Author Topic: Help understanding current draw from center tapped transformer...  (Read 3292 times)

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Shredhead

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Help understanding current draw from center tapped transformer...
« on: July 08, 2016, 08:35:09 pm »
I'm trying to find out the total power output of the secondary of this transformer.  The Bridge rectifier is not shown here.  I measured the currents out of all 3 secondary leads.  I don't think it's right to add both rails readings together for total secondary output power because that would indicate that the transformer has no loss from input to output.  I thought transformers were quite inefficient.  I don't then understand why both rails wouldn't be even in power draw.

The regulated supply is not connected to anything other than 4.7k bleeder resistors after the regulators and LED's before the regulators.

Are the currents circulating on each half of the transformer using the center tap as a common path in which case there is some kind of phase cancellation going on?

IanB

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Re: Help understanding current draw from center tapped transformer...
« Reply #1 on: July 08, 2016, 09:05:03 pm »
You have two secondary windings that just happen to be connected in series. So the right analysis is to look at each secondary winding individually and then add the results together.

For example, if the total voltage across both secondaries is 34.55 and they are symmetrical then you have 17.27 V per secondary (measure them individually to get a more accurate figure).

Taking #1: 33 mA at 17.27 V is 570 mW.
Taking #2: 28.36 mA at 17.17 V is 487 mW.

Add them together and the total output power is 570 + 487 = 1057 mW.

(Strictly speaking this is not power but VA, and the actual power will depend on the power factor, but the principle is close enough.)

« Last Edit: July 08, 2016, 09:37:30 pm by IanB »
I'm not an EE--what am I doing here?

Shredhead

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Re: Help understanding current draw from center tapped transformer...
« Reply #2 on: July 08, 2016, 09:17:03 pm »
Measure both secondaries in relation to the center tap and ignore the measured current flowing in the center tap then?

Ian.M

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Re: Help understanding current draw from center tapped transformer...
« Reply #3 on: July 08, 2016, 09:28:31 pm »
You'd need either a very good true RMS meter that's accurate at high crest factors, or a DSO that can do waveform arithmetic + a calibrated current probe to be certain what's going on, as when the bridge rectifier and filter caps are very lightly loaded, the conduction angle of the diodes in the bridge is only a few degrees.

However, I suspect what you are seeing is a very small imbalance between the output voltages of the two halves of the secondary.  At low loads, the one with the fractionally greater voltage supplies more current.  As the load increases it will become better balanced.

MosherIV

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Re: Help understanding current draw from center tapped transformer...
« Reply #4 on: July 08, 2016, 09:40:24 pm »
Quote
I thought transformers were quite inefficient.  I don't then understand why both rails wouldn't be even in power draw
Actually, the transformer on its own is very efficient. Just do search for "transformer efficiency" and you will see pages which explain the calculations and some pages quoting efficiency between 95 to 98%.
That is for the transformer and does not include the voltage regulators, their job is to actually dissipate power and therefore reduce the efficiency of the power supply.

Transformer efficiency is also effected by the type, core material and windings.
Generally speaking, the most efficient type are torroid transformer, they have best magnetic coupling (or least magnetic leakage).

You can draw uneven power (current) from dual secondary outputs, though it is best not to.

bktemp

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Re: Help understanding current draw from center tapped transformer...
« Reply #5 on: July 09, 2016, 01:02:51 pm »
Quote
I thought transformers were quite inefficient.  I don't then understand why both rails wouldn't be even in power draw
Actually, the transformer on its own is very efficient. Just do search for "transformer efficiency" and you will see pages which explain the calculations and some pages quoting efficiency between 95 to 98%.
It depends on the size. Large transformers are very efficient, but small transformers can be really bad, especially when loaded with a rectifier + capacitor. For a small 5-20VA transformer you can expect around 60-80% efficiency. Very small transformers (<1VA) often barely reach 50% even when operated at the best operating point.

Shredhead

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Round 2...
« Reply #6 on: July 09, 2016, 09:59:21 pm »
Would this breakdown be correct?  I'm using a Fluke 87-V to measure RMS current and voltage.  "A" and "B" are the two taps coming off of the transformer in relation to the center tap ground.  I realize that this is only measuring apparent power but I suppose that's all that I can measure unless there is a way to calculate phase angles with a digital scope and shunt resistors or something like that.  Corrections are welcomed and thanks for the guidance so far.

