Voltage clamp question.

[Denver]

Motor Control.png (18.45 KiB) Viewed 6229 times

Hi I have a Roboclaw 2 x 45 to run a 17 amp motor that peaks under load at 50 amps with the motor bridge. Could someone please tell me if the wiring and components I will be using are correct for running on a 80 amp 24 volt power supply. I would like to have a voltage clamp. I want to be sure I will not damage anything. Design attached
Thanks

[Basicmicro Support]

I answered your question via email but here is the gist. We recommend you use the BUK965R8 N-Channel mosfet. This mosfet has a much lower onvoltage than the IRF1405 which means it will turn on fully.

Also the 1ohm power resistor may be a little low for your voltage. In most cases regen is going to generate a couple amps(a fraction of the motors stall current), say 5 amps in the case of your motors. At 24v a 3 to 4 ohm resistor will dissipate 6 to 8amps. Going with a smaller ohm resistor will work but it may over heat. Your 100watt is big so you may be fine as is. Note that the load on the motor will also dictate if you need more dissipation or not.

[Denver]

If I wanted to clamp more do I go up or down in ohms of the resistor ? Also what are the pros and cons of having a resistor that clamps more amps than required ? I'm using a resistor of 100w and a temperature controlled fan for cooling but I was thinking that clamping more may be give extra safety just in case something causes a increase above what I have allowed for.
Thanks

[Basicmicro Support]

A lower resistance will clamp with a higher current(i = V/R). The primary drawback to clamping at a higher current is power disipation(p = I^2*r). It will increase with the square of the current.

[Denver]

Thanks for your reply. One more question, would 2 resistors in parallel be a good idea to share the load/heat allowing me to have a higher clamp potential for added safety ?
Thanks again

[Basicmicro Support]

Yes, paralleling multiple resistors is a way to spread the thermal dissipation. However keep in mind that paralleling ressitors reduces the effective resistance. new resistance = 1 / (1/r1+1/r2+...+1/rn).

If the resistance of each resistor is the same then the power is dissipated equally among the resistors and the new effective resistance is r/n(eg two 1ohm resistors look like .5ohms, 3 in parrallel look like .3ohms etc...).