I am working on a thyristor (in my case: Semikron SKKT 253/08) circuit operating at 400 V mains (3 phase). What I'm wondering is how one can calculate a thyristor gate resistor so 12 VDC applied at pulse transformer can't damage the thyristor gate as in the following circuit:

Driving a thyristor using pulse transformer

In SKKT 253's datasheet, it states that minimum gate voltage is 3V and minimum gate current is 200 mA is required to turn on the thyristor. So I decided to drive a 1A pulse into its gate using factor of 5 according to Semikron manual page 152. Now back to my circuit:

    Ig = Ir2 = 1A (1)
    Vg + Vr2 + Vd2 = 12V -> Vg + (Ir2 x R2) + 0.7V = 12V (2)
    From equation (1) and (2), we have:
    R2 = (11.3 - Vg)/(Ir2) = (11.3 - Vg)/1
->  R2 = 11.3 - Vg (Ohm) (3)

Now I'm stuck at equation (3) because I can't decide which voltage Vg will be in case Ig = 1A. According to SKKT 253's gate trigger characteristic:

enter image description here

when Ig = 1A, Vg can be from 1V to 6V. Could anyone please enlighten me on this problem ?

  • 1
    \$\begingroup\$ There is no thyristor in you schematic and there is nothing labelled SKKT 253. \$\endgroup\$
    – Transistor
    Jan 1 '16 at 10:27
  • \$\begingroup\$ I dont like the slow mains 1N4007 for D2 .Use something better like a 40V 1 amp schottky .Then your circuit will perform .Try 10 ohms for your resister .It will get you in the ballpark.The source impedence of your driver is not stated so an accurate calc is not possible. \$\endgroup\$
    – Autistic
    Jan 1 '16 at 10:35
  • \$\begingroup\$ Please link the datasheet \$\endgroup\$
    – jippie
    Jan 1 '16 at 10:42
  • \$\begingroup\$ hi guys, the link for SKKT 253/08 is here: datasheet.directory/semikron/SKKT213-08E.pdf In my circuit, G1 and K1 is to be connected to gate and kathode of the thyristor. Are you sure the diode SS14 (40V-1A) good enough to replace 1N4007, it's connected to mains which has voltage as high as 400V (rms) ? For the IGBT driver, i use TLP152, which can source and sink maximum current of 2.5A. \$\endgroup\$ Jan 1 '16 at 11:05
  • \$\begingroup\$ Are you using the gate driver from your Previous Question ? \$\endgroup\$
    – Marla
    Jan 1 '16 at 16:20

Your pulse transformer is integral part of the design, so I will answer with the assumption that the pulse transformer is ideal. Obviously you don't want to saturate your pulse transformer with too long of a pulse width.

Using the circuit I have shown below : You don't have to use R2, but could be helpful when prototyping your first unit.

Secondary of transformer will give 12 volts out. Lets go for gate voltage of 4 volts. Voltage drop across R1 would be 8 volts, giving 0.8 amp gate current. (You won't hurt the SCR gate even with 12 volts, a very transient 12 volts.

Diode D1 is to prevent reverse voltage on SCR gate. Yes a 1N4007 works in this application. This circuit has been used on 300 KW systems operating from mains.

The gate cathode snubber circuit (which is another question/answer all of it's own) is to prevent the SCR from turning back on when voltage is reapplied to the SCR anode and cathode (high dv/dt).

You can test your driver circuit for functionality on an SCR without powering the SCR at anode cathode. For preliminary tests, the SCR gate behaves similarly whether the SCR is powered or not powered at anode and cathode. Now you can adjust R1 for your design needs.

enter image description here

  • 1
    \$\begingroup\$ Personally I like to put a couple of high-pass capacitors to facilitate a higher leading edge current pulse. This assists in the diffusion of the minority carriers into the gate region to ensure uniform turn-on (and thus mitigates hot-spots) \$\endgroup\$
    – JonRB
    Jan 1 '16 at 18:59
  • \$\begingroup\$ Actually, this is the first time I've seen a circuit like this with gate-cathode snubber and a diode placed like that to prevent reserve voltage on scr gate. For the gate-cathode snubber circuit: I thought to prevent false triggering caused by high dv/dt, one uses a rc snuber across a thyristor's cathode and anode. Do you also have another rc snuber across scr1's cathode and anode or using this one is enough ? For the diode: gate's voltage is always 0.7 volt lower than cathode's voltage, which means it's always in reserved voltage stage. Is it correct ? \$\endgroup\$ Jan 2 '16 at 13:08
  • \$\begingroup\$ An RC snubber across anode to cathode should still be used. In the induction heating industry, when SCR inverters were used, all manufacturers used RC across gate cathode (systems up to 5 MW were built) \$\endgroup\$
    – Marla
    Jan 2 '16 at 14:25
  • \$\begingroup\$ I can't say whether other manufacturers used the diode (my D1). Gate cathode leads could be quite long in our machines, which means added inductance, and I used the D1 because it was cheap insurance. \$\endgroup\$
    – Marla
    Jan 2 '16 at 14:29
  • \$\begingroup\$ Hi Marla, thank you for your expert comment on this problem. After some further research, I now understand that a capacitor placed across gate and cathode will improve a SCR's dv/dt withstanding capacity (especially with a sensitive gate SCR). For diode D1, i think it will clamp gate-cathode voltage to never goes below -0.7 V to prevent sudden turn-off when a gate pulse is too short (approximately 1 us). I will report if my circuit work. Thank you again. \$\endgroup\$ Jan 3 '16 at 16:08

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