How do I put an LED in series with an optotriac?

Question

I have a set of SSRs (Vf=1.3V, If=20mA) that I am using to switch low-voltage AC loads, and I would like to have a set of indicator lights, one for each switch. Is it possible to simply put another LED in series with the SSR input? This LED would certainly not match the LED inside the optotriac, and I may not even have precise information on this LED, so how would I go about calcuating the resistor I need? Is this even possible or advisable?

Answer 1

You can put a LED in series with the LED in the solid state relay. That will make the total voltage you need to drive the combination higher, but as long as you have the voltage available that's no problem.

Let's say you want to use a green LED to indicate the SSR is being driven on, and you have a 5V output to turn on the SSR. Green LEDs drop about 2.1V. You should be easily able to find one rated for 20mA, as most are. Probably the 20mA spec is a maximum for the SSR (for you to check), so let's say we want 10mA going thru the string. The LEDs will drop 1.3V + 2.1V = 3.4V, leaving 1.6V for the resistor. 1.6V / 10mA = 160Ω

So, put the 160Ω resistor, LED, and SSR input all in series. Now you can drive the whole thing with 5V to turn on the SSR and light the indicator, and 0V or open for off. When on, it will draw about 10mA.

Added:

In the description above the same current is flowing thru the indicator LED as the SSR LED. It looks like this SSR wants fairly large drive current of around 20mA. While you can certainly find a LED that can handle that, most "normal" and cheap LEDs are rated for 20mA max. They will also be plenty bright at much less current.

As Russell mentioned, you can put a resistor accross the LED to split the SSR drive current between the resistor and the LED. With a total of two resistors, you can set the SSR drive current and the external LED current separately. Russell made it sound a bit mysterious, but it is actually very easy. Here is a circuit:

Let's say we're using a common green LED for the external indicator. As I said above, figure about 2.1V drop for one of those. The max drop of the SSR LED is 1.4V, so that leaves 5V - 2.1V - 1.4V = 1.5V accross R2. This resistor sets the SSR current, which we want to make 20mA in this example. 1.5V / 20mA = 75Ω

D1 will have about 2.1V accross it roughly independent of the actual current we put thru it. R1 will therefore pass 2.1V / 150Ω = 14mA. That's how much of the total current will NOT run thru D1. The D1 current will therefore be about 6mA, which is plenty to light up most normal LEDs for indicator use.

Anyway, the point is not these particular values, but how you can use two resistors to set the currents thru both LEDs independently but still predictably enough for this application.

Note that you can use the same idea if you want the indicator LED to take the higher current. In that case you put the parallel resistor over the SSR's LED. This could be useful if your indicator LED has to be visible in daylight and the SSR's LED has limited current capability.

Answer 2

Olin's solution will work well, but depends to some extent on matching the LED you choose as an indicator with the current drain of the LED inside the indicator. Also, in some cases only you may have problems with having enough drive voltage to operate both LEDs. LEDs of different current requirement can be run in series, with th lower current LED having the "unwanted" current shunted around it with a resistor, but it makes life harder.

Depending on the parameters of the solid state input circuit a safer and more defined alternative MAY be to drive the indicator in parallel with the SSR input.

This has the advantages of

• Being able to select indicator LED current independent of the SSR LED current. This is probably the main advantage

• Being able to drive from a lower voltage. Usually not an issue except in extreme cases.

• Failure of indicator LED circuit will not affect SSR operation.

It has the disadvantage that,

• in the (unlikely) event that the SSR LED fails the indicator LED will still (incorrectly) show that the SSR is activated.

• More overall drive current required (2 LEDs). Not usually a problem.

Method:

Resistor from Vin to LEd to ground.

R = (Vin - V_LED)/I_LED.

Vin is SSR operating signal. I_LED is independent of SSR current. Vin must have enough current capacity to operate SSR LED and indicator LED. Vin must be > V_LED in SSR and > V_LED in indicator, but need NOT be > sum of two LED voltages, as is the case where LEDs are in series.

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Added:

Specific example

Your SSR LED is rated at 1.4V 20 mA.(Data sheet says 1.2 V typical, 1.4V max, and worst case MUST always be used for design. )

If you wanted to operate this from 3V you would be hard pressed to operate two LEDs in series. 3V - 1.4V - 1.4V = 0.2V "headroom". Using a red LED with Driving the two in series with no resistor from a processor pin "may" would but would be undesignable.

In such a case the two LEDs in parallel would be a much better solution, at the cost of needing current for two LED's, so probably not possible using a single processor pin for drive. However, many processors do not provide 20 mA drive fro all pins (or any in some cases) so a driver transistor or similar would be needed.

Note that red LEDs are typically shown with Vf's of 1.6V to 2V typical at 10 mA or 20 mA. In which case, "designed" 3V series drive would be essentially impossible.

Note that this insistence on using worst case datasheet figures is not just a designer's aberration that can be overlooked by mere mortals. Real world examples sometimes approach worst case close enough that a design based on "typical" parameters will work poorly or not at all. I have seen TRIAC outout zero crossing photo couplers that specified apparently unreasonably high LED currents but which not operate at all on less than about 80% of rated drive, and then questionably below 90%+.

If 5V is available for drive the series arrangement becomes doable.

Say external LED = 2V worst case. R = (5v - 2V - 1.4V)/20 mA = 1.6/20 mA = 80R ~= 82R.

= OK.

If a white LED (Vf ~= 3V up) is used the series arrangement is rather marginal at 5V drive. V resistor = 5 - 3+ - 1.6 = <= 0.6V

The 0.6V or less drop across the resistor is small wrt supply voltage. Slight changes in LED forward voltages or supply voltage will significantly affect LED current. So at 5V drive the series arrangement is marginal for white LEds but OK enough for eg red. .

At drive voltages of say 10V up the series arrangement works well.

Power required overall is about half with the series drive as drive current is shared by both LEDs. At lower drive voltages the power used in minimal.