Bridge Rectifiers

[Roger HeadParticipant @rogerhead16992]

Posted by John Stevenson on 17/03/2016 10:52:56:

Dress circle in Ilkeston ???

Are you real man ??

Ilkeston ??? …..hang on …… Google Maps ……. jeez, that's 12000 miles away. Do they have moving-picture houses over there too?

[Howard LewisParticipant @howardlewis46836]

Second for comments about voltage regulation. A supposedly 12V power supply for a OO gauge train was 18V open circuit, did no good to the electronic controller connected to it!

Is there a high resistance (poor/corroded) connection any where in the circuit? The current draw will cause a voltage drop through that, although it ought to start feeling warm, if that is the case.

Just a thought

Howard

[AjohnwParticipant @ajohnw51620]

It's possible to buy battery replacement power supplies Howard. Amateur radio people use them. The regulation is pretty good on those. These days though there is a tendency to use switch mode supplies.

If I want a transformer for something where it matters I buy from RS Components. There will be a full spec for the better ones including regulation and temperature rise over ambient when fully loaded. These will be more expensive than the usual types for sale on ebay for instance. The full load temperature rise can be awful on some.

Just add if it matters I am more inclined to use wanted volts 0 wanted volts transformer or a twin primary one and use 2 diodes for rectification.

John

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Edited By Ajohnw on 17/03/2016 12:34:04

[John FieldingParticipant @johnfielding34086]

OK you guys have all taken this thread way off track.

If I may help the original poster.

A half wave or full wave rectifier produces an undulating quasi dc output voltage which varies to zero to some peak voltage. It will read lower than the text book maths tell us it should because the meter you are using is reading the average voltage and not the peak voltage. To get the peak voltage you need a storage device, like a battery or a capacitor to charge up to the peak voltage.

So what you were reading was quite correct when the load is a simple resistor. However, if you connect a suitable capacitor across the output the dc voltage will now be the peak.

This capacitor is called a "reservoir" or "smoothing" capacitor. If you do the maths the secondary voltage ac rating will increase by 1.414 times when you measure at the peak voltage for a true sinewave. So if the secondary voltage is, say 10V ac, which you can measure with a DVM, then the smoothed dc voltage will be 14.14V.

The reason they don't fit smoothing capacitors to simple battery chargers is because it is not necessary, in any case most batteries seem to prefer the pulsing voltage. There have been many studies done over the years on this topic.

Transformer used in battery chargers are also deliberately made with a "Flux Limited Core" this means under a heavy current load they collapse in output voltage, which limits the charging current into the battery.  Normal transformers for power supplies use a different method where they want the secondary voltage to remain as constant as possible.  An electric arc welder of the old fashioned type has a movable core piece to adjust the current, it is also a type of flux limited core for a special reason.

 

Edited By John Fielding on 17/03/2016 13:15:04

[John FieldingParticipant @johnfielding34086]

At the risk of sounding pompous!

For those readers who are interested in power supply design for the home constructor, I wrote a comprehensive sort of textbook a while back. I don't normally advertise to boost sales as they sell well enough already.

But if you should wish my book can be purchased through any good bookshop for a reasonable cost.

It is published jointly by the Radio Society of Great Britain (RSGB) and its American counterpart the ARRL and is entitled "The Power Supply Handbook" – what else could I name it? And it is sold under my name above.

[AjohnwParticipant @ajohnw51620]

Not so sure about dvm's and reading averages John – not at 50 or 60 Hz anyway. It depends on how the quantify it. That can be dual slope or peak with a correction factor but the reading will only be correct if derived from a sine wave. An analogue meter is a better option as that will average if not or true rms of course.

Out of curiosity did you write the original power supply handbook or has there been a more recent version? In my late teens I used the two caps with an R in between to get rid of ripple. It did get rid of the ripple but the resistor burnt out.

I had already mentioned lossy transformers but not how they do it. One other way of looking at it is poorer coupling between primary and secondary. The same thing is done in stick welders etc that aren't based on an inverter but can probably be done with those too. It gives a high striking voltage to get the arc going and then regulates current. I've lost count of the number of cheaper units I looked at to find one with a decent striking voltage but they were never as good as industrial units. I did find one that was usable eventually.

