0
\$\begingroup\$

So I've been messing around with DC motors and electrical braking. BY placing a variable resistor across the motor terminals of the motor I've been able to vary the braking effect and observe the changes in voltage and current across the resistor, i.e power extracted. During my testing I implemented dynamic braking at 105 rads to ensure transparency and for the majority of my results I found that the voltage across the resistor is consistently 7V upon the instant braking occurs, similar to what I applied to drive the motor.

However, as the resistance is decreased, as is the voltage. It would be expected that the current would be increased, which it is (i=V/R), however I'm unsure why the voltage drops as the resistance across the motor terminals approaches zero. Could anybody explain this?

Thanks for your time in advance,

\$\endgroup\$

1 Answer 1

4
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Dynamic braking circuit.

Don't forget that your motor has internal series resistance. As R2 approaches a short-circuit more and more of the power will be dissipated in the internal resistance of the motor.

Try measuring the motor terminal resistance and then do a voltage divider calculation using that number and your external resistor. Hopefully the maths will make sense of the situation.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.