Six-pulse full-bridge rectifier: firing angle vs output voltage

[omglol]

Hello!

I need a little (theoretical) help regarding six-pulse full-bridge rectifier. I would prefer if someone can also suggest me some literature/good book about the problem I am to describe:


Here we have standard 3-phase rectifier with 6 diodes:




Now we all know in this configuration, each diode conducts for 120º of complete 360º cycle.


I have a similar circuit, but with transistors instead of diodes, so I can control firing angle. I can't lower conduction degree below 120º, because the current is flowing through the bypass diode of the transistor, but I can increase it. For example, I can increase conduction degree of each transistor from 120º to 130º. This, of course, causes overlapping. But, I would like to know (and understand) how this effects output voltage.  Thank you very much in advance!

[SeanB]

Normally in these situations the standard ( aside from the horrid PFC and RFI it causes without a decent filter on the input side) is to have a half bridge, with regular diodes on the negative lead, and have 3 thyristors on the positive side, with the gate anode being driven by a pulse transformer, so that you can control the point where they start to conduct on each mains cycle. This allows you to drop the voltage on the output capacitor, or at least control the peak voltage it will be charged to, by triggering the thyristors later on in the mains cycle, so as to reduce the conduction angle and thus the current into the capacitor/load side.

Using transistors and making them conduct early means that you are basically feeding one phase back to another, which will lead to problems of high circulating currents in the transistors, and will not really give you any voltage control as the BE junctions of the transistors break down in reverse.

[oldway]

First, you should investigate better what is the rms value of rectified value.....0,707 x peak value is the rms value of 1 phase bridge rectifier, not of 3 phases bridge rectifier....

[technogeeky]

First, you should investigate better what is the rms value of rectified value.....0,707 x peak value is the rms value of 1 phase bridge rectifier, not of 3 phases bridge rectifier....

Is it just sqrt(n) where N is the number of legs of the phases? (e.g. 2 for 1-phase, 3 for 2-phase)?

[Zero999]

Ignoring diode losses, the RMS output voltage on a single phase rectifier, is the same as the RMS input voltage. This isn't the case with a three phase rectifier. Do the maths.

How about simulating it in SPICE?

Here's a plot of the DC voltage vs the phase to neutral voltage. Try plotting the phase-to-phase voltage vs the DC voltage. LTSpice is freely downloadable and works on Linux via WINE.

[omglol]

Well I have built the hardware that does exactly what I described ... I can set conduction angle via microprocessor. However, I do not know where are the upper limits and what is the theoretical background behind this. Can somebody point me in the right direction where to start? I need maths behind it. Thanks.

[sibeen]

The ideal no load direct voltage (non-filtered) is (3/pi) x Vp, where Vp = peak voltage.

As an example: The supplying voltage is 415 volts line to line.

V_p = sqrt(2) x 415 = 587.

V_dio = 3/pi x 587 = 560.4.

So a simple way to calculate is V_dio = 1.35 x V_L-L (voltage line to line).

Now, if you examine the waveform closely you can see that each diode begins to conduct when its associated sine wave reaches 60 degrees. This is called the natural delay angle and is normally designated alpha'. If we place a thyristor in the circuit instead of a diode we can then delay the firing and conduction of the device to sometime after the alpha' point which we then call the alpha delay.

The equation then becomes:

V_dio = 1.35 x V_L-L  x cos(alpha). When alpha = alpha' then cos(0) = 1 and we have the original equation.

When alpha = 90 degrees then cos(90) = 0 and the output is 0 volts DC. So you can adjust alpha between 0 and 90 degrees to adjust the output DC voltage.

[omglol]

Now, if you examine the waveform closely you can see that each diode begins to conduct when its associated sine wave reaches 60 degrees. This is called the natural delay angle and is normally designated alpha'. If we place a thyristor in the circuit instead of a diode we can then delay the firing and conduction of the device to sometime after the alpha' point which we then call the alpha delay.

The equation then becomes:

V_dio = 1.35 x V_L-L  x cos(alpha). When alpha = alpha' then cos(0) = 1 and we have the original equation.

When alpha = 90 degrees then cos(90) = 0 and the output is 0 volts DC. So you can adjust alpha between 0 and 90 degrees to adjust the output DC voltage.

Thank you, I understand that. However, I have a different case. I am interested in what happens if I turn ON the thyristor (or MOSFET in my case) BEFORE the natural delay angle. Can you help me understand it? Thanks.

[sibeen]

I am interested in what happens if I turn ON the thyristor (or MOSFET in my case) BEFORE the natural delay angle. Can you help me understand it? Thanks.

Nothing. Unless you turn off the device that is originally conducting, something that you cannot do with a thyristor.  Take the case of the mosfet where you have one already on. Without turning this one off you then apply a gate voltage to the next in line before the alpha'. As the original has a larger voltage across it there will be no current flow in the one just gated until you reach alpha'.

[omglol]

Well maybe that is the case with pure resistive load/source, but because of inductances in both, this is not true. At least not according to my measurements. When I increase angle of conduction beyond 120º, output voltage rises. Until, at some point, the system goes into short-circuit. This point is dependant on load type. I would like to know how this works in theory.

[Wolfram]

If the top devices have a conduction angle wider than 120 degrees, then by definition two devices will be on at the same time for a part of the cycle. If the devices conduct in both directions when on, then the devices will short across two phases of the input for the part of the cycle when more than one device is on. This will act like a a boost converter together with the source inductance, hence the higher voltage.

[oldway]

http://minitorn.tlu.ee/~jaagup/kool/java/kursused/15/robootika/elektriopik.pdf

See page 205 : Three phase controlled rectifiers

[omglol]

http://minitorn.tlu.ee/~jaagup/kool/java/kursused/15/robootika/elektriopik.pdf

See page 205 : Three phase controlled rectifiers

Thank you very very much! I will try to study this and come back if I will still have some questions.

[omglol]

Ignoring diode losses, the RMS output voltage on a single phase rectifier, is the same as the RMS input voltage. This isn't the case with a three phase rectifier. Do the maths.

How about simulating it in SPICE?

Here's a plot of the DC voltage vs the phase to neutral voltage. Try plotting the phase-to-phase voltage vs the DC voltage. LTSpice is freely downloadable and works on Linux via WINE.

Is it possible to create a simulation with MOSFETs instead of diodes and then adjust firing angle from 0 to 135 degrees? I would like to see what happens. Thanks.

[Dave]

Of course it's possible, you just need to add some impedance to the source to make it more realistic, because you will be essentially shorting out phases like that (conduction angle > 120°).

[omglol]

Thanks!

I am trying to build a simulation. I have replaced diodes with MOSFETs and added source inductance (real values, measured on a real alternator). I also added load, with which I would like to simulate output current around 10A at 14-15V output DC voltage.

How can I add logic for MOSFET triggering?