# How to choose a flyback diode for a relay?

A diode is put in parallel with a relay coil (with opposite polarity) to prevent damage to other components when the relay is turned off.

Here's an example schematic I found online: I'm planning on using a relay with a coil voltage of 5V and contact rating of 10A.

How do I determine the required specifications for the diode, such as voltage, current, and switching time?

First determine the coil current when the coil is on. This is the current that will flow through the diode when the coil is switched off. In your relay, the coil current is shown as 79.4 mA. Specify a diode for at least 79.4 mA current. In your case, a 1N4001 current rating far exceeds the requirement.

The diode reverse voltage rating should be at least the voltage applied to the relay coil. Normally a designer puts in plenty of reserve in the reverse rating. A diode in your application having 50 volts would be more than adequate. Again 1N4001 will do the job.

Additionally, the 1N4007 (in single purchase quantities) costs the same but has 1000 volt rating.

• Nice tip about the 1N4007. May 16 '14 at 6:12
• For such a type of relay, even a 1N4148 would do the job (Vrrm = 100V, If = 200mA, Ifsm = 1A for 1 second). This diode might be faster and will most probably be cheaper too, although that's not important if used for low quantities of course. Jul 26 '15 at 8:56
1. Voltage required is the nominal coil voltage, since that is what will be applied. Give it a factor of 2 for safety.

2. Current requirement is the nominal coil current.

3. Speed is probably not a consideration for relay coils, since they are not turned on/off very often, as compared to, for instance, a PWM motor drive.

In your case, a 1N4001 will probably work just fine.

• Speed is important, so Shottkys are actually preferred. Not for switching frequency, obviously, but for low-delay fast action, in order to clip the transient peak as early as possible, right when it is the highest.
– Sz.
Jul 12 '18 at 20:41

Things aren't always as simple as they seem, though in the case of relays it's highly application dependent. While the diode provides a safe discharge path that preserves your switching transistor and power supply, it can cause a few issues in certain applications. Relays on closure can form a small weld at the contacts, and by placing the diode there you are essentially preventing the relay from opening with its full force. This can cause the contacts to 'stick' together slightly longer, and overall is bad for the relay.

A trick I learned a few years ago to prevent that from happening was to put a zener diode in series (obviously in different direction) with the the regular diode, this allows you to control the maximum voltage and allows the relay's coil to discharge in a slightly better way.

Page 13 of this Panasonic application guide contains the zener diode and other options for contact protection: https://www.panasonic-electric-works.com/pew/eu/downloads/technical_information_relay_en.pdf

• This can also be achieved with a resistor instead of a zener, with potentially even better results. You could pick a resistor value that allows the voltage to rise as high as your components can tolerate, for maximum energy absorption. Nov 24 '17 at 10:29
• Huge thanks for this tip. Also I have learned that electro-clutches work better with that series zener. Check this link for more info Aug 5 '20 at 12:53

Question: What size of fly-back diode do I need for my inductive load?

My Answer: Fly-back diodes are sized based on power dissipation

$$\P = 1/10(I^2)R\$$

P: power dissipated in fly-back diode

I: steady state current flowing through the inductor (fly-back diode not conducting)

R: resistance of the fly-back diode in conduction

Proof:

The fly-back diode will be held at a constant temperature; diodes have a constant resistance in conduction when held at a constant temperature. (if the temperature changes, so does the diodes resistance)

Now the conducting diode behaves as a resistor so the question becomes: How much power do I need to dissipate in my diode's internal resistance?

By observing a series RL curve, we know that the inductor discharges or charges in 5 time constants and one time constant is equal to the inductance divided by the series resistance ($$\T = L/R\$$).

Some math people told us that the energy stored in an inductor is:
$$\E = (1/2)L(I^2)\$$. Here E is in joules, L is in Henrys. They also said that power is energy per second ($$\P = E/time\$$). Here, power is in watts.

So... if our understanding of physics is working... the time in which the inductor discharges is: $$\5(L/R)\$$ seconds, and a stored energy of $$\(1/2)L(I^2)\$$ joules is released in that time. Here R is the resistance of the fly-back diode in conduction, I is the current flowing through the fly-back diode and L is the inductance supplying the current.

If we solve for the power, something very interesting happens... $$\P = ((1/2)L(I^2)R) / (5L)\$$ Here, L cancels out and $$\P = 1/10(I^2)R\$$. We know that R is the resistance of the diode in conduction and I is the current flowing through the diode during the discharge. But now, what is the diode current during discharge?

Consider a circuit as such: simulate this circuit – Schematic created using CircuitLab

R1 is the internal resistance of L1, and R2 is our charging resistance. D1 functions as the fly-back diode, and R3 is the resistance of D1 in conduction.

If the switch is closed and we wait forever, a current of 10mA flows through the circuit, and the inductor stores an energy of 50μJ (50 micro Joules).

Using conservation of energy theory:

If the switch is opened, the inductor reverses polarity to try to maintain the 10mA current. The fly-back diode is biased into conduction, and an energy of 50μJ is dissipated through the diode resistance in $$\5(L/R) = 500\mathrm{ms}\$$. The power dissipated in the diode is 50μJ / 500ms = 100μW (100 micro watts).

$$\(1/10) (10\mathrm{mA} ^2) (10\mathrm{ohms}) = 100\mathrm{\mu W}\$$

So to answer the last question: the diode current during discharge can be thought of as equal to the steady state charging current of 10mA when using the equation: $$\P = 1/10(I^2)R\$$. While the current during the inductive discharge actually decreases exponentially and is not a steady 10mA, this simplification will allow for quick computations of the required diode power in a circuit by knowing the initial conditions.

Best of luck with your designs and never use technology for evil purposes.

• Alternatively, would it also be reasonable to approximate this as P = 1/10 * I * Vf@I? (Since it's common to think about Vf rather than equivalent resistance.) For example, if the diode drops 0.5V at 1A and the stready-state inductor current is 1A then is P = 50mW saying the same thing? Apr 1 '20 at 4:04