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I have breadboarded a simple circuit using a 2N6395 thyristor to drive a resistive load. In simulation all works well, and even the physical implementation works sometimes.

schematic

simulate this circuit – Schematic created using CircuitLab

The emitter of the transistor should be approximately 800mV and the SCR gate should be approximately 700mV by design with an emitter current of 10mA. The transistor is thus in saturation and all is well. A SPICE simulation reveals all this to be as expected.

After physically building this circuit I found that nothing happened when triggered despite the expected voltages/currents being present with a nominal 10mA current draw. After tinkering with the circuit a bit I found that if I applied a larger pulse voltage to the gate of the SCR it would trigger and latch.

So, what I found seemed to require a component redesign, right? Actually, after the circuit initially conducts in the manner described above it will trigger thereafter by the transistor when an appropriate pulse is applied to the base. In summary:

  • Circuit is 'cold', active-low base pulse has no effect
  • Provide higher voltage/high current pulse to thyristor gate (load/thyristor heat up)
  • Remove power from circuit, reapply power - circuit is off (no load conduction) as expected
  • Apply active-low base pulse, thyristor triggers as expected

Trigger and resetting the circuit will continue to work as expected from this point forth until I allow the circuitry to cool down before trying again.

I thought it may have something to do with the given gate current not allowing sufficient charge to be stored quickly enough at the thyristor gate, so I tried leaving the circuit on for a sufficient amount of time for the charge to build; this had no effect.

Looking at the gate current/pulse duration curve given in the datasheet, 10mA should be sufficient. This is especially true considering the pulse is done physically and not controlled by logic.

enter image description here

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  • \$\begingroup\$ The continuous DC Gate trigger voltage maximum is listed at 2 VDC for that device. Can you see if it triggers reliably with a higher GK voltage? \$\endgroup\$ – R Drast Apr 15 '15 at 14:30
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    \$\begingroup\$ Try moving your 10R load to "above" the thyristor instead of "below" it. As you've got it connected, as soon as the thyristor begins to conduct you're reducing the GK voltage by whatever the drop is across the 10R. Could possibly end up with a bit of a race between the thyristor latching on and the gate pulse effectively being cut short. \$\endgroup\$ – brhans Apr 15 '15 at 14:32
  • \$\begingroup\$ The figures in that curve are "typical" not definitive. Look for a worst case guaranteed switching current in the datasheet : I'm guessing it's a bit higher. (datasheet now linked : looks like 30mA to me) \$\endgroup\$ – Brian Drummond Apr 15 '15 at 14:32
  • \$\begingroup\$ As @BrianDrummond says.. and taking temperature into account.. it looks to me like you need to give the gate about 5x as much current (50mA) if you want it to be guaranteed to work over temperature. \$\endgroup\$ – Spehro Pefhany Apr 15 '15 at 15:22
  • \$\begingroup\$ I believe if you add a 5k to 10k resistor on the collector of the transistor to negative (Ground) it will help to stabilize the circuit. You are depending totally on the current draw through the thyristor, as the temperature fluctuates the voltage require to trigger the thyristor may change. \$\endgroup\$ – user66377 Apr 16 '15 at 23:58
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The gate trigger pulse needs to be referred to ground (i.e. the cathode potential), not to the supply rail. You will need to redesign the gate driver accordingly.

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