# Back EMF & Current, how can we make the current $I$ stable?

Let's assume that we have a wire with $10 \, \mathrm{V}$ across it and $1 \, \mathrm{A}$ flowing through it. Then if this conductor is introduced to a changing magnetic field, inducing $-EMF$, can we control our voltage to increase it and allow current to be stable at $1 \, \mathrm{A}$?

Thus more power is required correct?

Generally, in any situation we can keep the current stable at the same value by increasing the voltage?

Yes this is common practice for stepper motors only that the strategy is a little bit different. Instead of the voltage value one controls the time average of the voltage through a pulse-width modulation. This works because the coil works like an integrator: $$i_L(t) = i_0+\frac1{L}\int_{0}^t v_L\left(\bar t\right) d\bar t$$ $v_L$ would be $$v_L(t) = V_{\rm bat} - Ri_L(t)$$ in the on-phase and $$v_L(t) = - Ri_L(t)$$ in the off-phase. Thereby, $V_{\rm bat}$ is the battery voltage and $R$ is the resistance of the coil. Sometimes one further decreases the voltage in the off-phase via diodes to get a faster discharge of the coil.

Specific for motor controllers one also has to account for the induction voltage caused by the motion of the rotor. (This is not regarded above.)

There are more refined kinds of PWM, so called full bridges that can also drive with $$v_L(t) = -V_{\rm bat}- Ri_L(t)$$ in the off-phase.

• What above the idea of increasing power so that voltage is increase to oppose the back $EMF$ and allow current to stabilize at $1$ $A$ Jun 14 '14 at 19:58
Correspondingly, given that the current does not change, the output power of the supply,$$P_{\text{out}}=V_{\text{out}} \, I \,,$$will either increase or decrease.