# Voltage of inductor/transformer with a square wave input

I'm kind of having trouble thinking of what the response of an inductor to a square wave input would be, as well as the response of a transformer (I found this answer for a square wave input to a transformer a little incomplete or unclear: Square Wave input into Transformer). It's my understanding that the graphs for the current and voltage of an inductor when a circuit is closed are exponential shaped, so for a transformer with an idea square wave input (modeling the primary as an RL circuit) what will the output waveform be? Would the output voltage be a exponential triangle wave, a square wave with the rising portion exponential portion, something else or am I not thinking of this right?

• You should stipulate whether you're asking about ideal inductor/transformer circuits or not. It's true that the current through a non-ideal inductor, for a square-wave voltage across, is segments of an exponential curve but that isn't the case for an ideal inductor. Jun 8, 2018 at 3:10

Would the output voltage be a exponential triangle wave, a square wave with the rising portion exponential portion, something else or am I not thinking of this right?

If you ignored the resistance of the primary winding and used the standard formula for an inductor: -

$$V=L\dfrac{di}{dt}$$

And then applied a positive step voltage, you would get a rising ramp of current whose slope is V/L (as per the above formula).

When the square wave goes negative you get a falling ramp of current and the cycle repeats. That rising and falling current produces a rising and falling flux in the core.

Then, using the other well-known formula for transformers: -

$$V = N\dfrac{d\Phi}{dt}$$

We see that the output waveform from the secondary is also a square wave because the rate of change of flux is either a positive constant value or a negative constant value.

• Oh, I didn't know that formula. That makes sense. Thanks.
– Tom
Jun 8, 2018 at 16:49

An inductor is integrating the applied voltage over time.

So, if the voltage is a positive constant, the current is a upwards ramp with constant grade. If the voltage is a negative constant, the current is a downwards ramp with constant grade. The current cannot "jump", it's continous.

• Oh, that makes sense. But they'll be exponential ramps?
– Tom
Jun 8, 2018 at 3:10
• @Tom, what is an exponential ramp? Typically, a ramp is, e.g., a segment of a linear function of time. Jun 8, 2018 at 3:11
• Voltage where Vl=V(e^-(rt/l))
– Tom
Jun 8, 2018 at 3:16
• @Tom, for a square-wave voltage across an ideal inductor, the current through is triangular, not exponential. See, e.g., this -- Image credit Jun 8, 2018 at 3:24
• I added this to my answer. Jun 8, 2018 at 3:29