0
\$\begingroup\$

I'm having some trouble to understand the thyristor voltage waveform for the full wave bridge controlled rectifier. I plotted the Voltage at the secondary, the current at the load, the voltage at the load and the voltage at the thyristor T1.

Full Wave Bridge Rectifier with RL Load

enter image description here

When the current at the load drops to zero, there's no voltage drop at the load. So why the voltage at the thyristor 1 (T1) is half the source voltage before the thyristor T2 and T3 are fired (red circles)?

\$\endgroup\$

2 Answers 2

1
\$\begingroup\$

Thyristors as an assembly often have snubber's across anode and cathode to protect them from dv/dt, as in voltage spokes due to inductive loads. The snubber is normally a capacitor less than 0.5 uF in value in series with a relatively low ohm 1 to 10 watt resistor. Capacitor is often rated 3 times line voltage, or more.

This causes a substantial leakage current to flow to the output stage at ALL times AC power is ON, perhaps 1 percent of total output current. Thus you may need a substantial load to bring the output voltage down close to zero, but it will never zero as long as power is ON. This will show up as voltage on the output, possibly enough to be a serious shock hazard or seem confusing to test equipment.

Proper testng means having a substantial load on the outputs, at least 50% of maximum load, preferably a resistive load to avoid back EMF. Leakage current in the snubbers and a light load can make "modes" seem irrational in their behavior.

This means that thyristor-based power controllers do leak enough current to be a shock hazard, especially if output has no load. In which case it appears to not shut off by using pulse-width control, as the snubbers can leak milliamps of current as long as AC power is ON. Best to work on these devices with AC power disconnected, and a minimum load to discharge snubber capacitors.

If the load is inductive then back EMF can be suppressed with series resistors of 5 ohms to 15 ohms WW at 300 watts, but wattage should be at least what the power controller can put out.

\$\endgroup\$
0
\$\begingroup\$

When thyristors are off (all four of them), voltage of the source comes across two thyristors in series (T1, T2) in parallel with two other thyristors also in series (T3, T4). Putting the reference (0 V) in the lower output of the transformer, anode of T1 will have maximum positive voltage, cathode of T2 will have 0 V. Both divide total voltage of source. Ideally, 50% share of voltage for both of them, but that will depend on many other factors. This accounts for the second circle (red, in the right) of the thyristor voltage (zero current, all four thyristors off). When two thyristors are on, the voltage applied on the load is also applied (reversely) on both thyristors that are off, which hold full negative voltage. Then, when all thyristors go off again (zero current on load), the shared voltage between two series thyristors appear on them, wich accounts for the first red circle in the graph.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.