Which full-wave rectifier scheme should I prefer for obtaining symmetric DC? - Page 1
Forum section: electronics / projects

author: hkBattousai
author_rank: Regular Contributor
timestamp: July 27, 2014, 02:18:20 am
content: 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.

author: IanB
author_rank: Super Contributor
timestamp: July 27, 2014, 02:24:31 am
content: Why do you think the first one is a full wave rectifier?

author: hkBattousai
author_rank: Regular Contributor
timestamp: July 27, 2014, 02:27:16 am
content: Quote from: IanB on July 27, 2014, 02:24:31 amHmm. I wasn't sure either. Now that you pointed it out, it looks like half-wave rectifier to me. Correct?

author: IanB
author_rank: Super Contributor
timestamp: July 27, 2014, 02:33:48 am
content: Quote from: hkBattousai on July 27, 2014, 02:27:16 amWell consider the half cycle where the . end of the secondary is -ve and the other end is +ve. Will any current flow?

author: tautech
author_rank: Super Contributor
timestamp: July 27, 2014, 02:37:46 am
content: The diode symbols you have used represent Schottky diodes.Is this what you intended?

author: hkBattousai
author_rank: Regular Contributor
timestamp: July 27, 2014, 02:42:42 am
content: Quote from: IanB on July 27, 2014, 02:33:48 amI understand. So, I must use the second circuit (the one with four diodes).Thank you.Quote from: tautech on July 27, 2014, 02:37:46 amYes, 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?

author: IanB
author_rank: Super Contributor
timestamp: July 27, 2014, 03:03:56 am
content: Quote from: hkBattousai on July 27, 2014, 02:42:42 amWell 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?

author: hkBattousai
author_rank: Regular Contributor
timestamp: July 27, 2014, 03:06:52 am
content: Quote from: IanB on July 27, 2014, 03:03:56 amIf the dots weren't like in that picture, the diodes wouldn't conduct, right? I understood it. Thank you IanB.

author: onlooker
author_rank: Frequent Contributor
timestamp: July 27, 2014, 03:10:32 am
content: 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.

author: hkBattousai
author_rank: Regular Contributor
timestamp: July 27, 2014, 03:12:17 am
content: Quote from: onlooker on July 27, 2014, 03:10:32 amI got it. Thank you very much.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 27, 2014, 04:04:07 am
content: What's powering the transformer?

author: hkBattousai
author_rank: Regular Contributor
timestamp: July 27, 2014, 04:10:06 am
content: Quote from: T3sl4co1l on July 27, 2014, 04:04:07 amAn H-bridge.Actually, the transformer's main purpose is something else. I will add additional secondary windings to obtain these symmetrical voltages.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 27, 2014, 07:18:21 am
content: Constant voltage H bridge?Any regulation? Current limiting? Soft start?Tim

author: hkBattousai
author_rank: Regular Contributor
timestamp: July 27, 2014, 12:29:22 pm
content: 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.

author: David Hess
author_rank: Super Contributor
timestamp: July 27, 2014, 12:48:57 pm
content: 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.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 27, 2014, 04:43:51 pm
content: 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

author: hkBattousai
author_rank: Regular Contributor
timestamp: July 27, 2014, 04:55:57 pm
content: Quote from: T3sl4co1l on July 27, 2014, 04:43:51 pmThe 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?

author: SeanB
author_rank: Super Contributor
timestamp: July 27, 2014, 05:22:03 pm
content: 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.

author: IanB
author_rank: Super Contributor
timestamp: July 27, 2014, 06:30:39 pm
content: 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?

author: hkBattousai
author_rank: Regular Contributor
timestamp: July 27, 2014, 06:44:35 pm
content: Quote from: IanB on July 27, 2014, 06:30:39 pmThe transformer is driven by PWM. The frequency of the PWM is 50kHz. The PWM timed so that the output will form a 50Hz sinusoidal.

