Or, conversely, if, 40 V and 20 V voltages are applied at the anode and cathode terminals respectively?
Semiconductor material is having one to two electron volt energy band for electron to come into conduction band to form the conduction, in material. Which needs the potential difference, between ends, to energized, not the polarity respect to ground zero. Hence thiristor will worked normaly, as the needed potential difference, according to manual are achieved( anode is at 20 volt higher than cathode raferance.).
When a thyristor turns on the voltage across it become zero. But when you apply negative voltage across it, it goes into reverse biasing mode. The outer junctions J1 and J3 become reverse biased mode.... current falls abrupty and device gets turned off. you can verify this using simulation.
It's not simple because thyristors respond to both voltage and rate of change of voltage. The "on" voltage drop is not zero. Each junction drops a voltage roughly proportional to the log of the current - typically 1/2 to 3/4 volt. Since I don't know what you plan to do with the device, it's ratings, etc, best read the manual. Radio Shack used to carry some decent simplified books on the devices.
If you apply a voltage source of -20V at the anode of a SCR with respect to the circuit ground, and also apply a voltage source of -40V at its cathode, then heavy current will be drawn from the sources through the SCR since the SCR voltage drop will be low in the present ON condition, somewhere around 1.5V and hence cannot support a 20V difference without drawing heavy current (theoretically till its forward drop builds up to 20V or till the applied 20V voltage drops down due to its finite source impedance !) Undoubtedly the SCR will remain ON as it is still forward biased, irrespective of its voltages with respect to circuit ground, since the SCR appears to be floating with respect to the circuit ground potential.
Dear Sirs,
It is said that, for a thyristor to conduct (after it is triggered), the anode must be at a higher potential, or it must be more positive, than the cathode.
Relating to the above fact, will not the thyristor work normally? or is there something else to be considered?
Dear Kunal, that statement does not give the total facts. The anode must be positive biased with respect to the cathode in order to conduct. However, one cannot dictate the voltage drop in the device from outside: it depends on the device characteristics. One can apply 40V to a circuit comprising of a SCR and a resistor in series and if forward biased and triggered, the SCR will conduct current, dropping majority of the voltage across the resistor and only a small portion across itself. Otherwise, the efficiency of such circuits would be very poor if the SCR itself was to have a high voltage drop and consequent high power dissipation ! Thus you cannot apply 40V across the SCR in the forward biased mode and expect the SCR to sustain the 40V drop. The same thing is true for a diode rectifier. If the applied input is say 40V ac, dont you assume that the diode drops are negligible and you get the rectified ac waveform at the output dc terminals ? Now if I short the dc terminals, will the applied 40V ac be dropped across the diodes ? Never ! It will draw huge current from the input. Your SCR problem is similar.
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