout. Then measure the voltage at MCU.

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  • \$\begingroup\$ Q1 is not backwards. MCU line was 2V. After increasing that to 3.3V, the M1 gate went down to 30mV. But even at 30mV, I see some light coming out from the LEDs. But they are almost invisible now. \$\endgroup\$
    – nagylzs
    Commented Mar 6, 2019 at 17:40
  • \$\begingroup\$ What is the voltage at the source of M1 when it should be off? \$\endgroup\$
    – TimWescott
    Commented Mar 6, 2019 at 17:43
  • \$\begingroup\$ At the gate of m1: 34mV when off, 9.3V when on. \$\endgroup\$
    – nagylzs
    Commented Mar 6, 2019 at 18:07
  • \$\begingroup\$ At the source of m1 when on: 13 to 19mV (depending on the color) \$\endgroup\$
    – nagylzs
    Commented Mar 6, 2019 at 18:24
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    \$\begingroup\$ Can you either easily measure the current, or put some modest resistance (like \$1\mathrm{k}\Omega\$ in parallel with your LED string? There should be a minuscule current going through M1. It could be that you cooked it before -- if it's easy to swap out you may want to do so. \$\endgroup\$
    – TimWescott
    Commented Mar 6, 2019 at 21:15

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