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In a video lecture a professor is explaining how to prevent an electric arc caused by the switching operation in an RL circuit. Here is the section (at around 18:26) where he explains it: https://youtu.be/FYDLG_A2P1I?t=1106

I know that a flyback diode is used for this purpose: enter image description here

... yet in his example he uses a very large resistor to create a path for the reverse current.

My questions are:

  1. If a very large resistor is used, wouldn't it be a problem if a continuous switching is very fast comparing to LR time constant?

  2. If a diode is used, would the inductor burn if the back emf current is too high? In almost all examples they just use a diode without mentioning any possibility about this.

user16307
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    Don't expect people to watch the video. Include all of the required information in the question. – Eugene Sh. Jul 06 '16 at 19:52
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    A "feature" of using the resistor is that it dissipates the energy more rapidly than the diode. Diode loss initially = Vf_diode x Iinitial whereas Resistor = I^2_initial x R. For more than about 1 Volt drop the resistor energy dissipation rate starts higher and above a few volts its higher or much higher overall. In some case this matters - eg relat release times are "slugged" by a diode and may be adversely extended. A compromise is a resistor in series with a diode so dissipation is higher/faster, Vmax can be designed and back current is blocked in the on state. – Russell McMahon Jul 06 '16 at 22:41
  • @Russell McMahon: another method to speed up the energy dissipation rate by increasing the voltage is to put a Z-diode in series and opposite direction to the normal diode. The Z-diode voltage can be choosen to be just below the maximum tolerable voltage across the L. – Curd Jul 07 '16 at 09:24

5 Answers5

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You need to be aware that the professor in the video is skipping over a few things. Note that at about 22 minutes, he writes the equation for the current through the resistor $$i = I_0 e^{\frac{R}{L}t}$$ but conveniently fails to evaluate $$ I_0= iR $$ In other words, for his proposed 10,000 ohm resistor, a 1 amp current will provide a 10 kV voltage spike.

Using a resistor instead of a diode is one of those entertaining tradeoffs. A diode will limit the switch voltage to only a bit over the source voltage. This is extremely useful when the switch is a solid-state device such as a MOSFET which absolutely cannot handle kV spikes. On the other hand, the voltage limit restricts dI/dt, so it takes a long time for the diode to bleed off the current. A resistor will allow much higher pulse voltages, but they will typically be much shorter. So:

1-) If a very large resistor is used, wouldn't it be a problem if a continuous switching is very fast comparing to LR time constant?

Yup. You don't do it with switching regulators. The professor was referring to very low switch rates, such as applying and removing power from a circuit. Very old school.

2-) If a diode is used, would the inductor burn if the back emf current is too high? In almost all examples they just use a diode without mentioning any possibility about this.

Nope. The "back emf current" is simply the value of the current which existed when the switch was opened. The diode needs to adequately sized, though - it must be able to carry whatever current the inductor was carrying.

Dave Tweed
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WhatRoughBeast
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This is an addition to what others have said - not a complete answer by itself:

A "feature" of using the resistor is that it can dissipate the stored energy more rapidly than the diode.

Diode loss initially = Vf_diode x Iinitial whereas
Resistor initial loss = I^2_initial x R.
For more than about 1 Volt drop the resistor energy dissipation rate starts higher and above a few volts its higher or much higher overall.

In some case this matters - eg relay release times are "slugged" by a diode and may be adversely extended. A compromise is a resistor in series with a diode so dissipation is higher/faster, Vmax can be designed and back current is blocked in the on state.

_____________________________________

Added:

I was thinking using the resistor instead of diode would bring a longer exponential decay so it would be problem if the switching speed is much higher than LR time constant. Don't you think a large R would increase the time constant?

No. This is not a "normal" situation.

In an inductor, at turn off I_existing WILL continue to flow.

If there is NO series path then 1/2Li^2 energy will convert to 1/2CV^2
where C is available capacitance - added or stray.
If only stray capacitance is present it is often small .
For small C V must be very large - you can get kiloVolts of spike from a low voltage supply worst case.

The time constant for an LR combination is tc = (L/R).
ie as R goes up tc goes DOWN.
This is because Iexisting MUST flow so
V = Iexisting x R.
For large R you get large V and larger power as
V^2/R or I^2R.

ie increasing the value of a shunting resistor
will dissipate more power per time period
(and more peak power)
and take less time for Vspike to decay
but voltage increases with R
and energy dissipation rate with R squared.

