Full wave rectifier CT secondary current

[mpegjohn]

Dear all,

If I have a full wave rectifier with a centre tapped transformer, feeding a capacative input filter, what is the relationship between the sec Ac rms current and the DC current.

I made up a circuit at the weekend with 2 diodes and a 12V- 0 12V transformer, 1000uF cap and a resistive load . The current into the load was 95mA, measuring one of the diode legs, I read 51mA on a Fluke true RMS meter.


So I reckon that each half of the sec is Idc/2. Is that correct?

Any good books on the maths behind this?

In the past I have always used bridges, and a single sec, so it was all so simple, like me.

Many thanks,
John

[Al (astral highway)]

Originally Posted by mpegjohn

If I have a full wave rectifier with a centre tapped transformer, feeding a capacative input filter, what is the relationship between the sec Ac rms current and the DC current.

Hello John, it's a little bit unusual to talk about the current in any configuration of rectifier, from half-wave, to bridge, to half-wave doubled etc. It's normally the voltage that is theorised and graphed.

That's because the current in the circuit depends on the load... which you haven't described or determined.

Of course, sometimes it's helpful to know the behaviour of current, for example to see how inrush current (while capacitors are charging for the first few cycles) might overload a rectifier/rectifiers...

Is this what you're asking?

You are asking for a relationship but it seems complicated. You have the tools there to directly measure DC current through your load resistor... You can also calculate it by Ohm's law if you want to check.

What are you really trying to establish here and why?

[AC/HL]

I think this little chart came from an old RS catalogue if it helps.

[Radio Wrangler]

A good fallback is to enter the circuit into LTspice. It will plot voltage and current waveforms for you and calculate RMS values for you.

David

[daviddeakin]

Originally Posted by mpegjohn

So I reckon that each half of the sec is Idc/2. Is that correct?

Not quite. It depends on power factor, which is typically around 0.66 in an 'ordinary' rectifier circuit near full load. This means the RMS current coming out of the rectifier would be about 1/0.66 = 1.5 times larger than the DC current drawn from the capacitor. But! The RMS current in one-half of the transformer winding is not half but 1/(sqrt2) times the total RMS current.

Your measurement of the RMS current looks too low, possibly because the fluke is struggling to read the peaky waveform. I would expect closer to 100mArms in each half of the transformer winding.
EDIT: Are you sure you were reading the load current and not the total current into the cap and load? The numbers would make a bit more sense if your 95mA was for the cap+load current.

Any good books on the maths behind this?

The maths is non-linear. It will make your head hurt even if there is an analytical solution! You really need a graphical solution:
http://valvewizard.co.uk/psu.html

[kalee20]

The transformer will be seeing a peaky current, with a capacitor input filter, so although the total number of electrons flowing will be exactly right, the RMS value of the current will be greater that you'd expect. It will certainly be greater than Idc/2, and likely to be greater than Idc/(root-2). The 51mA you measure does look a bit light.

Many small transformers have quite high winding resistances, which act to reduce the peaks and widen the duration, which does give the diodes, the transformer, the mains wiring, etc an easier life. It hurts regulation though, but that might be tolerable.

Ultimately, if you have the transformer and have got it working, why are you worried? Keep an eye on the transformer's temperature, and if that is not excessive, you will be fine.

[G8HQP Dave]

The full-wave rectifier with CT secondary is not inherently much more complicated than a single secondary feeding a bridge. You get the same currents (so the same maths) but some of them are in different places. Like others, I believe the 51mA you measured in the CT to be wrong. 95mA DC should lead to something like 120-180mA RMS from the secondary. Calculating it is messy (even using approximations) because it involves calculus so most people either use a rule-of-thumb factor or simulate it using PSUD2 or some version of Spice.

[mpegjohn]

Thank you everyone for your help.

I looked in the Big Red Book, chapter 30.

In there the sec RMS current in each half of the sec is said to be approx = 2.2 x Load current/2. ie 1.1 x Load current.

[G8HQP Dave]

So you should see 2.2 x DC at the CT.

[daviddeakin]

Originally Posted by mpegjohn

In there the sec RMS current in each half of the sec is said to be approx = 2.2 x Load current/2. ie 1.1 x Load current.

Yes, that's because the RMS current in each half is 1/(sqrt2) times the total RMS current, which is typically about 1.54 times the DC load current (i.e power factor of 0.65).
Or in other words, every milliamp of DC load current needs 1.54mA of RMS current from the rectifier, which amounts to 1.54/(sqrt2) = 1.1mA in each transformer half.

[joebog1]

Easy method is to use .62 as power factor.
One amp transformer can supply 620 mA.

Simple and mostly adequate if not so accurate.
Joe

[Craig Sawyers]

That factor of 0.62 looks about right as a good rule of thumb.

The original paper on all this was by Otto Schade, "Analysis of Rectifier Operation", Proc IRE July 1943, p341 onwards.

[Refugee]

I once had the same problem with a battery charger.
The thing was used indoors and my boss told me what transformer to use. I just told him that it would not work for long and he told me to use it anyway.
He had to sort the customer out after it burned a hole and caused an insurance claim.
I had seen the de-rating charts and he was an ex banker.

[trobbins]

John,
Rectifier Applications Handbook, HB214/D, ON Semiconductor, Rev 2, 2001 is a relatively easy read for describing the basic maths, and presents Schade's basic tables:
http://www.introni.it/pdf/Motorola%2...20Handbook.pdf
PSUD2 is easy for setting up your exact circuit and looking for waveforms and mean, rms, and peak values in windings, and confirm parameters used in Schade's tables.

[Argus25]

Not to go off topic here, but it is interesting that Otto Schade, who was from the Tube Department at RCA Victor division, Harrison N.J. wrote what I believe to be the definitive technical article on Magnetic Deflection Circuits for Cathode Ray tubes. RCA publication Television Volume V 1947-48. In this series of articles he gave credit for the basic idea of energy recovery scanning to Blumlein and his 1936 patent for the basic idea.

Otto appears to have been a Guru level expert in circuits with pulsed currents, so it doesn't surprise me he is behind the great tables you have cited.