Author Topic: Which full-wave rectifier scheme should I prefer for obtaining symmetric DC?  (Read 8496 times)

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Offline hkBattousaiTopic starter

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Which of these two full-wave rectifiers are best for obtaining symmetric DC voltages? Is there a better third configuration for this purpose (I don't prefer using two bridge rectifiers for each side, because more diodes will be used and there will be more voltage lost on the diodes)?

Total harmonic distortion must be minimal at the AC grid side (especially when positive source current is not equal to negative sink current). And less importantly, minimum number of diodes should be used.

 

Online IanB

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Why do you think the first one is a full wave rectifier?
 

Offline hkBattousaiTopic starter

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Why do you think the first one is a full wave rectifier?

Hmm. I wasn't sure either. Now that you pointed it out, it looks like half-wave rectifier to me. Correct?
 

Online IanB

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Why do you think the first one is a full wave rectifier?

Hmm. I wasn't sure either. Now that you pointed it out, it looks like half-wave rectifier to me. Correct?

Well consider the half cycle where the . end of the secondary is -ve and the other end is +ve. Will any current flow?
 

Offline tautech

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The diode symbols you have used represent Schottky diodes.
Is this what you intended?
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Offline hkBattousaiTopic starter

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Why do you think the first one is a full wave rectifier?

Hmm. I wasn't sure either. Now that you pointed it out, it looks like half-wave rectifier to me. Correct?

Well consider the half cycle where the . end of the secondary is -ve and the other end is +ve. Will any current flow?

I understand. So, I must use the second circuit (the one with four diodes).
Thank you.

The diode symbols you have used represent Schottky diodes.
Is this what you intended?

Yes, the AC voltage will be at 50kHz frequency. That's why I prefer Schottky diodes.

A second question:

If there are two isolated secondary windings, is the dot-convention important when I build up my circuit as in the schematics below? If important, how must it be, and what happens if I do the connection wrongly?

 

Online IanB

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If there are two isolated secondary windings, is the dot-convention important when I build up my circuit as in the schematics below? If important, how must it be, and what happens if I do the connection wrongly?

Well think of what happens if you put two batteries in series. If you put them both pointing the same way the voltages will add up. If you reverse one of the batteries, what will happen then?
 

Offline hkBattousaiTopic starter

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If there are two isolated secondary windings, is the dot-convention important when I build up my circuit as in the schematics below? If important, how must it be, and what happens if I do the connection wrongly?

Well think of what happens if you put two batteries in series. If you put them both pointing the same way the voltages will add up. If you reverse one of the batteries, what will happen then?

If the dots weren't like in that picture, the diodes wouldn't conduct, right? I understood it. Thank you IanB.
 

Offline onlooker

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The winding polarity (the dot) is important. If you do it the other way, your secondary is "equivalent" to a simple single winding and   D7 can be combined to D8 as a single diode (and the same for D9 and D10). And you now end up to having 2  half-wave rectifications, one for each of your 2 rails.
« Last Edit: July 27, 2014, 03:35:15 am by onlooker »
 

Offline hkBattousaiTopic starter

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The winding polarity (the dot) is important. If you do it the other way, your secondary is "equivalent" to a simple single winding and   D7 can be combined to D8 as a single diode (and the same for D9 and D10). And you now end up to have 2  half-wave rectifications, one for each of your 2 rails.

I got it. Thank you very much.
 

Offline T3sl4co1l

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What's powering the transformer?
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Offline hkBattousaiTopic starter

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What's powering the transformer?

An H-bridge.

Actually, the transformer's main purpose is something else. I will add additional secondary windings to obtain these symmetrical voltages.
 

Offline T3sl4co1l

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Constant voltage H bridge?

Any regulation? Current limiting?  Soft start?

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Offline hkBattousaiTopic starter

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Supply voltage of the H-bridge is constant. The H-bridge is driven by PWM pulses. The PWM pulses are timed to form 50Hz sinusoidal. Yes, there is a current limiting control and soft start.

Even if the H-bridge is turned-off, there will be a secondary supply (a linear regulator from a battery) to obtain these symmetric voltages.
 

Offline David Hess

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The only reason to use the half wave rectifier is economy when performance may be compromised.  The full wave rectifier will provide better regulation and lower ripple.  The only time I have seen the half wave rectifier used is when the output current is low.
 