Russell McMahon
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  • I was thinking using the resistor instead of diode would bring a longer exponential decay so it would be problem if the switching speed is much higher than LR time constant. Don't you think a large R would increase the time constant? – user16307 Jul 06 '16 at 22:55
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    @user16307 No. This is not a "normal" situation. At turn off I_existing **WILL** continue to flow though all the forces of nature oppose it. If there is NO series path then 1/2Li^2 energy will convert to 1/2CV^2 where C is available capacitance - added or stray. If only stray capacitance is present and small then V must be very large - you can get kiloVolts of spike from a low voltage supply worst case. **IF** you use a resistor the time constant = (L/R). As R goes up tc goes DOWN. This is because Iexisting MUST flow so V = Iexisting x R. For large R you get large V and power V^2/R or I^2R. – Russell McMahon Jul 07 '16 at 00:00
  • Alright I understand thanks. But one last question. Can we say according to what you say, by using large R we can obtain very small time constants; hence we can also use this in switching applications such as pwm motor speed control instead of flyback diodes? Why is diode preferred in general then? Maybe I might be repeating my question but I just couldnt get that point. – user16307 Jul 07 '16 at 00:26
  • @user16307 Drivers are often / almost usually rated for voltages not vastly n excess of operating max. Max spike they will tolerate is V_driver_max so Vspike above Vdd supply is (Vdrivermax-Vdd). eg a 12V supply with 1 20V max driver (FET or IC) allows a 20-12 = 8V spike max. If energy dissipation time is not crucial then a diode only gives a 1 to 2V max spike, you can get large peak current diodes in air cooled 2 lead packages and they are well priced compared to other components in the system. A resistor may allow say 6V spike in the above case = 3 x diode. ... – Russell McMahon Jul 07 '16 at 03:32
  • ... Max ratio peak/avg may not be as good and size is generally larger than equivalent diode. SO a resistor may be a better solution in critical cases but diodes work well, minimise voltage spike are well understood, ... . | If you have lots of voltage headroom and want to minimise dissipation time a resistor allows this but MUST be designed. – Russell McMahon Jul 07 '16 at 03:34
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V=IR. This is something you cannot work around.

With the switch closed such that the inductor is charging or has charged up, a parallel resistor will also provide a "leakage" path. If the R is large enough this might not be a concern.

But when you open the switch the inductor is going to want to keep the current flowing and now the path is via the resistor: V=IR and thus the voltage across the inductor will rise to satisfy this.

You might be able to tolerate this in which case ... Sure and its a balance between the losses due to the leak path and the resultant voltage to maintain current flow.

Or you could just use a diode

  • Note that once the inductor current has stabilized, the voltage across it is zero, and so is the current through the parallel resistor. Under these conditions, all of the voltage is dropped across the series resistor. It's only after the switch opens that current again flows through the parallel resistor. – Dave Tweed Jul 06 '16 at 21:12
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In general, the use of a resistor alone for flyback suppression is frowned upon due to the conflict between making the resistor large (to avoid excessive steady-state dissipation when the coil is being driven) and making the resistor small (to limit the back-EMF voltage in a meaningful way as opposed to the uselessness of the 10kOhm resistor given in the question). A more practical non-diode-based snubber network uses a capacitor in series with the resistor -- this allows a small resistor (say on the order of 30-100 ohms instead of 10k ohms) to be used without dissipating any power in the resistor under steady-state conditions.

ThreePhaseEel
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That professor is talking out of his a... No current through the resistor when the switch is closed, and the resistor will take up the current when open.

Well, the first statement is just dumb, and the second is correct - but he fails to mention the size of the voltage across that resistor. Let's say the current through the coil is 100 mA, the resistor is of 1 kOhm - then the voltage across the resistor will peak at 100 V when the switch is opened, which again means that the resistor has to dissipate 10 W. That's one expensive component compared to a diode.

You might want to use a bleed resistor in series with a flyback diode, but only to expedite the dissipation of the power stored in the coil - ie. to make a relay switch faster. But that's another story.

Peter R.
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  • L/R time constant will be very low if R is huge. So maybe the resistor can handle that power for a very short amount of time i.e transient. Heat is related to time anyway. – user16307 Feb 25 '18 at 21:05