Planning to measure voltage (back EMF) caused by opening a switch in series with an inductor

[David lopez]

i am planning to measure the back emf produced by inductor when you open a switch. i know it is very hard to predict the voltage. but is there any
way to narrow the possibilities?

 

[Borek]

Try to simulate the circuit first with something like (free) LTSpice.

 

[David lopez]

i am thinking i will leave the switch open for a short period of time. then i will close the switch. is the back emf easy to predict then?

 

[berkeman]

i am thinking i will leave the switch open for a short period of time. then i will close the switch. is the back emf easy to predict then?

Yeah, it's zero when you do that, so much easier.

For the case of opening a switch in series with an inductor, many things come into play. There is switch bounce (what kind of switch?), and if the voltage can rise high enough, you have to account for arcing.

A better approach is to use a semiconductor switch to eliminate bouncing, and use a small capacitance to give you a predictable peak voltage from the back EMF. That's how flyback transformer circuits for CRTs work, for example.

 

[alan123hk]

i am planning to measure the back emf produced by inductor when you open a switch. i know it is very hard to predict the voltage. but is there any
way to narrow the possibilities?

In principle, current of inductor cannot be changed instantaneously.

The inductor equation V = L*dI/dt means that if the current changes suddenly, a very large voltage will be generated.

This means that after the switch is turned off, the inductor current will try hard to flow continuously at the same amplitude.

In short, let Ia be the current just before turning off the switch, then the instantaneous voltage of the inductor just after turning off the switch is equal to Ia multiplied by Za (loading Impedance).

Namely Va = Ia * Za, where Va is instantaneous voltage of the inductor just after the switch is turned off.

Prediction Examples : -
Va will be very large if Za is a very high resistance
Va will be extremely large if Za is another high inductance inductor with zero initial current
Va will be nearly zero if Za is a high capacitance capacitor with zero initial voltage
Va will be very difficult to be predicted if Za is a complex network which containing L, C , R and other power sources.
...etc

Of course, the long-term varying of the inductor voltage/current after the switch is turned off is another story.

All in all, in today's times, using simulation to predict circuit behavior is much easier than analysis.