Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In some applications a flyback diode is added parallel to the load. Is that because the switching causes reverse voltage for inductive loads? What if we have a resistive load instead? Do we still need to take care of switching? In the below figure there is a read switch operating on an inductive load. Would we still need it if the load were nothing but a resistor? enter image description here

share|improve this question
There is always parasitic inductance. – Matt Young Jun 16 '14 at 15:58
Can you really make a load that's just a resistor? What about the wires connecting the resistor to the switch? They look awfully like a single-turn inductor to me... – Phil Frost Jun 16 '14 at 18:25
the question is would it be negligible? – user16307 Jun 16 '14 at 18:27
@MattYoung: There's also always parasitic capacitance. Combinations of parasitic inductance and capacitance will cause ringing when a circuit disconnects. If inductance dominates, the peak voltage amplitude on open may be very high; if capacitance dominates, the peak current on close may be very high. If neither dominates, voltage and current will stay within reasonable bounds even in the absence of protection. – supercat Jun 16 '14 at 20:33
up vote 8 down vote accepted

If the load is a resistor you should be safe without a diode. Resistors have a parasitic inductance of some sort but the inductance value is usualy very, very small unless you are dealing with high power wire wound resistors that usually have a (somewhat) high parasitic inductance.

Have a look here too.

So the answer is no, you won't need it in almost all practical cases.

share|improve this answer

I know this has been answered already but thought I'd throw this in there as well--

You only need a diode on inductive loads, such as a transformer or a motor. This is because when the magnetic field collapses, it causes back-EMF that can damage your circuitry. The diode provides a safe path so that the EMF can dissipate without causing damage, but will only conduct when back-EMF is present.

Purely resistive loads do not generate any back-EMF, though in reality every load has some inductance and capacitance as well as resistance. However, for all intents and purposes it is generally safe to leave out the protection diode unless you have a significant inductance for the load.

share|improve this answer

Generally speaking, at least in an industrial environment, the only time shunt diodes are necessary is when a relay coil is being driven by a Solid State component. Diodes are there to prevent spikes created by the collapsing field on the relay coil when it is switched off. I cant for the life of me see any reason to protect a resistive load.

Well maybe, if its hooked to a very high impedance solid state output, a.k.a. A CMOS driver driving through a low resistance. But generally new CMOS outputs have protection diodes internally. In the old days of CMOS merely handling the chips without grounding the arm you pick it up with could damage the chips outputs (and inputs) from static electricity.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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