I need to eliminate main switch bouncing, which badly influences my device. In order to achieve that, I want to use a delayed start with N-MOSFET an RC circuit at the gate. MOSFET I've chosen have a very low gate threshold voltage of 1.05V and very low Rds (csat Vgs 2.5V it is less then 50 mOhm). My calculations suggest, that at the selected values for R1 and C1, 1.05V at the gate will appear at approximately 90 ms, which is well beyond any bouncing of the main switch. I'm using this formula: \$t = -R\cdot\!C\cdot\ln(1-\frac{V_0}{V})\$

At start up load draws no more then 150 mA. Did I miss something?

EDIT: I was suggested to add a discharge path for the C1 in order to ensure that gate is not driven high when the circuit is off. I planned to add a 12V Zener at the M1 gate, will it be sufficient to discharge C1?


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ How is the capacitor discharged? To slowly switch on you need to be sure to start with a discharged capacitor so that the MOSFET is off. \$\endgroup\$ Commented Jan 10, 2020 at 10:05
  • \$\begingroup\$ Device is started, then operated for several ours, then switched off for a significant time. Minimum time between turning off and on is I guess at least 30 minutes. I think 10 uF can be safely considered discharged in much lower time. \$\endgroup\$
    – Zhenek
    Commented Jan 10, 2020 at 10:11
  • 1
    \$\begingroup\$ Unless there is a resistor or a resistive path shorting the capactor, you can never assume that a capacitor will be fully discharged. In your circuit as shown, the supply is disconnected so there is no resistive path for the capacitor to discharge. Relying on "thinking it will be discharged after a long time" is a recipe for disaster. Add a 1 M ohm resistor in parallel with the capacitor and the problem is solved. \$\endgroup\$ Commented Jan 10, 2020 at 10:20
  • \$\begingroup\$ Or put 10 kohm across the switched rails. \$\endgroup\$
    – Andy aka
    Commented Jan 10, 2020 at 10:43
  • \$\begingroup\$ @Andyaka you mean after the SW1, but before RC? Which is better? \$\endgroup\$
    – Zhenek
    Commented Jan 10, 2020 at 10:49

2 Answers 2


Your circuit is correct in general. Though you need to note a couple of things

  1. The way a zener diode works, it starts to conduct as early as you apply a small forward voltage or a larger reverse voltage(known as the zener voltage). In your application the Zener diode prevents the mosfet Vgs from rising beyond 5.1V. It does not help discharge c1. To fully discharge C1 when your switch has been powered off, it would be better to put a resistor in parallel to C1 so it is effectively discharged when you switch off the button. Remember that this resistor would create a potential divider with R1 so you would have to choose resistances wisely. A value equal to R1 could work aswell.

  2. While the solution you suggested would work, your mosfet would always operate as partially turned on during on or off cycle for a small period of time and energy would lost as heat from the mosfet when operating in such condition. The amount of current through your mosfet means that this is not a major concern, but if the load could be higher or switching would be more frequent you would have to add a trigger circuit to handle the debouncing from the switch. Such a trigger could involve the same RC circuit followed by a comparator(op amp) that triggers on the mosfet rapidly once the output voltage from your RC circuit reaches a certain value. This way your Mosfet would always be fully ON or OFF.

Hope this helps.

  • 1
    \$\begingroup\$ Thanks! MOSFET is chosen with a good margin, I don't ever expect current to exceed it's max value (about 5 amps). SW1 is a solid rocker switch, even on purpose one couldn't switch it more than maybe 5-7 times per second, so I'm safe here as well. \$\endgroup\$
    – Zhenek
    Commented Jan 10, 2020 at 11:30
  • \$\begingroup\$ Yes you are right, just note while mosfets can handle a lot of current while fully ON due to low Rds value in ON state, the resistance and hence the heat dissipation during a slow turn on might heat it up. \$\endgroup\$
    – NoumanQ
    Commented Jan 10, 2020 at 13:47
  • \$\begingroup\$ This I fully understand! But as I said, that's not the case here. Transistor won't be PWM'ed, it's only bouncing from the big switch. It lasts for only about 100 us, as I measured \$\endgroup\$
    – Zhenek
    Commented Jan 10, 2020 at 15:13

Did I miss something?

Yes, the capacitor will remain partially charged once the switch is opened causing the circuit to activate significantly quicker the next time the switch is closed. This may or may not be problematic but, a solution is to ensure the switched supply properly collapses.

I planned to add a 12V Zener at the M1 gate, will it be sufficient to discharge C1?

No, because the zener may not conduct much current significantly below its zener point and this may still leave a high enough voltage to short-cut the delay at start up.

What engineers usually do is put a reverse diode across R1 and provide an unambiguous discharge path on the switched supply: -

enter image description here

But it all depends how critical the timing must be.

  • \$\begingroup\$ Thanks! I guess I can safely increase 10k to 20k in order to save current during normal operation. Load is actually a 5V LDO, does it need D1 as well? I thought it is only required for inductive lead. \$\endgroup\$
    – Zhenek
    Commented Jan 10, 2020 at 11:27
  • \$\begingroup\$ D1 ensures that the discharge path via R1 (100 kohm) is short circuited via the diode. Belt and braces. \$\endgroup\$
    – Andy aka
    Commented Jan 10, 2020 at 11:57

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