Hello, I have a project where I apply a voltage of 8.4 volts with 2 MOSFETs in it and a 1 ohm resistor in order to fix the current at about 8 amps. Basically my 2 MOSFETs have, when fully on a resistance (Rds) of 0.0078 ohm, so together the circuit has a resistance of 1.0156 so 8.4/1.0156 amps flow, so the power-loss at the MOSFETs is very low but I fear that the temperature of the 1 ohm resistor will be fricking high => (8.4/1.0156)^2 * 1.0156 watts. Is this calculation true and if yes what could I do to have my circuit working WITHOUT heatsink, thank´s in advance!
Yeah, your calculation is about right. 64W dissipated in the resistor. The rule-of-thumb is to use a resistor with double the power rating so a 100W resistor is probably close enough. Those are available.
So... What does this circuit do besides generate heat? The current will only be "fixed" if all 8V is dropped across the resistor and there is no other current path.
It's an h bridge for control of a 8 ampere dc motor, I have absolutely no clue how the motor influences the circuit, above all how it changes the power dissipation at the 1 ohm resistor, probably it lowers it but how can I calculate this?
Even a "standard" 100watt resistor will require a heat sink. However if you use a grossly under-rated one (such as 500watt) then no heat sink required. Alternatively if you use a fan to blow air over a 100watt resistor you may get away without a heat sink.
You may find that a couple of 12v 50watt halogen lamps (in parallel) would provide a suitable load. The lamps may have a reduced life if run for too long as they are dependent upon regeneration of the element. I've often used such lamps as high power low ohm resistors. The purists will of course say they have a non-linear resistance as they heat up but if you are working in fairly steady state conditions they may well be adequate for your purpose
Are you sure it's 1 ohm and not 0.1ohm for current sensing.
Yeah, if it is in series with the motor, it is almost certainly a current sense element. Which brings up another point: The resistor won't have 8.4 volts across it.
I think that 8A is the current drawn by the motor at full load, and that will reduce when run lightly loaded.
You sound as though you are trying to force 8A through the motor.
It is more normal to apply a voltage to the motor, and allow it to draw whatever current it needs depending on how heavily/lightly it is loaded.
So would you say I should just let the motor draw current without resistor? By the way, at maximum the motor has 70 watts of power.
Yes.... Why was the resistor added in the first place?
My thought was basically to add the resistor in order to have a calculable maximum current. When I have the voltage and restistance fixed, no matter how much current the motor "could" draw through the circuit a maximum flow of V/R is possible.
Well, it doesn't really work that way. I suppose you were applying the idea of the current-limiting resistor in an LED circuit to your motor.
LED != motor
What voltage is the motor supposed to run off of?
Assuming you're running it at the correct voltage, you shouldn't need to do anything like that (maybe you'd want a fuse though?)
I'd also recommend you get an idea of the actual current in opperating conditions, too - that 8 amps is under heavy load, possibly stall.
Another XY problem 
DrAzzy:
I'd also recommend you get an idea of the actual current in opperating conditions, too - that 8 amps is under heavy load, possibly stall.
While it might be interesting to know these values, when designing motor drive, you would use locked rotor values from the data sheet and add 50%.
My thought was basically to add the resistor in order to have a calculable maximum current.
Ah - then you really want a shunt resistor - a low value resistor that you measure voltage drop across to calculate current:
If you put it between the motor and ground, you only need to measure one side.
And you are not wasting 8 W of power and dropping all your voltage across - you need that voltage to be dropped across the motor.
Say you had the 10mOhm part, then with 8A thru the voltage would be
V=IR = 8A * .01ohm = 80mV, and power P=IV = 8A * .08V = 640mW. Still a lot, but way better than 8W. Use 2 in parallel to cut that in half even, as the transistors will have voltage drop too:
Say you were using MOSFETs with Rds of 10mOhm - now you have the shunt and 2 transistors, and 0.24V lost across the switching components, each dissipating 640mW, 1.92W total, that's all heat you have to get rid of in your system that, power that is not going into turning the motor.
Your resistor would be a good dummy load for testing the H-bridge.
Real DC motors are not just resistance of course, they are voltage sources in series
with a resistance, where the voltage depends on the speed of rotation - this is
normally called back-EMF.
They also have inductance and are a source of switching noise (from the commutator).
You would never(*) add a resistor in series with a motor because that simply wastes power,
its bad enough power is dissipated in the winding resistance...
(*) These days with cheap semiconductor devices available for power conversion.
So Back-EMF reduces the voltage, since an opposite voltage in the motor is induced, hence also the current is reduced, but doesn´t this also reduce the power of the motor then, I mean it should have a higher power, when you start it and should then fall to a lower 1 due to the induced voltage, right? Is this the case in reality but so short that it´s unnoticeable?
There are valid motor circuits that end up using depressingly-large resistors to limit current.
IIRC, this used to be common in stepper-motor circuits where you'd use a higher voltage to help defeat winding inductance and permit higher stepping rates (because calculus), while having the big resistor to limit current. (So you'd operate a 10V 1A stepper (~10 ohm motor winding) off a 20V supply with a 10 ohm resistor (retaining the 1A max - motors being essentially a current-driven device despite being rated by voltage.)
(in modern times, you'd use a chopper-mode stepper motor driver, which would use a much smaller current sense resistor, and "smartness." Although the resistors can still be depressingly large compared to the chips...)
So Back-EMF reduces the voltage, since an opposite voltage in the motor is induced, hence also the current is reduced, but doesn´t this also reduce the power of the motor then, I mean it should have a higher power, when you start it and should then fall to a lower 1 due to the induced voltage, right? Is this the case in reality but so short that it´s unnoticeable?
Why don't you tell us:
-
What the motor is for?
-
What is the main power supply (the motor one, I mean)?
-
What are the plate values: voltage, current & power.
-
What is the d....... resistor for?
Regards
jackrae:
Another XY problem
Now you're getting the hang of it!
All this idle discussion and no circuit schematic and specifications.