3
\$\begingroup\$

Usually, power lines over large distances use alternating current (AC). However, in some cases, direct current (DC) is also used, as high voltage direct current (HVDC), where the high voltage is used in order to minimize power losses.

In order to convert AC from the power plant into DC, converter stations use thyristors in a diode bridge circuit. A diode bridge with regular diodes already performs full-wave rectification, converting AC into DC. On the other hand, a thyristor is like a transistor, but is bistable, meaning it does not require a constant secondary supply to stay on: when on, it acts as a diode.

Thyristor bridge converter

Why do converter stations use thyristors instead of plain diodes? In other words, why is it desirable to be able to switch diodes on and off, when plain diodes already perform AC to DC conversion?

\$\endgroup\$
7
  • \$\begingroup\$ Could you add an circuit diagram example. \$\endgroup\$
    – TonyM
    Nov 16, 2021 at 14:20
  • \$\begingroup\$ I've added the circuit diagram, but note that even though the fact that thyristors and not diodes are used, I still have no clue what the thyristor gates are connected to... \$\endgroup\$
    – David Cian
    Nov 16, 2021 at 14:34
  • \$\begingroup\$ Thinking out loud here.... what's the forward voltage drop of the thyristor, maybe lower conduction losses than a diode? What's the voltage rating needed, maybe easier to achieve with a thyristor? \$\endgroup\$ Nov 16, 2021 at 14:38
  • 1
    \$\begingroup\$ Anything with large power these days doesn't do plain rectification that you could achieve with a diode bridge rectifier. The reason is the extremely poor power factor of this method. For chopping at high frequency one can use GTO thyristors, their advantage being that they can block very high voltages (higher than IGBT I guess) but I think they are also falling out of fashion because WBG MOSFETs are also targeting this sector. \$\endgroup\$
    – tobalt
    Nov 16, 2021 at 14:51
  • \$\begingroup\$ Power Products International mentions that the power output needs to be be controlled to adjust to the fluctuation in demand. They say this is achieved using thyristors, which allow this control by being able to change their conduction angle. \$\endgroup\$
    – David Cian
    Nov 16, 2021 at 15:12

2 Answers 2

3
\$\begingroup\$

You can conduct for only part of the LATTER half of each half-cycle for voltage modulation.

Also note: Thyristors cannot interrupt current themselves. You can trigger them to conduct They latch on and you have to wait until the current through them drops to "zero" by some external means (usually a current zero cross) in the absence of a gate trigger signal for them to turn off.

So when entering a half-cycle, you first have to wait some amount of time with the thyristor not conducting until the remaining time will produce what you desired if it was conducting. Then trigger and latch the thyristor into conduction, allowing the zero cross to switch it off naturally.

Like a light dimmer.

\$\endgroup\$
8
  • 2
    \$\begingroup\$ Actually you can switch a thyristor off actively. It involves a second circuit that's specifically designed to momentarily reverse the current in the "main" thyristor. I can never hold in my head exactly how it's done -- but thyristors have been used in DC traction service for decades. Some web-spelunking should get you some answers. \$\endgroup\$
    – TimWescott
    Nov 16, 2021 at 16:28
  • 1
    \$\begingroup\$ @TimWescott That still seems to qualify as thyristors being unable to interrupt current themselves. But I will reword to thyristors can't interrupt current themselves. \$\endgroup\$
    – DKNguyen
    Nov 16, 2021 at 16:29
  • \$\begingroup\$ Thyristors can turn themselves off ....there are GTO (gate turn off) thyristors available where you must maintain gate current to be ON, remove the gate current and they turn off. electronicshub.org/gate-turn-off-thyristor ...so it all depends on the device type you select. \$\endgroup\$ Nov 16, 2021 at 17:00
  • \$\begingroup\$ @JackCreasey GTOT are what Tim describes IMO. They must be actively switched off (according to Wiki, never used one) \$\endgroup\$
    – tobalt
    Nov 16, 2021 at 17:47
  • \$\begingroup\$ @DKNguyen GTOs are not what Tim describes. GTOs con in various flavors, on some you can simply short the gate to cathode, on some you put a reverse current into the gate (depends then on what quadrant you are operating in), some you simply stop providing gate current. So in the end the answer becomes ..it depends on the device you select from the range of thyristors. For the OPs question of course the answer is that you use thyristors because they can be switched on demand. \$\endgroup\$ Nov 16, 2021 at 18:08
2
\$\begingroup\$

On the "receiving" side of the HVDC line you need DC-to-AC conversion.

Therefore, you have at point 1 a thyristor converter operating in rectifier mode AC-to-DC (thyristor control angle close to zero, behaving like a diode rectifier), and at point 2 you have an identical thyristor converter operating in inverter mode DC-to-AC (thyristor control angle close to 180°). Energy flows from point 1 to point 2.

You can invert the energy flow in the HVDC line any time by employing 180° for the thyristor control angle of converter 1 (now operating as DC-to-AC inverter), and by employing 0° for the thyristor control angle of converter 2 (now behaving like a diode rectifier).

The DC current flows always in the same direction. Therefore, the DC voltage of the whole line has to be inverted in case energy flows from 2 to 1. This voltage inversion puts a lot of stress on the cable insulation.

That's one of the reasons why there are also more recently HVDC lines operating with active devices (e.g. the "HVDC Light" product line from ABB, recently sold to Hitachi). They can do both DC current directions without inversion of the DC voltage. So you could think about future HVDC grids instead of connecting just two points. Other benefits of active devices are the ability to reduce harmonics and var compensation, but cost and losses are higher.

\$\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.