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I have a Crydom D1D07 solid state relay and the control side is a +5 V DC on terminal 3 and a +3.4 V DC (don't ask) on terminal 4.

When terminal 4 is switched to ground (from +3.4 V DC) the SSR closes and the output side closes and passes the voltage as expected.

However, when the control side goes back to +3.4 V DC from ground the SSR does not open but remains closed and so the output side continues to pass voltage.

SSR:

  • The minimum on voltage is 3.5 V DC
  • Minimum input current is 10 mA for on state
  • Must turn off voltage is 1 V DC
  • Impedance is current regulated

It seems the must turn off voltage is not being met.

Adding a 370 Ω resistor to the terminal 4 side corrects this failure above but I still do not see how it does. The math does not compute. I think the voltage goes from 3.4 V DC to 3.7 V DC with the resistor added.

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    \$\begingroup\$ What is the nature of your 3.4 V source? Is it able to sink current? What other loads are attached to it? \$\endgroup\$ – The Photon Apr 23 at 20:19
  • \$\begingroup\$ 1. it just ties that lead to ground when told to. 2. the resistor is 437 ohms, not 370. \$\endgroup\$ – oldo Apr 23 at 20:25
  • \$\begingroup\$ How does it produce 3.4 V when it needs to do that? \$\endgroup\$ – The Photon Apr 23 at 20:27
  • \$\begingroup\$ it just does by default. Hopefully this detail does not matter to this situation... it just is what it is right now and I cannot change that input voltage except with the resistor. \$\endgroup\$ – oldo Apr 23 at 20:38
  • \$\begingroup\$ So what you connect to terminal 4 is either a short to ground, nothing at all (which results in 3.4 V on the terminal), or 437 ohms to +5 V? \$\endgroup\$ – The Photon Apr 23 at 20:45
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You are not using the input as specified.

Minimum Turn-On Voltage (6) 3.5 VDC
Must Turn-Off Voltage 1 VDC    ( your circuit is 5-3.4 = 1.6V too high )
Minimum Input Current (for on-state) 10 mA   (unknown)
Maximum Input Current 15 mA    ( your value is unknown)

You can specify your options available to you for driving or someone can tell you how to do it.

This is the input behaviour.

enter image description here

schematic

simulate this circuit – Schematic created using CircuitLab

These are a couple of options of many more.

enter image description here

This is exactly what I expect Crydom has on their inside design.

Since the current limiter needs 1.5V for IR LED and + 2V for current limiter, This is why you see 3.5V on an open circuit input but there must be enough leakage to trigger the MOSFET.

Using the pull-up resistor causes the input voltage difference to be zero ( both at 5V. ) and this draws little or no current and turns off the IR LED..

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  • \$\begingroup\$ I see the 5.0 - 3.4 is 1.6 but the 430ohm resistor changes it to 5.0 - 3.7 is 1.3 Still above the 1 vdc must turn off. YET it works this way! I do not know how to test the max current with my fluke 179, so can we assume that is OK? \$\endgroup\$ – oldo Apr 23 at 20:34
  • \$\begingroup\$ Can you say how your driver looks? \$\endgroup\$ – Sunnyskyguy EE75 Apr 23 at 20:38
  • \$\begingroup\$ sorry, and thank you, but unfortunately I do not know enough about circuit diagrams for these from the datasheet for the crydom to help me understand this problem. \$\endgroup\$ – oldo Apr 23 at 20:42
  • \$\begingroup\$ Just put a 10k resistor across the inputs to turn off. I see you are using a normally open switch to ground. \$\endgroup\$ – Sunnyskyguy EE75 Apr 23 at 20:44
  • \$\begingroup\$ i realize its probably atypical to have a unequal voltage on each of the two control terminals, but this is what I have to use at this point, and can only make the resistor change to make the setup work. I just do not understand why this makes it work yet, and would like to understand this or see the math prove it out. \$\endgroup\$ – oldo Apr 23 at 20:45
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You could do something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

The transistor provides the voltage gain and offset you need to keep the SSR off when the control voltage is higher than about 2.7VDC. It also does not pull the driving output high (like above 3.3V) , which may be a latent issue for you, and retains the same logic (low = ON).

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  • \$\begingroup\$ Common base will not pull up the input and this will result in the same voltage as no transistor. Into the leaky current source with 1mA @ 3.4V that pulls down to 10mA \$\endgroup\$ – Sunnyskyguy EE75 Apr 24 at 2:20

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