John

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Edited By Ajohnw on 17/03/2016 15:46:33

[MuzzerParticipant @muzzer]

A battery behaves like a giant zener diode across a tremendously large capacitor. Fitting an external cap across it will achieve precisely nothing, part from the impact on your wallet that is. When the transformer voltage exceeds a diode drop above the battery, it will conduct.

To do some very basic sums, if you want to limit the ripple to 1V p-p with a 10A load (typical for a charger?), you'd need a 0.1F capacitor rated at something like 25V (assuming 50Hz mains). That would be pretty big, physically speaking – and to what end? As soon as you connected the battery across it, you wouldn't know it's there.

Seems that all these years I've been barking up the wrong tree. But what would I know?

[norm nortonParticipant @normnorton75434]

Thank you for all the constructive comments, and even a thank you for the off beam comments

I used the diode test setting on one of my DVMs for the first time in my life (!). The new rectifier gave 0.570, 0.574, 0.567 and 0.578 v across the four diodes and OL in the other direction. The old four plate/ five tapping rectifier gave 0.185, 0.330, 0.221 and 0.185 v, and OL in reverse – so I guess that was a germanium one?

So my old rectifier was probably still working and I was wrong to assume it had failed. The assembled charger was connected to a 12v 12w bulb and it lit ok, but when connected to a 12v 60w bulb the overload trip kept triggering. Ah Ha I thought that is the problem, so I bypassed it with a cable. Now the charger will light the 12v 60w lamp, but it was only delivering 3A, and it should be 5A.

So, I put another ammeter in series and confirmed it was only delivering 3A. I put a voltmeter across the lamp and noted the following:

Switch 12v; switch Hi; voltage = 8.8v

Switch 12v; switch Lo; voltage = 7.3v

Switch 6v; switch Hi; voltage = 5.0v

Switch 6v; switch Lo; voltage = 4.1v

The voltages are in the right order but all about 60% of what they should be. Nothing was getting hot or smelly, except the heat sink on the rectifier that was warmish. Unless I have one plug socket in my workshop that delivers 140v AC I can only presume that a winding has failed in the transformer.

I don't really understand how a winding can fail yet I still have circuit continuity and the ability to deliver 3A. Perhaps a winding has blown and shorted to itself, thus reducing the number of turns on the output by 60%, thus reducing the voltage. Have I got that the right way round?

Looks like it will have to go to the great scrapyard in the sky and I will have to buy a new plastic one.

Edited By norm norton on 17/03/2016 17:11:40

[norm nortonParticipant @normnorton75434]

John

I have just had a longish think about what you said a few posts ago. Are you saying that my 'average' 7.3v from a DVM into a resistive load will be pulsing to over 13.6v and thus will charge a lead acid battery?

Now I need the oscilloscope that's in the loft.

Norm

[Neil WyattModerator @neilwyatt]

Posted by john swift 1 on 17/03/2016 11:13:21:

I've not seen a convincing reason why the mains frequency of 50 cps in the UK and 60Hz in the US

Wikipedia says "In the US at the end of the 19th century, Westinghouse decided on 60 Hz and AEG in Germany decided on 50 Hz eventually leading to the world being mostly divided into two frequency camps."

As there is nothing to gain and too much to lose by changing it, I don't suppose it will ever change without a massive change in the world order or how we use electrickery.

neil

[Neil WyattModerator @neilwyatt]

Posted by norm norton on 17/03/2016 17:23:51:

John

I have just had a longish think about what you said a few posts ago. Are you saying that my 'average' 7.3v from a DVM into a resistive load will be pulsing to over 13.6v and thus will charge a lead acid battery?

Now I need the oscilloscope that's in the loft.

Norm

No, 7.3 x 1.4 = 10.4v

[Neil WyattModerator @neilwyatt]

Good stuff on the history of mains frequencies

en.wikipedia.org/wiki/Utility_frequency

Neil.

[AjohnwParticipant @ajohnw51620]

With your voltmeter on the AC range what are the open circuit volts (no load readings) going into the rectifier at the various switch settings ?

John

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[norm nortonParticipant @normnorton75434]

Posted by Ajohnw on 17/03/2016 18:04:39:

With your voltmeter on the AC range what are the open circuit volts (no load readings) going into the rectifier at the various switch settings ?