author: IanB
author_rank: Super Contributor
timestamp: July 27, 2014, 07:31:45 pm
content: 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.

author: Mark Hennessy
author_rank: Frequent Contributor
timestamp: July 27, 2014, 07:38:11 pm
content: 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.

author: dannyf
author_rank: Super Contributor
timestamp: July 27, 2014, 07:42:13 pm
content: QuoteIf 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.

author: hkBattousai
author_rank: Regular Contributor
timestamp: July 27, 2014, 08:16:11 pm
content: Quote from: Mark Hennessy on July 27, 2014, 07:38:11 pmSignal timing diagrams for the PWM are as shown inthis post. The circuit structure has greatly changed since I started this project though, the PWM signal format is still the same.Quote from: Mark Hennessy on July 27, 2014, 07:38:11 pmI'm changing the output polarity to achieve the negative part withthis method.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 27, 2014, 08:29:07 pm
content: 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.Tim

author: dannyf
author_rank: Super Contributor
timestamp: July 27, 2014, 08:42:27 pm
content: QuoteIs it worth noting that the two sequences are the same?"1/4 up, 1/2 down, 1/4 up" + "1/4 up, 1/2 down, 1/4 up" + "1/4 up, 1/2 down, 1/4 up" + ...= "1/4 up, " + "1/2 down, 1/4 up + 1/4 up" + "1/2 down, 1/4 up + 1/4up" + "1/2 down, 1/4 up + 1/4up" + ...= "1/4 up, " + "1/2 down, 1/2 up" + "1/2 down, 1/2 up" + "1/2 down, 1/2 up" + ...Also, for what the OP is trying to do, a half bridge would work.

author: IanB
author_rank: Super Contributor
timestamp: July 27, 2014, 08:50:20 pm
content: Quote from: dannyf on July 27, 2014, 08:42:27 pmNot the same. Look at the sequence of values.In the first case the values are: 1/4, -1/4, 1/4, -1/4, ...In the second case the values are 1/2, 0, 1/2, 0, 1/2, ...In the first case the average (DC) current is 0.In the second case the average current is 1/4.A non-zero DC average current through a transformer core will magnetize it towards saturation.

author: dannyf
author_rank: Super Contributor
timestamp: July 27, 2014, 08:54:38 pm
content: QuoteIt depends on what kind of load it is powering.

author: David Hess
author_rank: Super Contributor
timestamp: July 27, 2014, 11:59:57 pm
content: Quote from: hkBattousai on July 27, 2014, 08:16:11 pmPage 11 of Linear Technology application note 18 shows this kind of output stage used to make an amplifier with a bipolar +/- 100 volt output. A pair of SCRs are used after the 4 diode bridge rectifier so only 2 diodes are active at a time making it effectively a pair of switched full wave center-tap rectifiers.

author: hkBattousai
author_rank: Regular Contributor
timestamp: July 28, 2014, 12:32:19 am
content: Quote from: hkBattousai on July 27, 2014, 08:16:11 pmSorry, I wrongly stated this detail. The signals are not the same. At the moment, the PWM signal is continuously generated by TL494 (TL494 is never turned off or disabled unlike in the block diagram I previously gave a link of), and the transformer is driven with that by using an H-bridge. The change in PWM duty cycle (50Hz) is much slower then the change in PWM frequency (50kHz), so the transformer is driven equal positive and negative pulses, so there is no DC flux inside the transformer core. The transformer output is rectified by a bridge rectifier, then filtered by an inductor and alternating polarities are formed by the polarity changer circuit. That's how I get the sine wave.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 28, 2014, 01:31:04 am
content: Alright...If that's how it is, then it's a proper forward converter, and as long as the PWM doesn't ramp up too quickly, you don't have to worry about transient imbalance. You do still have to worry about systematic / DC offset (pulse width imbalance), which can be helped with a nice beefy coupling capacitor (half bridge is usually done by splitting the bypass caps, which also serves as coupling, so you don't need much extra).Tim