I am using a DC/DC converter to charge a battery. The output current of the converter is about 6 Ampere at 41 Volts.

I'm trying to eliminate the standby leakage current of the output capacitors in order not to drain the battery (even mere milliamps are too much for this specific use case) while the DC/DC converter is not active.

Furthermore, any capacitor inrush current greater than 30 Ampere is too much for the battery. Thus, it is preferable for the capacitors to be charged through the DC/DC converter.

I have thought of the following approaches:

Using a diode to let current flow only into the battery and not back from the battery to the capacitors. (This would solve both leakage current and inrush current considerations) Thing is, that the diode dissipates a lot of power in this configuration.

Using an N-Channel MOSFET as a low side switch between the output capacitors and ground. This would eliminate the standby leakage current, but not the inrush current problem. I have thought of ramping V_GS of the MOSFET in order to "slowly" decrease the RDS during startup to charge the capacitor more slowly.

Is this a feasible approach? Are there better solutions to this problem?


I would be inclined to use a PMOS transistor as a high-side switch:


simulate this circuit – Schematic created using CircuitLab

Select D1 to set the turn-on threshold voltage at the output of the charger, and select R1 to set the base current of Q1 to about 1 mA at the maximum charger voltage.

R3 and R4 serve to limit the VGS that M1 sees, and C2 can be added to slow down the turn-on.

Note that the MOSFET is connected "backwards" — the source terminal is connected to the battery, not the charger. This is a common configuration used in protection circuits.

If the charger capacitor doesn't bleed down on its own, you might want to explicitly control Q1, perhaps using the same signal that enables the charger.

  • \$\begingroup\$ Originally, I had used a PMOS in my design. Wouldn't the body diode dissipate a lot of power at 6 Amps with a maximum body diode forward voltage of more than 6V (IRF9530)? \$\endgroup\$ – fscheidl Mar 23 '16 at 12:19
  • \$\begingroup\$ No, the MOSFET turns on and bypasses the body diode. \$\endgroup\$ – Dave Tweed Mar 23 '16 at 12:34

You could use a hot swap controller which has programmable inrush current.

You would need to use 2 back to back MOSFETS instead of a single device


The 1M resistor is there for DC balance. Note the direction of the body diode: when the gates are off, and charger disconnected, there is little reverse leakage from the battery.

  • 1
    \$\begingroup\$ Wouldn't the body diode still dissipate a lot of power at 6 Amps due to its forward voltage? \$\endgroup\$ – fscheidl Mar 23 '16 at 12:20
  • 1
    \$\begingroup\$ The body diode (when the FET is on) is bypassed as the FET drain to source voltage is close to zero. \$\endgroup\$ – Peter Smith Mar 23 '16 at 12:28

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