John

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AC volts across the rectifier AC inputs, no load on the output:

Switch 12v; Switch Hi; AC = 12.9v

Switch 12v; Switch Lo; AC = 11.1v

Switch 6v; Switch Hi; AC = 7.6v

Switch 6v; Switch Lo; AC = 6.5v

Norm

[john swift 1Participant @johnswift1]

Hi Norm

the reading are what I'd expect

if the battery is not charging its either fully charged or faulty

John

[AjohnwParticipant @ajohnw51620]

Those sound about right. You should get about 17 and a bit volts on the high range out open circuit and circa an old style float voltage of 14.7 on low – enough to keep the battery fizzing a bit when fully charged to equalise it. Not a good idea at all on modern low maintenance ones unless you can open them up to check and top up if needed. They have also done things in there make up that make a bit of a nonsense of this voltage anyway but it can still be used for equalising the charge in the cells.

You could try checking these on the output with the meter. I did know what went into meters some time ago and a method was used that takes care of ripple at mains frequency and multiples of it but I am totally out of touch with small hand held dvm's. It's easier to buy one.

You could check the two 12 volts settings with your 2 loads on ac into the rectifier.

John

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[norm nortonParticipant @normnorton75434]

Ok, the penny has dropped. The good thing about posing questions is when you think about the answers, and look at the problem again and learn something. My first error was in treating the charger as some kind of dynamo and thinking that its average voltage had to rise above the charging voltage of the battery. The second error was thinking that a resistive load like a bulb would behave like a battery as a test device, whereas a battery behaves like a capacitor. I guess quite a lot of people erroneously think that a 12v battery charger will drive a light bulb at 12v.

I now recall my AC theory from 30 years ago. It is only the tops of the voltage pulses from the rectified AC waveform that flow into the battery, and as soon as you start to charge the battery its voltage will rise to some value, and continue to rise as the plates convert, reaching a maximum when all the conversion has been done.

I have just put the charger on to a small, moderately charged battery that was reading 11.83v: on Hi it read 13.96v and on Lo it read 12.83v. So, yes it is working.

Odd thing is why the previous DVM readings were less than the 1.414 sine wave ratio; either the DVM does not integrate the waveform correctly or the output is not a sine wave?

Thanks again to all and especially john fielding, john swift1 and ajohnw.

Norm

Edited By norm norton on 18/03/2016 09:37:10

[Michael GilliganParticipant @michaelgilligan61133]

Posted by norm norton on 18/03/2016 09:33:55:

Odd thing is why the previous DVM readings were less than the 1.414 sine wave ratio; either the DVM does not integrate the waveform correctly or the output is not a sine wave?

.

Norm,

You're right in your deduction …

The output is most definitely not a sinewave.

MichaelG.

.

Example

Edited By Michael Gilligan on 18/03/2016 09:52:40

[John FieldingParticipant @johnfielding34086]

Hi Norm,

Yes, you got it spot on! It fools many people when the charger doesn't seem to work correctly and assume it puts out a smooth or pure dc voltage.

The reason the DVM gives an apparent wrong reading – it actually measures correctly with the waveform it is presented with – is because a half-wave sinusoidal pulse is not a sine wave. The same thing occurs with PWM square wave forms of varying duty cycle. If you use a moving coil meter and measure the voltage which is, say going from zero to 10V for example and the duty cycle is 20% on time and 80% off time, then it will read 2V. It averages the voltage because of the inertia in the movement. If you increase the duty cycle to 50% it will read 5V etc.

A DVM can give a totally different reading as it sample the waveform periodically and if the sampling rate is not the same as the applied waveform frequency then it gets confused. Suppose the sampling rate is exactly the same as the applied frequency, what will it measure? Well it could read 0V or it could read 10V, it depends where it catches the pulse!

I came across this when working on radar systems and a technician connected a RF power meter in line and was puzzled why it read so little power. I had to explain to him although it was a 10kW transmitter it was only transmitting for one micro-second every milli second, so the meter read virtually nothing, just a few watts indicated. When you calculated the duty cycle it was only 0.1% so the meter was correct.

So I am glad you have remembered your lessons from ancient history!

The other question I am often asked is "My battery I know is flat, but when I connect my charger it draws very little current. Surely if it is flat then it should draw maximum charge current?"

