8
\$\begingroup\$

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

\$\endgroup\$
5
  • 3
    \$\begingroup\$ Non-ideal diodes have leakage current. It, together with random noise, charge up the M4 gate capacitance. \$\endgroup\$
    – Long Pham
    Oct 20, 2018 at 12:31
  • \$\begingroup\$ That's in fact the answer. \$\endgroup\$
    – Janka
    Oct 20, 2018 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 support Ukraine
    Oct 20, 2018 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 support Ukraine
    Oct 20, 2018 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, 2018 at 20:07

4 Answers 4

Reset to default
16
\$\begingroup\$

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.
\$\endgroup\$
7
\$\begingroup\$

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.

\$\endgroup\$
2
  • \$\begingroup\$ Are you talking about M1 and M4, or only M4? And what would be the solution? Add a pull-down resistor? \$\endgroup\$
    – nettek
    Oct 21, 2018 at 6:52
  • \$\begingroup\$ M4, if we assume that the "+3v" point is never left floating itself. \$\endgroup\$
    – rackandboneman
    Oct 21, 2018 at 7:59
2
\$\begingroup\$

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.

\$\endgroup\$
2
  • \$\begingroup\$ So the resistor to the ground will basically be in parallel to the gate resistance? This is the idea? \$\endgroup\$
    – nettek
    Oct 21, 2018 at 6:55
  • 1
    \$\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, 2018 at 19:00
1
\$\begingroup\$

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.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.