How to determine Thyristor gate resistance?

Thread Starter

zazas321

Joined Nov 29, 2015
926
Hello guys. I am currently studying about ways to drive DC motors. I came across Thyristor and have been looking in to them for a while. I am a little confused about the resistor choises to limit the current to the gate.
https://ibb.co/cd2yuc
Lets analyse this circuit. How do I know what value of Rg and Rgk to pick. Lets say i need 10 mA of current to flow into the gate of thyristor and the supply is 9v. I can determine what current would flow through Rg but I cannot determine what current will flow to Rgk and gate. The current that was flowing to Rg will be split to Rgk current and Gate current but as we do not know internal resistance of Gate we cannot calculate the current that will be flowing to the gate. I might have confused everything but thats what I am confused about..
 
Last edited:

AlbertHall

Joined Jun 4, 2014
12,158
The gate-cathode of a thyristor behaves much like a diode. From the datasheet you should be able to get a reasonable value for the gate voltage when the gate current is 10mA, then you can solve the problem.
 

ebp

Joined Feb 8, 2018
2,332
A gate-cathode resistor may exist internal to the thyristor. Using one externally is typically done to reduce the sensitivity and help prevent problems with false triggering due to high die temperature or rate of rise of applied voltage. For many applications no external gate-cathode resistor is required.

In reality it is common to drive the gate with a current pulse much greater than the minimum trigger current, especially if the external circuit would allow a very high rate of rise of current, for example a capacitive load. Allowable rate of rise of on-state current is usually specified in the datasheet.

If only the minimal trigger current is applied, the initial turn-on of the thyristor happens in only a small area of the die, resulting in hot-spotting that can be destructive. Some devices have an "amplifying gate" so this issue is largely taken care of internally, but most don't. By using a pulse of several times the minimum gate current, initial turn-on over a large area is improved. The gate signal should not only be quite high in amplitude but have fast rise time. Care is required not to exceed the specifications for peak gate current and average gate power. A device with minimum gate trigger current of 50 mA might have a peak gate current limit of 2 amperes or more and a limit of half a watt of average gate power.

Sometimes inductance can be used to slow the rate of current rise in anode-cathode circuit, but then the energy stored in the inductor must be considered to prevent problems at turn-off.
 

Thread Starter

zazas321

Joined Nov 29, 2015
926
Okay thank you for both replies! Lets have a look at the datasheet then. http://www.dccomponents.com/products/Rectifiers/SCRs/2N6395-6398.pdf
1.So lets say that the forward trigger current is 20mA and the forward tirgger voltage is 2V. With the supply of 9v Rg=470 and Rgk=1k. so that creates a potential divider circuit and the voltage across Gate and Cathote will be V=9*1000/1470=6.1V(potential divider). and the rest of the voltage which is 5.9V will drop across Rg. Am I right here?

2. Though I am still unsure how to achieve 20mA flow into the Gate even though we know the voltage across Gate and Cathode because we need to know the resistance of the Gate.. Lets say if I put a gate resistor of 100Ohms and calculate current from there . I=6.1/100=60mA (without taking into account internal Gate resistance) . Would that be right?

3. As can be seen above, Voltage across Gate and Cathode was calculated to be 6.1V and in our case we need trigger voltage to be 2V so that would be too high wouldn it? We would need to decrease Rgk to reduce it.
Okay now I have a few more questions.
 

AlbertHall

Joined Jun 4, 2014
12,158
If the gate voltage is 2V then you know the voltages across the two resistors and therefore the current through each resistor. The difference between those currents goes into the gate (Kirchoff's first law).
 

Thread Starter

zazas321

Joined Nov 29, 2015
926
If the gate voltage is 2V then you know the voltages across the two resistors and therefore the current through each resistor. The difference between those currents goes into the gate (Kirchoff's first law).
Oh I get it. So what you are saying that if the Gate / cathode voltage is 2V that means that Voltage across Rgk would be 2V aswell because they are in parrallel and the remaining 7V will drop across Rg. Is that it?
 
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