Hi

I made a simple PCB (design A) with a high side switch using SI1553 N P mosfet like the attached schematic. I put a led on the load just to test and everything worked as expected, when i put a HIGH on the Arduino pin, led is turned on.

Later i made another PCB (design B) with the same circuit to drive a load but when i power it, no matter the arduino pin is HIGH or LOW, the load is always on. I tested a second PCB (design B), this time i put only the minimum components and one led at the load, i get the same behavior.

I checked my circuit several times for short / unconnected pins but everything is ok, so i have made some mistake on design B.

One thing i notice in design B, when arduino pin is not connected i get 0,9V at Vgs on N mosfet, should be 0V because of R2 pulldown.
It seems this 0,9V is turning ON N mosfet that turns ON the P mosfet, but why this voltage here? I tried changing resistor values to 100K but didn't made any difference.

Any ideas?

 

looks like your P Channel is connected backwards, and the body diode is conducting.

 

Yes, for a high-side switch the P-MOSFET source terminal must be connected to the plus supply.

I see no purpose for TR1 buffer except to add a 2mV/°C change in the reference voltage due to the transistor base-emitter junction.

 

Here's a diagram of what the others are talking about, it helps to use this symbol when you're doing your schematic capture.

 

I could not find a datasheet for the MOSFETs in your diagram. What is the required gate voltage to switch them on hard?
When the N MOSFET switches off, the only discharge path for the P MOSFET gate is through a 1 Megohm resistor. The gate capacitance can be several nF so the MOSFET will stay on for several seconds until Cgs discharges. The same applies to the N MOSFET when you disconnect it from the arduino pin. 1Kohm or less would be a better value for R2 and R3.

 

I could not find a datasheet for the MOSFETs in your diagram. What is the required gate voltage to switch them on hard?
When the N MOSFET switches off, the only discharge path for the P MOSFET gate is through a 1 Megohm resistor. The gate capacitance can be several nF so the MOSFET will stay on for several seconds until Cgs discharges. The same applies to the N MOSFET when you disconnect it from the arduino pin. 1Kohm or less would be a better value for R2 and R3.

Here's the datasheet: https://www.vishay.com/ppg?67693
It must turn ON with 3V3
I use 1M resistors because i want to power the circuit with a CR2032 coin cell, so when turned ON it must draw as little current as possible.
Anyway, i changed R3 to 10K and now the circuit works.
just don't understand why, i've use plenty of these circuits with 1M resistors and always worked. Bad luck with component tolerances?

Thank you all for your help, especially @KeithWalker who put me in the right direction.

 

Here's the datasheet: https://www.vishay.com/ppg?67693
It must turn ON with 3V3
I use 1M resistors because i want to power the circuit with a CR2032 coin cell, so when turned ON it must draw as little current as possible.
Anyway, i changed R3 to 10K and now the circuit works.
just don't understand why, i've use plenty of these circuits with 1M resistors and always worked. Bad luck with component tolerances?

Thank you all for your help, especially @KeithWalker who put me in the right direction.

Glad you shared your experience. It puts an upper limit on the pull up/pull down resistors for this common circuit. I’ve always used 10k resistors.

 

You flipped the mosfet around right?

 

You flipped the mosfet around right?

No, i use the circuit exactly as on the schematic. It works.

 

So it diverts enough current to turn it off, not sure if it’s the most efficient way to do this.

 

Do you have a better working circuit?
I need to turn on/off a load with a 3V3 HIGH signal, with the lowest current consumption. The load side will drive a few mA.

My circuit consumption with 1M resistors is about 9uA when ON

 

Does it have to be a high side switch? Low side would be more efficient.

Bob

 

Do you have a better working circuit?
I need to turn on/off a load with a 3V3 HIGH signal, with the lowest current consumption. The load side will drive a few mA.

My circuit consumption with 1M resistors is about 9uA when ON

Why stop at 9uA? Why not use a 100 Meg and get it to 90 nA?

 

Does it have to be a high side switch? Low side would be more efficient.

Bob

Well, correct me if i'm wrong but low side does not provide a "true" ground because of the voltage drop on the transistor.
Low side is ok for driving a led, but not ok for driving a micro-controller or other sensitive stuff.

 

Why stop at 9uA? Why not use a 100 Meg and get it to 90 nA?

To be honest, when doing some tests i only had up to 1M resistors and i was happy with 9uA.
But now that you mention, i'll give it a try with bigger values just to see if it works.

 

low side does not provide a "true" ground because of the voltage drop on the transistor.
Low side is ok for driving a led, but not ok for driving a micro-controller or other sensitive stuff.

Where did you get that idea?
You don't need a "true" ground for a micro-controller or other sensitive "stuff".
Most digital circuits can tolerate a 1V or more above ground and still recognize that as a low state.
A MOSFET to ground will provide a low signal just a few mV above ground for small loads.

 

If you are switching power to a microcontroller, be aware that you have to isolate all inputs since a voltage on an input leaks through the protection diodes.

Bo