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I have simulated the following circuit and am curious to the position of the load resistor. The circuit switches an ac signal using a pulse. It seems to be common to have the load in the position or R2 and the signal is as it should be. The blue plot represents thecurrent through R2. However the current through R1 is a pulse (same as firing pulse) and does not represent the ac signal at all. Why is this?When the triac is open the current should be the same through both.

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Edit 1: Added plot of voltage at V2:

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Edit 2:

Gate resistor has been added. Device switches as expected however there is a large voltage drop. Vout is now 200V instead of 400V. There is a lot of current flowing back out through the gate through R3.

If I change any of the component values the simulation freezes so I cannot even try anything else. Why is this happening?

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Thanks.

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  • \$\begingroup\$ Mind plotting the waveform of V2, too? \$\endgroup\$ – Jonathan S. Mar 7 '18 at 16:25
  • \$\begingroup\$ I have added V2. R1 basically follows what appears on the gate. \$\endgroup\$ – MXG123 Mar 8 '18 at 8:11
  • \$\begingroup\$ What does your SCR model look like? Can you probe across the TRIAC only, I bet you willvfind that the Ron of the TRIAC is about 5 ohms. Also your trigger voltage should go between the gate and the main terminal right above the resistor. \$\endgroup\$ – DIODEX Mar 9 '18 at 15:25
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The TRIAC has a diode junction between its gate and lower anode, so current flows from voltage source V2 through the triac down into R1. You could add a resistor in series with the gate to limit the gate current.

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  • \$\begingroup\$ I have added a resistor at the gate (shown above) and it works to a certain extent. There is a large voltage drop at the output. To make things worse when I change any parameters the simulation freezes. Do you know why this is happening? Thanks again. \$\endgroup\$ – MXG123 Mar 9 '18 at 12:20
  • \$\begingroup\$ @MXG123 The load has to stay in the position of R2. You're trying to trigger the TRIAC "across" the load (which is R1 now), effectively adding the load voltage to the trigger voltage, and that won't work. Increase R2 to 1k again. \$\endgroup\$ – Jonathan S. Mar 9 '18 at 12:35
  • \$\begingroup\$ @ Jonathan S. As you said it works fine when I put the load across R2. I did see in some circuit diagrams that had the load in the other position. So you're saying this is definitely a no no?? \$\endgroup\$ – MXG123 Mar 9 '18 at 13:40
  • \$\begingroup\$ @MXG123 Can you show the circuits where that's done? I'm sure the control circuitry and TRIAC still have a common potential somewhere. In general, it really is a "no no" - just think about what's going to happen if your load is capacitive or inductive. (The TRIAC might randomly trigger, fail to trigger, or just turn off in the middle of a cycle) \$\endgroup\$ – Jonathan S. Mar 9 '18 at 20:02
  • \$\begingroup\$ @ Jonathan S The circuit here has the load in that position. I thought a small gate current would have negligible affect on the load but it clearly does. Thanks for clarifying that. \$\endgroup\$ – MXG123 Mar 12 '18 at 12:28

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