# Parallel a diode to solid state relay load is a short circuit?

I'm trying to use a solid state relay control a 12v dc air pump, so I was reading the "Solid State Relays Common Precautions"(like below). http://omronfs.omron.com/en_US/ecb/products/pdf/precautions_ssr.pdf

When I read the paragraph shown in the screen, I'm wondering if I parallel a diode to the load, I'm I making a short circuit since there won't be much resistance in the dioade so the current will just bypass the load and go through the dioade?

• How the schematic is drawn makes it very confusing and makes it look like the diode is forward biased when it's in fact reverse biased. – Selvek Mar 1 '18 at 0:23

You need to connect the diode the correct way (reverse biased, as shown), otherwise it will indeed conduct when the SSR turns on and likely destroy the SSR and quite possibly the diode.

When connected as shown, it is reverse biased until the SSR turns off, upon which the inductive load will cause the diode to become briefly forward biased until the energy in the magnetic field is absorbed by the coil resistance and diode forward drop. Otherwise the voltage across the load would increase to potentially damaging (to the SSR) levels.

Consider the below schematic- the switches represent the SSR and the stuff in the boxes represents your pump motor.

simulate this circuit – Schematic created using CircuitLab

The switches open at t=0 and current is 1A through each circuit at that time. As you can see in the below simulation, the voltage across the switches increases to a bit over 12V and then drops back to 12V when the diode stops conducting about 2.5ms later.

If I remove D2, the difference is dramatic:

The voltage across SW1 spikes to about 900V, before dropping back.

• Thanks, yes I need to connect the diode the correct way, I think I have better understanding now. – Todd Liang Mar 1 '18 at 0:40

Diodes are applied in reverse of the applied voltage.

If you look closely, it is a ground side switch and probably an opto coupled N-FET.

So return or back EMF from stored energy when power shutoff is the only time when the diode becomes forward biased. The current rectifier acts as the 2nd "Throw" in a SPDT switch but in this case just "quenching" the same current and dissipated into all the series resistance for a short time, defined by the T= L/R ratio. where R includes the diode series bulk or ESR and the load series resistance. Motors are rated with DC resistance or DCR.

Since you are concerned about the diode, I thought I'd start with helping you get more comfortable with the circuit you posted. On the left, below as Figure 1, is the equivalent circuit -- just drawn a little differently, is all. If you remember about diode directions, the diode is arranged so that it is reverse-biased in this circuit. So it should not cause any troubles.

simulate this circuit – Schematic created using CircuitLab

On the right side, Figure 2, I show a way you can test this out in a safe way. Do NOT hook up the pump motor. Just apply the diode itself as you think it should be. Activate the SSR and then measure the voltage across $R_1$. If the voltage measures $5-6\:\text{V}$ then the diode is indeed reverse-biased. If you measure anything much more than $6.5\:\text{V}$, then you have the diode in the wrong orientation and should reverse it and take your measurement again. If the diode is good, then one of the two orientations will give $5-6\:\text{V}$. That's the arrangement you want in the end.

The article you cite also mentions the use of a zener diode. I'll draw that up as well as another possibility:

simulate this circuit

In Figure 3, I've added a zener with the correct orientation. It looks like it is arranged to be forward-biased when you activate the SSR. But diode $D_1$ prevents that. So the only time the zener's action occurs is when $D_1$ is forward-biased when the circuit turns off and the pump's inductance "kicks back." The reason for adding the zener is to provide a faster "turn off" time. If you don't need it to be faster, then you can avoid the zener.

In Figure 4, I've added another possibility. (And there are still many others.) Here, if you can work out about how much current your motor requires, you can size that resistor to use about the same current. This will help dissipate energy faster in the circuit and will also reduce the turn-off time.