There is no schematic needed. Just use a power resistor rated to for the power dissipation at the current/voltage you're trying to test at.
For the 5 amps at 10 volts you mentioned you will need a 2ohm resistor that can dissipate 50 watts of power, basic de-rating common sense says you should get a resistor that can dissipate 75-100 watts safely.
The best power load I've used was just a power transistor with a big heat sinc, and a power emitter resistor, connected to a power supply between the base and ground. Just connect the output of the power supply under test across the transistor/emitter resistor. The current will be
IL = (V(supply) - .7V)/RE. Just make sure your components can withstand the power dissapated.
Yea I know . But I have to use mosfets . So with a µC I have to select a channel that is used to work at that voltage with that current. And eg. after 5 minutes the current has to change from 1A to 5A . The programming is not a problem , its finding the correct schematic .
You're doling out the requirements little by little.... I lose interest when that happens. And BTW, I can't get anything with your link. Try the linking facility, the "globe and chain" at the top of the text window.
No Sneak, you're not fully specifying YOUR requirements. The link you provided doesn't tell us anything about what you actually want to do.
What are your input voltage range allowance? What is your power requirement? What range do you need and at what resolution? What other considerations might there be that you're not telling us?
Your first post didn't mention a micro controller or a mosfet, are we supposed to read your mind? We shouldn't have to spend the time to incremenetally become aware of your intent if you would properly describe what you ACTUALLY want to do properly the first time.
Just winging it but if you want to use a MOSFET to switch a load you will need logic level mosfets if you plan to turn them on using a PIC. That or additional circuit between PIC and MOSFET.
Years ago I built a small demo unit for load testing a PSU but used switches. The stuff from then can be seen here. You would replace the switch(s) with MOSFETS and calculate your resistances accordingly. Even if your max load will only be 5 Amps I would use good power resistors and I would use AWG 12 for all the load wiring to eliminate as much I*R loss as possible. Yeah, overkill.
What you are building is a "Load Bank". If the current needs to be very accurate I would measure it with a shunt and use a fixed and adjustable resistor to really trim the load. Your call on that.
You are using the MOSFET as a switch and that is about it.
You could provide more info as I see 10 Volts and you want 1.0 amp then advance to 5.0 amps.
Yeah, the circuit you posted does what was mentioned earlier using an active load. The darlington is the load. However if you want to view current you will need to add a current meter.
No Sneak, you're not fully specifying YOUR requirements. The link you provided doesn't tell us anything about what you actually want to do.
What are your input voltage range allowance? What is your power requirement? What range do you need and at what resolution? What other considerations might there be that you're not telling us?
Your first post didn't mention a micro controller or a mosfet, are we supposed to read your mind? We shouldn't have to spend the time to incremenetally become aware of your intent if you would properly describe what you ACTUALLY want to do properly the first time.
the voltages are 12 & 24V and the power levels must be from 60 - 100W . So i would have like 4 outputs that work at a different power level and simultaneously .
Thanks for the good info Reloadron !
I also found an interesting schematic at this website : 200Watt Active Load Unit . Point 4.3 . With the switch he was able to set the current. I would have to remove the resistor R3 & R4 and change the darlington to a Mosfet N channel with logic levels.
In that curcuit, the current is adjusted by the center tapped resistor connected to the op-amp input. The current is the voltage at the "+" input divided by the .1 ohm emmitter resistor. The power is that current times the volts meter shown in the drawing. There should be a second volt meter at the amplifier's input.