Offline T3sl4co1l

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Oh, so this is an iron core transformer?

You'll still have to worry about charging spikes into those capacitors.  If they're low level, you could simply put resistors in series with the diodes and call it good enough, I guess.

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Offline hkBattousaiTopic starter

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Oh, so this is an iron core transformer?

You'll still have to worry about charging spikes into those capacitors.  If they're low level, you could simply put resistors in series with the diodes and call it good enough, I guess.

Tim

The transformer is ferrite core.

Do I really need to put resistors? The positive source current will be 50mA and negative sink current will be 5mA at most. The voltage levels will be +/-18V. Would a 10uF capacitor require charging a resistor? I chose the capacitor type to be electrolytic on purpose, so that its internal ESR would limit the current a little. Wouldn't it suffice?
 

Offline SeanB

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Not for 10uF and 50mA. The ESR will be enough. You could use a second LC section if you want to remove more of the switching noise from the output.
 

Online IanB

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At first you said 50 kHz. Then you said 50 Hz. But 50 Hz would need an iron core transformer. So that was a mistype and it is really 50 kHz?
 

Offline hkBattousaiTopic starter

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At first you said 50 kHz. Then you said 50 Hz. But 50 Hz would need an iron core transformer. So that was a mistype and it is really 50 kHz?

The transformer is driven by PWM. The frequency of the PWM is 50kHz. The PWM timed so that the output will form a 50Hz sinusoidal.
 

Online IanB

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So your transformer is going to filter out the 50 kHz component and pass through the 50 Hz component. It seems to me (a non-EE) that in that case the transformer would need to be designed to operate at 50 Hz, and that would usually indicate an iron laminated core.
 

Offline Mark Hennessy

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No, I think this is OK. He's varying the duty cycle of the 50kHz signal to change the output voltage - that works with any switched-mode power supply. But in this case, it just so happens that the output voltage is being moved up and down 50 times a second.

I'm guess he's trying to build an inverter that outputs a nice sine wave. Though I can't immediately see how the negative going part of the 50Hz cycle will be achieved... Depending on the voltage and current levels required, it might be worth investigating class D amplifiers for this sort of thing.
« Last Edit: July 27, 2014, 07:39:54 pm by Mark Hennessy »
 

Offline dannyf

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Quote
Which of these two full-wave rectifiers are best for obtaining symmetric DC voltages?

If I were you, I would provide a full background on what you are trying to do with this setup. Based on what you have talked about so far, I would think the answer to your question ranges from "all" to "neither" or "doesn't matter".

It is usually far better for you that more information is provided so others can help you.
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Offline hkBattousaiTopic starter

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No, I think this is OK. He's varying the duty cycle of the 50kHz signal to change the output voltage - that works with any switched-mode power supply. But in this case, it just so happens that the output voltage is being moved up and down 50 times a second.
Signal timing diagrams for the PWM are as shown in this post. The circuit structure has greatly changed since I started this project though, the PWM signal format is still the same.

I'm guess he's trying to build an inverter that outputs a nice sine wave. Though I can't immediately see how the negative going part of the 50Hz cycle will be achieved... Depending on the voltage and current levels required, it might be worth investigating class D amplifiers for this sort of thing.
I'm changing the output polarity to achieve the negative part with this method.
 

Offline T3sl4co1l

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Hmm, interesting.  But note a C-input filter will charge instantly to the full peak voltage.  You need an L-input filter, typical of a forward converter, at each rectifier.  Followed by the polarity switching transistors.

The same should be true of any other load on the transformer. If you need auxiliary supplies, you're probably better off just adding another converter.  Bit of a pain, but it won't have 100Hz ripple on it either, and you can control it differently from the main output (e.g., it stays on when the output is disabled / faulted).

I'd suggest a sigma-delta style control, where you either pulse, or skip a pulse, according to whether the output voltage is above or below the ideal value.  This won't necessarily reduce the ripple, but most of it will be above the fixed 1kHz in your current approach.

Worth noting that, a pulse should be defined as 1/4 up, 1/2 down, 1/4 up.  Not 1/2 up, 1/2 down.  That way, the flux doesn't stay to one side, it remains balanced.  Also a bit of a pain for the added logic -- though if you've actually got an FPGA handy, it's not a big deal.

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