1
\$\begingroup\$

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?

\$\endgroup\$
  • \$\begingroup\$ 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. \$\endgroup\$ – Alfred Centauri Jun 8 '18 at 3:10
2
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ Oh, I didn't know that formula. That makes sense. Thanks. \$\endgroup\$ – Tom Jun 8 '18 at 16:49
2
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ Oh, that makes sense. But they'll be exponential ramps? \$\endgroup\$ – Tom Jun 8 '18 at 3:10
  • \$\begingroup\$ @Tom, what is an exponential ramp? Typically, a ramp is, e.g., a segment of a linear function of time. \$\endgroup\$ – Alfred Centauri Jun 8 '18 at 3:11
  • \$\begingroup\$ Voltage where Vl=V(e^-(rt/l)) \$\endgroup\$ – Tom Jun 8 '18 at 3:16
  • \$\begingroup\$ @Tom, for a square-wave voltage across an ideal inductor, the current through is triangular, not exponential. See, e.g., this -- Image credit \$\endgroup\$ – Alfred Centauri Jun 8 '18 at 3:24
  • \$\begingroup\$ I added this to my answer. \$\endgroup\$ – Janka Jun 8 '18 at 3:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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