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leonar40 (Electrical) (OP)
21 Feb 07 15:03
Hello. I am designing a circuit to perform a discharge test on a battery. The load profile consists of several discrete resistance values that I will switch in and out of the load path at set times using an analog output controller and some power MOSFETs. The load bank design is almost complete and has worked well so far in testing. I have just one problem left. During the initial voltage rise time period of the battery, I am required to limit the current of the battery to a set value, for a set amount of time. After that time, I need to let the battery continue to finish its voltage rise naturally. Is there a simple way to perform this current limit function? Here is the circuit that I have tried so far:

_________________
| |
| |
+ |
VBatt ----
- |
| |----- MOSFET Gate Voltage
| ----
| |
| |
| \
| /-------- Load Resistor
| \
| /
|________________

This circuit "almost" works. I can vary the amount of current passed through the MOSFET by adjusting the gate voltage. However, it is not working as a "current clamp" like I would like. The current passed through the MOSFET is varying as a function of the battery voltage, which is a problem for me. What I really need is a way to limit the current no matter what the battery voltage is. Any ideas? I should mention that I know the MOSFET will get hot using this method. However, the time that I have to limit the current is very short, so this will not be a problem. Thanks in advance.
Skogsgurra (Electrical)
21 Feb 07 16:42
You should get that "near constant" current function if you keep the gate voltage constant with regard to ground and also at a potential that is equal to current times resistor plus gate-source voltage.

Near constant because there will be a variation in gate-source voltage depending on current and, to some extent, battery voltage.

An increasing current will then reduce the gate-source voltage and reduce current. Or, rather, keep it constant.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

OutToLunch (Electrical)
21 Feb 07 16:49
drive the FET gate with the output of an op-amp. tie the source of the FET to the inverting input of the op amp. apply a voltage to the non-inverting input of the op-amp which is equal to the current desired divided by the load resistor value.
itsmoked (Electrical)
21 Feb 07 17:11
I would switch the positions of the resistor and the FET.

What are you controlling all this with?  Why not just monitor the current with your controller and PWM the FET accordingly?

Keith Cress
Flamin Systems, Inc.- http://www.flaminsystems.com

BobbyNewmark (Electrical)
22 Feb 07 1:15
Gunnar Englund's idea is OK, but I would be concerned about oscillation.  Power MOSFETs usually have gain to surprisingly high frequencies, and your load resistor is probably a wire-wound model with a lot of inductance.  That's a good set up for getting an oscillator.  Yeah, it can be made to work reliably, but life is short.

I would use a voltage regulator like the National Semiconductor LM338.  Run its input voltage from the battery's positive terminal.  Put the load resistor on the output:  since the output voltage is (nearly) constant, the current will be too.  Making it stable is just a matter of putting in capacitors the way the datasheet says.  (And just throw more capacitors at it, if that isn't enough.)  The National LM-series regulators are generally pretty forgiving when it comes to oscillation, and protect themselves from burning up too.  If one resistor/regulator can't deliver enough current, just put several in parallel.

The LM338 assumes you need a lot of current (5 amps or more).  If your battery is small and you need more accuracy, a smaller regulator like the LM317 might be better.

Try to arrange so that the regulator is running close to its drop-out voltage (minimum voltage drop from input to output) when the battery is at its lowest.  That way most of the voltage, and heat, goes to the resistor.
leonar40 (Electrical) (OP)
22 Feb 07 14:57
Thank you very much for all of your inputs.  I am beginning to think the problem might be the fact that my load resistance is so low.  Here is the rest of my values if it helps to figure out what is going on.

Current limit is 6 A for the first 125ms after battery connection.
Load resistance is 0.2 ohms.
After 125ms, the battery current can rise naturally on the 0.2 ohm load until the other steps of the load profile kick in.

So if you multiply 6 x 0.2, you get 1.2 V, plus the MOSFET turn on voltage of approx. 3.3 V means I should set the gate to approx. 4.5 V.  The MOSFET certainly does restrict current.  This is easily proven because without the MOSFET at a 10V test voltage I would be pulling 50 A (!), but with the MOSFET it's right around 6 A like I want.  The problem is that if I change the test voltage to 15V, 20V, 30V, etc, the current goes up (7A, 8A, etc.).

I like the idea of controlling the voltage using a regulator set at 1.2 V to limit the output current to 6 A.  It would require a bit more circuitry than just adjusting the gate voltage on the FET though.  

coconutalley (Electrical)
22 Feb 07 15:11
The problem is that a MOSFET is not a very good linear device.  An NPN transistor would be best but you might have
a hard time finding one that can handle 50A.
I would suggest a better way but not as simple:
Switch the MOSFET gate with pwm.  Put a current sensor in line with the MOSFET, feed this into a PI with a setpoint (current level) control.  Feed the PI output into a duty cycle generator (compare voltage with a sawtooth waveform)and drive the MOSFET gate with that.  Actually, you can probably buy switching power IC controllers that are used for switching power supplies to replace the PI and duty cycle generator.
leonar40 (Electrical) (OP)
22 Feb 07 15:51
Thanks for your advice on the BJT.  I had also thought of that since they have a much larger linear region.  The problem is that they are current driven instead of voltage driven like  MOSFET.  I'm not sure how much current my voltage source can sink, but it bet it's not much.  I guess I could use a driver circuit before hand.  So far I'm thinking the voltage regulator is the simplest way around this problem if I can't get the MOSFET to do exactly what I want.

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