Dave's current load - 2.8v on gate? - Page 1
Forum section: electronics / beginners

author: ikrel
author_rank: Contributor
timestamp: December 01, 2012, 08:34:08 pm
content: I've setup Daves dummy load(same circuit), but I've used an LM358 op-amp and a RFP50N06 MOSFET.What i see on my oscilloscope is 2.88v on my mosfet's gate pin when i connect my 12v power supply to it(0v when i remove the supply).There is no current draw, and the curcuit does work as expected, I just want to know why this happens.Edit: The 12v Supply I refer to above is what I am generating the load for, the actual circuit is powered by the same 12v supply

author: ikrel
author_rank: Contributor
timestamp: December 01, 2012, 10:45:11 pm
content: Ok, I've changed some things, now I'm powering the circuit from a 12v battery, and loading a second 12v battery with the dummy load.With nothing connected to the dummy load(12v power/circuit is on/ mosfet Drain is floating) i get 11.5v at my gate.connecting the 2nd 12v battery to the load drops the gate voltage to 2.88v. everything works as it should.Why is there 11.5v on the gate when there is nothing on the drain.And why is there 2.88v on the gate when the second battery is connected?

author: dr_p
author_rank: Frequent Contributor
timestamp: December 01, 2012, 11:37:09 pm
content: When the drain is floating (no load connected) there is no current, so the op-amp tries to increase the current by opening the FET more and more. This obviously has no effect, since there is nothing connected to the drain. So the op-amp output goes to maximum, in this case dropping some 0.5V from it's power supply of 12V and thus outputting 11.5V. This is what you measure on the FET's gate.When you connect the second battery (to be loaded), current starts to flow, voltage starts to drop on the current sense resistor. This is compared by the op-amp to the positive input (from the potentiometer). When it reaches equilibrium (actual current is equal to what you set it from the potentiometer), the op-amp will lower it's output just so that it stays there. In your particular case, this is 2.88V that causes the FET to partially open, in turn causing exactly the current flow that you have set from the potentiometer.

author: ikrel
author_rank: Contributor
timestamp: December 01, 2012, 11:51:09 pm
content: Would I be correct in saying the the 2.88v would be the mosfets threshhold voltage? Since my pot is set to zero, it would still take 2.88v to reach the point where the mosfet is just about to turn on, but hasn't yet?Secondly, this circuit does not work when powered from 5v. the mosfet barely turn on, and the load only draws 250mA. and the gate voltage tops out at 3.6v. why is this? The op-amp should output 0-2.5v just like daves circuit but mine doesn't work.(note: when powered from 12v I've adjusted the voltage divider so i still get 0-2.5v - on 5v i use 2 10k resistors, on 12v i use a 10k + 2.2k)

author: dr_p
author_rank: Frequent Contributor
timestamp: December 02, 2012, 12:27:20 am
content: I now see that you had the pot to zero.If you look in theFET datasheet, the threshhold voltage is between 2 and 4v, so 2.88V would make sense.Theop-amp datasheetI think says that maximum output is 1.5V lower than power supply, but I don't get it why it went only 0.5V below when powered from 12V. If that's the case, 5V-1.5V=aprox. 3.6V which is only enough to open the FET for 250mA to flow.Other than too low voltage (5V), your circuit seems to work fine.

author: ikrel
author_rank: Contributor
timestamp: December 02, 2012, 12:49:17 am
content: Thats what i thought. Thanks for the help