Hello,

I faced a really weird problem with a rather simple transistor - switch circuit (picture in the attachments). A bit of background... The switch is lock type one, meaning it stays in its last state, and it is rated for only 100 mA. I have 2 18650 batteries which I want to switch in on-off mode. The maximum possible current of the application will be 1 A, so I wanted to use a mechanical switch to switch an appropriate transistor for that (I can not change the mechanical switch, because it is the part of my product design and the enclosure is already made.) Whenever I press the switch, I get only around 2.2V on the output (which should be slightly lower than the fully charged 18650 battery, so lets say no less than 3.8 volts.) I checked the wiring, I am getting around 3.7V at the gate when I press the switch, there are no shorts or anything like that, first the design was made to use a specified GaN mosfet, then I tried other mosfets, after that I passed to BJT, I used BC141, 2SD1207 and SS8050, every case was the same, I am getting 2.2V at the output. I don`t have much experience In electronics, but transistors were the thing that I was pretty confident with. Any ideas how to fix it or replace the design without too much excess.

 

The circuit you have drawn will not do that. There is no latching action from the components you have drawn.

 

The circuit you have drawn will not do that. There is no latching action from the components you have drawn.

Could you explain it a little bit more? What would you suggest for a substitution, granted I still have use the mechanical switch as a primary actuator (I have a little bit of space on the pcb, so I can work something out)

 

If I understand correctly you want to switch the BAT1 to the Batt_IN.
If so then you need to use a P channel mosfet and wire it like this:

 

Could you explain it a little bit more? What would you suggest for a substitution, granted I still have use the mechanical switch as a primary actuator (I have a little bit of space on the pcb, so I can work something out)

I'm still not quite clear on what functionality you are looking for.

 

If I understand correctly you want to switch the BAT1 to the Batt_IN.
If so then you need to use a P channel mosfet and wire it like this:

I think the TS was under the impression that such a circuit would have a "latching" action which IMHO is not possible with a single switching device.

 

I think the TS was under the impression that such a circuit would have a "latching" action which IMHO is not possible with a single switching device.

I was under the impression that the mechanical switch was a latching push button type.

 

yes.. proper term is maintained switch as opposed to momentary.
if i understand 3.7V battery is used as an energy source and when switch is active it is supposed to supply power to load connected to BAT_IN and GND. so transistor is used as switch (on the positive end, aka high side switch).

if using N-channel MOSFET or NPN transistor, there will be huge voltage drop no matter what. (couple of volts)
if using PMOS or PNP transistor, voltage drop can be tiny (some 100mV) but for that, stitch would need to be on the low side.

but maybe i am also interpreting TS intent incorrectly.

 

yes.. proper term is maintained switch as opposed to momentary.
if i understand 3.7V battery is used as an energy source and when switch is active it is supposed to supply power to load connected to BAT_IN and GND. so transistor is used as switch (on the positive end, aka high side switch).

if using N-channel MOSFET or NPN transistor, there will be huge voltage drop no matter what. (couple of volts)
if using PMOS or PNP transistor, voltage drop can be tiny (some 100mV) but for that, stitch would need to be on the low side.

but maybe i am also interpreting TS intent incorrectly.

Well, my task is the following: I want to have a kill switch for the 18650 batteries. I can not use the switch directly for that because of the current rating. However, I still have to use the switch to initiate that (like i mentioned, for the enclosure reason). So the only available latching soursce are the batteries, and I need to incoporate some kind of transistor configuration making or breaking the circuit when the buttin is on or off. Just like it was mentioned, to provide voltage to the Bat_in terminal, or to cut it off respectively. Could you explain a bit more, why NPN and Nmos have such a large voltage drop in this case? I want to thak you in advance.

 

The gate of a N-MOSFET must be several volts higher than the source in order to be on. Look at your cicuit. What is the voltage on the gate? What must the voltage must the source be then?

 

exactly... to turn transistor on there need to be sufficient voltage difference between gate and source. NMOS requires positive voltage at the gate. so if drain is at 3.7V, and you want source to reach some 3.6V, gate voltage need to be couple of volts higher than this 3.6V (7-8V would be nice). so if you are using NMOS, and do not have the extra voltage available, do not try to switch positive side of the load, switch negative. then you can still use the same 3.7V source.

the exactly the same goes on for PMOS but the polarities are reversed.

 

If I understand correctly you want to switch the BAT1 to the Batt_IN.
If so then you need to use a P channel mosfet and wire it like this:
View attachment 281315

I built exactly the same circuit like you suggested. I used AOD423 P MOSFET with a max. gate threshold voltage of 3.5V (in order to fully open mosfet from the battery voltage). Battery voltage is exactly 4V. When the button is pressed, I get 4V at the output just as expected. However, when the switch is open, I still get around 3.3V at the output. I kept my voltmeter for some 10 seconds after the switch was disconnected, the voltage was stable at 3.3V, so any parasitic capacitance is not the reason for that. Could it be that the mosfet is not fully closing due to a relatively large pull-up resistor? Would 1k resistor help instead?

 

the resistor is not large. have you considered possibility that battery may have some charge and hold the voltage even after button is released?

 

the resistor is not large. have you considered possibility that battery may have some charge and hold the voltage even after button is released?

I am not quite sure what do you mean by that? When the button is released switch becomes an open circuit ideally and gate voltage becomes same as the source voltage. At least in my understanding... I also read that including a series diode between the switch and a resistor could help, any observations on this idea?

 

Could it be that the mosfet is not fully closing due to a relatively large pull-up resistor? Would 1k resistor help instead?

Did you have a load connected at BAT_IN?
If not, connect a1K resistor from BAT_IN to ground.