# Back-Emf vs. Voltage

Discussion in 'Physics' started by shespuzzling, Jan 5, 2010.

1. ### crutschow Expert

Mar 14, 2008
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6,852
Not exactly.
The back-emf from an inductor always has to equal the source voltage (the V in V=L di/dt) as there is nothing else in an ideal inductor to generate the back voltage (which must equal the source voltage).
Thus the current linearly builds up according to the equation without limit, in a circuit with an ideal voltage source and ideal inductor.
Any constant current limit reached is due to some non-ideal circuit element such as wire resistance or current limit in the voltage source.

When the voltage source is removed, the inductor magnetic energy tries to keep the inductor current moving in the same direction until the energy is dissipated.
This means the voltage across the inductor has to change so the end of the inductor with the current flowing out will become positive with respect to the other end.
This causes the commonly noted back-emf polarity reversal when a switch in series with a inductor is opened.
At that point the inductor now looks like a voltage source, driving the current.

Below is an LTspice demo of an ideal inductor with a switched source.
As you can see, the inductor current (yellow trace) linearly increases with time until the source (green trace) is switched off at 1 second.
The inductor then keeps the current flowing through R1, which generates the initial negative spike of about -24V (blue trace) until the current decays back to zero as the inductive energy is dissipated in R1.

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2. ### Vinyasi Qx New Member

May 8, 2017
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I didn't get the exact same results... Although close.
My output implies that the reverse voltage of the inductor (in red) plays no part in the drop in current across the inductor (in yellow) since it reverses throughout the entire simulation. Only the switch shut-off (in green) drops the current over time.

This implies, to me, that the reverse voltage was fighting the source voltage throughout this demonstration, but losing up until the voltage was cut-off at the source (since -.03mV in absolute magnitude hardly compares with 10V) at which point there was no voltage source for the reverse voltage to fight anymore. So at that point, only the accumulation of current remains to gradually dissipate.

My reverse voltage does not equal my forward (source) voltage comparing their differences in absolute magnitudes.

How did I err?

• ###### demonstration of back EMF.asc
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Last edited: Dec 27, 2017
3. ### crutschow Expert

Mar 14, 2008
23,108
6,852
Look at the input control polarity of S1.

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4. ### Vinyasi Qx New Member

May 8, 2017
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Got it. Thanks.

5. ### crutschow Expert

Mar 14, 2008
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6,852
Just noticed an error in the first simulation.
Here's a corrected version.
The original simulation had an incorrect switch trigger voltage (Vt) and the switch was not fully turning off, thus the reverse voltage V(1) spike was 24V instead of the 30V it should be with 10A initially going through 3 ohms.

Last edited: Dec 28, 2017
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6. ### Vinyasi Qx New Member

May 8, 2017
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What do you make of this?
The Basics - Bypass Capacitors
http://www.seattlerobotics.org/encoder/jun97/basics.html

How is it that the amperage of the source voltage significantly travels in reverse direction to its supply voltage while the switch is closed?

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