If I supply 5-15V to the gate, the mosfet allows the current to flow from the (D to S). Using a 1.1 A load, the voltage at the load is 11.4, while battery V is 12.55. If I increase the load(Amps), then the voltage at the load is 5.5V with a 4.5A current. More current drawn by the load is equal to a less open mosfet at a given gate Voltage? Makes no sense. Is this common? I know the mosfet has an internal resistance, but this is overkill. I varied the gate voltage between 5 and 15V, there was no increase in current flow. Mosfet rated at 6.5A. I also have a 2M resistor across the gate and source to prevent gradual turning on and off.


Thanks

 

If I supply 5-15V to the gate, the mosfet allows the current to flow from the (D to S). Using a 1.1 A load, the voltage at the load is 11.4, while battery V is 12.55. If I increase the load(Amps), then the voltage at the load is 5.5V with a 4.5A current. More current drawn by the load is equal to a less open mosfet at a given gate Voltage? Makes no sense. Is this common? I know the mosfet has an internal resistance, but this is overkill. I varied the gate voltage between 5 and 15V, there was no increase in current flow. Mosfet rated at 6.5A. I also have a 2M resistor across the gate and source to prevent gradual turning on and off.


Thanks

Post your schematic, and the MOSFET part number.

 

Are you measuring the voltage at the battery while driving the load? Or is that the voltage of the battery before hooking it to your circuit?

 

Nothing special. Very simple mosfet switch setup. +12V to load, from load to drain of mosfet, from mosfet source to ground. 2M resistor between S & G. Part # SSH10N90A. Just one I had laying around. It is high voltage rated. Is it possible this mosfet requires higher voltage flowing from D to S to operate properly?

 

Not at the battery.

 

Does Rdson mean anything to you? It is 1.2ohm, so the on resistance will never get better than that.

 

Using random parts you happen to have lying around will give random results.

 

I read that. It actually states .93 ohm. BTW.....Don't have to be a wise guy about it. It still does not explain why the voltage is SO LOW. Only 1 ohm will not cause a huge drop like that.

 

The 0.93ohm is for approx 2.5A and 20Vgs.
1.1amp,1V drop gives around 1ohm resistance
4.5amp, 7.5V drop gives around 1.6ohm resistance, and if you account for a bit of sag in the battery voltage, then the numbers work out quite well.
Can you measure it again with measuring the actual battery voltage under load? Also what Vgs are you using?

 

OK. It just did not seem like only 1 ohm would drop the voltage that much under that load, but as I now see, under a high current draw, that 1 ohm causes a big drop. So in order to make up for the large voltage drop, I will have to use a higher voltage supplying the load feeding into the drain, or leave it as is and use the same load with a lower voltage lamp. The supply to the gate made no difference from 5V to 18V.



Thanks.

 

Or just use a better mosfet, for example IRF540N has just 0.044ohm Rdson. Then you will not be wasting almost any power in the mosfet and won´t need to change anything else in the circuit.

 

OK. Thanks.

 

You could also use a MOSFET with a lower ON resistance. They are commonly available with much less than 1 ohm resistance. That will also reduce the dissipation since, 4.5A current and a 7.5V drop causes a dissipation of 33W, which will rapidly smoke the transistor unless it is on a good heat sink.