The answer is NO. And you need to understand how a cell or battery works. A cell can be modeled as a constant voltage source which can supply infinite current with no drop in voltage. The ideal cell in other words. But no cell works like that, as you load up the cell discharge current the voltage begins to drop. To explain this effect we need to include a "loss resistor" which is between the perfect voltage source and the terminals. As the discharge current rises, more and more voltage is dropped across the resistor, which we call the ESR (Equivalent Series Resistance&quot. This cell model covers all the operating parameters of any cell type. So when the cell is fully discharged the ESR has risen to a high value and it can no longer allow sufficient current to flow to the load, most of the voltage is being dropped across the ESR.

Now when you now connect the cell onto a charger, the ESR is still a high value but after a little charge current has flowed into the cell it begins to drop as the cell recovers its chemical balance and so the charge current climbs to some maximum. This is known as the "Bulk Charge" region. When the cell is almost fully charged its voltage rises and equals the charger output voltage and the charge current falls towards zero. When a cell or battery is fully charged the ESR has reached its minimum value and the cell can deliver maximum current to a load.

Plenty of cells and batteries have been scrapped because of this effect. If they had waited an hour or so they would have seen the charge current begin to rise!

[AjohnwParticipant @ajohnw51620]

I'd still beg to differ on mains ripple and DVM but as I pointed out I can't be totally sure what hand held dvm's do. However as mains ripple at some level is always present they do go to some trouble to include it sensibly in dc readings by using a method that will account for it.

It should explain that here

**LINK**

John

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[Tim StevensParticipant @timstevens64731]

The output of a 12v system which delivers unsmoothed full-wave current (ie like you've got, Norm) is indeed variable. But it does not vary from zero to 12 volts, as you imply, Steambuff. It rises to more than 12volts so that the average is 12 volts. So, the max voltage is likely to be about 16V – if you measure with an oscilloscope, eg.

Neither is the output a sine wave or a folded sine, as the transformer won't be perfect, and the rectifier puts a kink in the wave because of the characteristics of the diodes.

For info, the 'voltage' and 'current' of a rectifier as stated in its specification can be misleading. The current is the maximum it will stand under set conditions, continuously, as you would expect, but the voltage is the maximum the diodes will stand in the wrong direction. Any diode will give in, eventually, as the reverse voltage rises. Most then produce a puff of smoke and an open circuit. Special diodes can be made which will stand this reverse voltage, generating heat but not failing (as long as the current is moderate). They are called Zener Diodes and are very useful as voltage control standards.

Many power sources can include spikes of much higher voltage. A good example is a vehicle electrical system with coil ignition, which can easily give 200V plus for an instant as the contacts open. This is why a rectifier needs a high-seeming voltage in the spec. The rectifier you have should be OK for your use, Norm, but for a motor bicycle (for example) I would consider it a bit low.

Hope this helps someone …

Tim

Edited By Tim Stevens on 18/03/2016 10:22:39

Edited By Tim Stevens on 18/03/2016 10:24:48

[Mike PooleParticipant @mikepoole82104]

A digital multimeter will need to have the specification read to establish what the measurements actually mean, modern Flukes claim to read true rms but only between 5% and 100% of range. Some meters specify the form factor of the wave that they will read correctly, so often waves other than pure sine will not read accurately. Reading a half wave rectified wave on rhe ac scale of a true rms meter may give a more accurate reading than the dc scale as it is not really dc.

Mike

Edited By Michael Poole on 18/03/2016 10:53:51

[AjohnwParticipant @ajohnw51620]

Must admit I can't say that I have ever stuck a decent meter on a rectified but not smoothed supply but that link should explain why the technique will take account of ripple. It will reject it completely and give a true reading at a specific frequency. It does it by applying the measured voltage to a capacitor so effectively measures the charge that the voltage it's measuring provides – that effectively integrates it and as ripple in the normal sense is either side of a steady value effectively measures it's actual value.

It will do the same thing to an unsmoothed supply too much thought needed at the moment to what effect that would have.

The catch is that the sampling period needs to relate precisely to the frequency of the ripple for it to work very accurately.

John

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[Russell EberhardtParticipant @russelleberhardt48058]

Most cheap DVMs respond to the mean of the modulus of the input waveform when on AC ranges. They are however calibrated to read the result as the rms value when given a true sine wave. So they are neither one thing or another.

Better DVMs have a true rms reading.

Russell.

[AjohnwParticipant @ajohnw51620]

Maybe I should have mention on the DC ranges in huge capitol letters in my last post. I believe there are other ways of achieving the same thing but accounting for it is more or less essential when taking accurate dc readings.

John

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