12
\$\begingroup\$

I'm looking to use several (12-18) solid state relays to control a set of water valves (24V AC solenoids, 20ma holding current, 40ma inrush current), and I'm having trouble finding suitable parts, probably because I don't understand the datasheets I'm reading very well.

Most of the low-current (<=150mA), inexpensive (~$1) SSRs that I've found have maximum "input forward voltage" ratings in the 1.0v-1.5v range (see here for a typical example). Does this simply mean that I need a 38 ohm resistor between my MCU (3.3v) and the SSR?

What do the other ratings mean, such as these:

  • Repetitive peak OFF-state current (output)
  • ON-state voltage (output)
  • Holding current (output)
  • Minimum trigger current (transfer)

I assume that ON-state voltage is the minimum voltage required over the output of the SSR for it to turn on, so at 24v, I'm well above the minimum, correct?

Minimum trigger current and holding current are the amounts of current across the output required to turn and keep the SSR on? Wouldn't the current drop to zero at the zero voltage crossing? I'm not sure what these ratings mean.

\$\endgroup\$
12
\$\begingroup\$

This type of opto-triac is mostly used in mains voltage applications. Due to the limited current capabilities it's often used as a driver for a triac which is the actual switching device. Your requirements are modest, so you won't need that, and you can use the opto-triac to switch your load directly. The opto-triac is a cheaper solution than an electromechanical relay then, so at first sight looks like a better choice.
An important difference between electronic and electromechanical switches, however, is that the latter have a very low on-resistance, while the former always will have a voltage drop when switched on. That's the on-state voltage mentioned in the datasheet. This can be up to 3V, which in a 230V application won't matter much, but if your supply voltage is only 24V AC that's more than 10%. Your load will probably work at 21V, but you'll have to check it.

Repetitive peak off-state current is the leakage current when the triac is switched off. 2\$\mu\$A is a safe value.

Holding current is the minimum load current the triac needs to remain on when the gate is no longer driven. For an average triac your 20mA may be a bit low, but again the opto-triac's 3.5mA is a safe value. (Besides, the gate will be continuously driven, so it's a moot point. It is important in four-component dimmers, where the diac gives a pulse to switch on the triac, after which the triac is on its own.)

Then there is the minimum trigger current. That's the minimum current you have to supply to the LED to switch the triac on, and we'll have to calculate the series resistor accordingly.
Where did you get that 38\$\Omega\$ resistor value? You need figures 3 and 4 to calculate the value for the LED resistor. Figure 4 shows that 10mA is a safe value, and figure 3 shows that at 10mA the LED voltage will be maximum 1.3V. So \$R=\frac{3.3V - 1.3V}{10mA}=200\Omega\$ maximum. Your 38\$\Omega\$ would result in more than 50mA, which is not only more than Absolute Maximum Ratings (page 4), but also more than your microcontroller will be able to supply. So don't exaggerate, and pick a 180 \$\Omega\$ resistor. At lower resistances the current may become too much for your microcontroller's output. If you want more current through the LED (no more than 20mA, never use the Absolute Maximum Ratings!) you may want to use a transistor. Since you'd need a lot of them, consider a driver IC like an ULN2803.

In conclusion I think this opto-triac is a good choice. Alternatively, you may have a look at the MOCxxx series, for instance the MOC3012 needs only half of the LED current, which your microcontroller would appreciate. It doesn't give a nominal value for triac current directly, but from maximum power dissipation (300mW) we can derive that this should be 100mA. (It says peak repetitive surge current is 1A, 120pps, 1ms pulse width.)

\$\endgroup\$
  • \$\begingroup\$ I used \$R=\frac{3.3V - 1.4V}{50mA}=38\Omega\$, so that should've been the absolute smallest resistor possible. I'd have used something larger in practice (I always like to err on the side of too little current, rather than too much, since those errors are less permanant...) \$\endgroup\$ – Mark Jun 30 '11 at 18:07
  • 1
    \$\begingroup\$ @Mark - then you should have calculated the largest resistor value allowed, for the smallest current! \$\endgroup\$ – stevenvh Jul 1 '11 at 11:22
6
\$\begingroup\$

I'm not sure your understanding what an AC SSR is.

Internally the input your driving on the SSR is connected to an LED, the "Input Forward Voltage" is the voltage drop across that diode. Just like driving any LED you need to use a resistor to control the current through the LED (see stevenvh's answer for the math).

The LED shines on a photo-diode which generates current in response. The photo-diode drives current to a triac (two back to back SCRs) which controls the output. With that in mind the values should make sense, if not, read up on Triacs.

Repetitive peak OFF-state current (output)

This is how much current leaks through the output terminals when the relay is OFF. This is really the leakage current of the output triac. In the data sheet you linked this is rated at the max OFF state voltage. (400-600V)

ON-state voltage (output)

This is voltage drop across the output when in the ON state. The output is controlled by passing current through the triac which has a voltage drop, so basically if you put 24V on the output IN terminal you'd see 21V on the output OUT terminal for the device you linked. Well, not quite as this is a non-RMS value where as your 24V AC is probably an RMS value, so you have to deduct this from the peak-peak AC not the RMS AC.

Holding current (output)

This is the minimum current that has to flow through the switch to keep it in the ON state. The device will remain ON until the current drops back below this value regardless of the state of the input pin. Since we're working with AC the current will drop below this value the next time the AC wave heads back toward a zero crossing. If the input is high when this occurs the output will remain ON, more or less, it will turn off briefly until the current swing around the zero crossing and again passes the holding current value on the other size. Effectively holding current is the minimum load you can switch with the SSR.

Minimum trigger current (transfer)

This is the minimum current that you need to apply to the input photo-diode to switch the SSR ON. This is where the 10mA in stevenvh's math came from as a minimum input current.

\$\endgroup\$
  • \$\begingroup\$ When you say, "effectively this is the minimum load you can switch with the SSR", doesn't that contradict the statement that "if the input is high when this occurs the output will remain ON, if not it will turn off"? Wouldn't the SSR stay on for a smaller load than this, assuming the input was driven constantly? \$\endgroup\$ – Mark Jun 30 '11 at 19:47
  • \$\begingroup\$ @Mark Yes, I was speaking high level. Technically the triac is actually off when the load is less than the the holding current no matter what, that is even if the input remains driven, the triac turns off when the load < holding current so it really goes on-off-on at every current zero crossing, twice an AC cycle. Because this region is small you can think of it as "staying on" even tho is actually goes off for a short period of time. \$\endgroup\$ – Mark Jun 30 '11 at 20:12
  • \$\begingroup\$ @Mark one consequence of this is that the load current has to be above the holding current at trigger time, which can be an issue with highly inductive loads. \$\endgroup\$ – Mark Jun 30 '11 at 20:15
  • \$\begingroup\$ @Mark such as a solenoid? That's exactly what I'm planning to drive with the SSRs. The solenoids are rated at 40mA inrush, 20mA holding, so I would be OK with this part, that only requires 3mA? \$\endgroup\$ – Mark Jul 1 '11 at 2:02
  • \$\begingroup\$ @Mark doubt a solenoid of that size would be an issue. \$\endgroup\$ – Mark Jul 1 '11 at 3:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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