0
\$\begingroup\$

I am determining the reverse leakage of a Schottky diode in a circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

The intention here is to bias the P-FET gate so that it conducts when the switch is open (technically when V1 is absent). When the switch closes, it should stop conducting and isolate the battery. (There is a body diode in the P-FET not shown by the schematic.) The purpose of the Schottky diode is to prevent current flow from the battery to ground through the gate bias resistor. However some leakage will of course be unavoidable.

The datasheet for a diode under consideration indicates that approximately 15µA of leakage should be expected for 3V reverse bias.

My question is how will this leakage be affected by the bias resistor? With the switch open, the potential at D1 anode should be 0V, but would the resistor not also further reduce this current? How do I calculate that?

This related question was not very helpful because I am unsure if the OP's formulas are remotely applicable or correct.

Edit:

This device is battery powered with a non-rechargeable battery. V1 represents an optional 3V source which, if used, should not be connected to the battery. The purpose of the circuit is to allow the device to be powered with the battery under normal circumstances, but if V1 is connected, it powers the device instead and the battery is disconnected.

The diode is meant to prevent the battery voltage from affecting the gate, while the FET (and its body diode) is meant to disconnect the battery when V1 is present.

Judging from comments and answers, this intention was not clear. With that explained, I am also sensing that there are bigger issues with this design, on which I would appreciate any constructive criticism.

\$\endgroup\$
0
3
\$\begingroup\$

Current will not flow through the 33K resistor anyway since the gate is insulated from the source (and drain).

Anyway, leakage current through the diode will cause the voltage at the gate to rise. If it's 15uA it will increase to almost 0.5V. Of course that particular diode can leak 5mA when hot, so it might be quite a bit higher than the "typical" value, depending on unit-to-unit variations and temperature. You should not depend on "typical" values if you want a reliable design.

\$\endgroup\$
5
  • \$\begingroup\$ Does this mean the FET could be slightly turned off by leakage current, causing excess power to be wasted on battery power? Proposed alternative for me to investigate? \$\endgroup\$
    – JYelton
    Mar 12 at 20:27
  • \$\begingroup\$ I don't see why the diode is there in the first place. \$\endgroup\$ Mar 12 at 20:39
  • \$\begingroup\$ See my comment on Kevin's answer. Isn't it needed to prevent the battery from raising the FET gate voltage? \$\endgroup\$
    – JYelton
    Mar 12 at 20:43
  • \$\begingroup\$ With or without the battery, and without the diode, the gate voltage will be approximately 0V with the switch open. No current to speak of flows to or from the gate in the situation shown. Often spec'd at +/-100nA but reality it's usually much, much lower, just costs money to test such low currents and is unnecessary. 100nA * 33K = 3.3mV. \$\endgroup\$ Mar 12 at 20:46
  • \$\begingroup\$ It may be worth saying that this is a battery powered device normally. If it is powered with the optional source, the non-rechargeable battery should be isolated. So the main operating case is with the switch open as shown. \$\endgroup\$
    – JYelton
    Mar 12 at 20:49
0
\$\begingroup\$

When V1 is present the FET will be turn-off off but current from V1 will flow through the diode to the output. Which is the desired effect.

When V1 is not present the gate will be pulled to ground by the 33k resistor; there will be 3V gate to source boas and the output will be turned on.

If there is leakage in the diode it will cause a small voltage across R1. With 15uA R1 there will be 33k*15uA = 500mV. This will reduce the gate to source voltage of the FET to 2.5V.

Provided the threshold voltage of Q1 is low enough that 2.5V will still make it conduct enough to drive the load there will be no problem.

If this is an issue the simplest change is to reduce the value of R1 at the expense of a slightly higher current from V1 when powered from that source. If that is an AC power supply that is probably not an issue so R1 could be much lower.

One problem with the circuit is that V1 has to match the voltage from the battery. If the battery is significantly higher than V1 it will turn on (ie if Bat-V1 > Vth).

\$\endgroup\$
6
  • \$\begingroup\$ The battery needs to be isolated when V1 is present. Without the diode the battery would raise the gate voltage and defeat the purpose of the FET. I may he missing something. \$\endgroup\$
    – JYelton
    Mar 12 at 20:20
  • \$\begingroup\$ @JYelton - no it doesn't. The gate is only connected to the ground via the resistor - there is negligible leakage from the gate. It will stay at ground within a few microvolts. \$\endgroup\$ Mar 12 at 21:15
  • \$\begingroup\$ In the case where V1 is absent (switch open), if the diode were replaced with a short, what is the outcome? \$\endgroup\$
    – JYelton
    Mar 12 at 21:38
  • \$\begingroup\$ @JYelton - No, don't replace the diode with a short, just leave it out. It is not needed. If you replace it with short the load will be powered by V1 when the switch is closed and the FET will partially power the load when the switch is open (actual voltage depends upon the Vth of the FET). \$\endgroup\$ Mar 12 at 22:35
  • \$\begingroup\$ Sorry to extend discussion in comments, but one of us is missing something. If I leave the diode out completely (replace with open), how is V1 supposed to power the device? I'll update the question with these details I feel are being missed. \$\endgroup\$
    – JYelton
    Mar 12 at 23:10
0
\$\begingroup\$

Other answers I think were confused about the purpose of the schematic posted and didn't focus on the root question. So here is what I wound up doing:

Given a leakage current at a specified voltage, determine the approximate resistance that the diode is presenting. For example, with 15 µA leakage at 3 V, the equivalent resistance is 200 kΩ. Add additional resistance from resistors and recalculate: 3 V / 233 kΩ = ~12.9 µA.

\$\endgroup\$

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.