How to reduce the leakage voltage?

[joshuacp]

Dear Friend，

I designed one portable device and battery is used to power the portable device. I choose one transistor, NSS20500UW3 which is from On Semi, to cut off the power supply from the battery when the portable device is in off work state.

The schematic is shown in the following picture. X3101 is a switch. When the portable device is in On work state, pin 2 and 3 of X3101 are connected to pin4 of X3101,ie. Ground. Then Transistor Q3101 is on, and e and c of Q3101 are shorted.

When the portable device is Off work state, pin2 and pin3 of X3101 are connected to pin 1of X3101,ie b and c of Q3101 are connected to Powere from Battery. And Q3101 is in off State.

But I found that when the portable device is in Off work state. There exists on leakage voltage, about 0.5V on c of Q3101, resulting 50mA leakage current in the total system. The total leakage power consumption is so high. Could anyone give me some advices to reduce the total leakage power in the total system when the portable device is in the Off work state?

Thank you very much in advance!

 

[Audioguru]

If the output leakage of the transistor is 50mA when it is turned off then the voltage across R3128 must be 500V!
Maybe your switch IC is drawing a lot of current.

 

[KlausST]

Hi,

10k for R3128 is too high. Use a low value resistor.

It seems you use a simulation tool.
If you need more help: Please show some voltages. And the device types. What switch is it?

Klaus

 

[joshuacp]

Yes, I made one mistake with the above shematic. The following picture is the real work schematic.

 

[KlausST]

Hi,

If you need more help: Please show some voltages. And the device types. What switch is it?

Still missing.

Try to place 1k resistor across BE of bjt.

Klaus