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When running this test circuit on CircuitLab and my expected result was that only M1 would open but then i noticed that the M4 mosfet is also open even though there is a diode between it and the the 3V the Gate threshold on both is set to 1.5v

i also noticed some strange spikes appearing on the current

Does it mean a diode blocks current but not voltage ?

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Non-ideal diodes have leakage current. It, together with random noise, charge up the M4 gate capacitance. \$\endgroup\$ – Long Pham Oct 20 '18 at 12:31
  • \$\begingroup\$ That's in fact the answer. \$\endgroup\$ – Janka Oct 20 '18 at 12:34
  • \$\begingroup\$ If you want M4 not to turn on because of D3 leakage current, add a a pull-down resistor to M4's gate. If its resistance times the leakage current is much lower than M4 threshold voltage, M4 won't turn on. It will also keep random noise from charging-up M4's gate. Something like 100k should be OK. \$\endgroup\$ – Lorenzo Donati Oct 20 '18 at 18:19
  • \$\begingroup\$ BTW, what is this circuit for? M4 seems to have no purpose. Did you simply play with the simulator or did you have specific functionality in mind? \$\endgroup\$ – Lorenzo Donati Oct 20 '18 at 18:20
  • \$\begingroup\$ @LorenzoDonati Yes, adding a 100k pull-down resistor is exactly what i ended up doing. the circuit that you see here is just to test what went wrong with the original circuit , which was a circuit with 2 on\off switches, one switch opening the M1 and the second switch opening both M1 and M4, the M4 labeling instead of M2 is just a result of deleting and adding mosfets. \$\endgroup\$ – soundslikefiziks Oct 20 '18 at 20:07
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enter image description here

Figure 3 from the 1N4148 datasheet.

  • Diodes have very small leakage current. At 3 V this will be between 3.5 and 10 nA.
  • The 2N7000 has a gate−body leakage current, Forward of -10 nA max.
  • The diode also has about 4 pF capacitance. When the supply voltage jumps from zero to +3 V on power-up the diode capacitance will cause the gate of M4 to jump up too.
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The effect of voltage that you care about in most electronics is current. When a device blocks current, it makes voltage inconsequential, save electrostatic effects.

The problem in the circuit shown is that it essentially leaves an unshielded MOSFET gate floating (no current path to or from it). A floating MOSFET gate is almost always a design error, leading to undefined behaviour.

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  • \$\begingroup\$ Are you talking about M1 and M4, or only M4? And what would be the solution? Add a pull-down resistor? \$\endgroup\$ – Eran Oct 21 '18 at 6:52
  • \$\begingroup\$ M4, if we assume that the "+3v" point is never left floating itself. \$\endgroup\$ – rackandboneman Oct 21 '18 at 7:59
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The idea of blocking voltage is a nonsense. You need to think the problem in terms of "resistance" and voltage divider (I'm putting quotes around «resistance», because it's not reactive but it's heavily non linear). Since the transistor is a MOSFET, it's gate has a much higher resistance than the resistance offered by the reversed diode (because of the leaking current) and the divider made from the diode D3 and the M4's gate transmit nearly all the voltage to the gate.

So the solution is simple : Just put a high value resistor between M4's gate and the chassis ground.

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  • \$\begingroup\$ So the resistor to the ground will basically be in parallel to the gate resistance? This is the idea? \$\endgroup\$ – Eran Oct 21 '18 at 6:55
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    \$\begingroup\$ Yes, in order to reduce the low (i.e. to chassisground) resistance of the voltage divider, which otherwise is high above 10⁶ ohms (which is the reason why those components are sensitive to electrostatic discharges). \$\endgroup\$ – Camion Oct 21 '18 at 19:00
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Let's look at what "there is a Voltage" actually means:

The presence of a voltage between two points of a circuit means we measure a difference in electric potential with the value \$U\$. So, if there was a conducting path between these points with \$R<\infty\$ there would flow a current \$U=RI\$. The presence of a potential difference (\$U\$) doesn't mean any electrons could get from A to B, but that one spot would accept the "surplus" from the other.

A diode is, in theory, \$R=\infty\$ in one way and \$R=0\$ in the other. However, those approximation isn't true in reality. This can lead to a small current flowing "backwards" through a diode.

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