Round 2 no load
Mains                                               A                         B                   Positive rail                      Negative rail
119.2VAC    34.68VAC split to:    17.35VAC              17.35VAC         21.61VDC(to regulator)    -21.71VDCV(to regulator)
17.01mA                                     33.64mA               29.03mA          15.01VDC(regulated)       -15.04VDC(regulated)
2.0276W                                     .5837W                .5037W
=1.0874W

2.0276W to 1.0874W 53.6% Efficiency primary to secondary with only LEDs, regulators and 4.7k bleeder R's as load.

With 323.8 ohms & 324.5 ohms load on positive and negative rail after regulator
Mains                      A                        B                      Positive rail                      Negative rail
118.8VAC           16.38VAC            16.35VAC            19.39VDC(to regulator)    -19.45VDC(to regulator)
36.34mA               99.7mA              85.6mA              15.01VDC(regulated)       -15.04VDC(regulated)
4.317W                1.633W                1.4W                46.3mA(after regulator)     46.35mA(after regulator)
=3.033W                         .695W                                .6971W
=1.3921W

4.317W to 3.033W =70.3% efficient primary to secondary
3.033W to 1.3921W =45.9% efficient secondary to regulated output
4.317W to 1.3921W =32.2% efficient primary to regulated output

Note: each of the 2 LED's after the bridge rectifier draws 11.83mA @21.68VDC =.2565W
« Last Edit: July 10, 2016, 10:00:41 am by Shredhead »

IanB

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Re: Help understanding current draw from center tapped transformer...
« Reply #7 on: July 09, 2016, 11:37:06 pm »
The primary winding of the transformer will act like an inductor and there will be a certain reactive current flowing even when the secondary windings are open circuit. You could get a slightly better estimate of the efficiency if you measure the current in the primary with the secondaries disconnected and subtract this current from your subsequent readings.

By the way, what is the VA rating of the transformer? Are your loads large or small relative to the size of the transformer?
I'm not an EE--what am I doing here?

Shredhead

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Re: Help understanding current draw from center tapped transformer...
« Reply #8 on: July 10, 2016, 09:55:56 am »
Wow, never thought of that.  Thanks IanB.

bktemp

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Re: Round 2...
« Reply #9 on: July 10, 2016, 11:34:51 am »
Would this breakdown be correct?  I'm using a Fluke 87-V to measure RMS current and voltage.  "A" and "B" are the two taps coming off of the transformer in relation to the center tap ground.  I realize that this is only measuring apparent power but I suppose that's all that I can measure unless there is a way to calculate phase angles with a digital scope and shunt resistors or something like that.  Corrections are welcomed and thanks for the guidance so far.
Exactly, you only measure the real power.
Measuring the phase angle only works if the load does not produce harmonicd, in other words does not distort the waveform. Most transformers distort the waveform heavily because the core starts to saturate at the end of each halfcycle. You need to measure both voltage and current at the same time and multiply the waveforms before calculating the avarage value.
For a small 6VA transformer my guess would be around 1W no load losses, but the value deps on the quality of the transformer.

By measuring the input current at no load and measuing the resistance of the winding you can calculate the losses in the winding (P=I²*R), but there are also losses in the iron core. So without being able to actually measure the real input power and the real power going into the load, it is impossible to measure efficiency.

Ian.M

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Re: Help understanding current draw from center tapped transformer...
« Reply #10 on: July 10, 2016, 12:45:40 pm »
See Fluke 87V specs in: http://media.fluke.com/documents/8xv_____cmeng0100.pdf
Quote from: page 13
For Model 87 in the 4 ½-digit mode, multiply the number of least significant digits (counts) by 10. AC conversions are ac-coupled and valid from 3 % to 100 % of range.  Model 87 is true rms responding. AC crest factor can be up to 3 at full scale, 6 at half scale. For non-sinusoidal wave forms add -(2 % Rdg + 2 % full scale) typical, for a crest factor up to 3.
Taking measurements at a higher Crest Factor than specified.is not likely to give a useful result as your meter's true RMS converter circuit is probably clipping or distorting the current peak.

As the load on a bridge rectifier tends to zero, the conduction angle becomes vanishingly small and as the current tends to zero, its crest factor tends to infinity.
As I said earlier, you need a *VERY* good True RMS bench meter (preferably using a thermo-resistive true power measurement RMS converter), or a scope that does waveform arithmetic + a calibrated current probe to take meaningful measurements of the current into a lightly loaded bridge rectifier + reservoir cap.

As such, small differences in terminal voltage of the two halves of the secondary, or variations in DC resistance will cause an imbalance at low load currents.  If the winding is wound with a tap midway, the increased diameter of the outer half of the winding means it has more resistance for the same number of turns, thus the peak current will be lower even if its unloaded voltage is identical.  If the transformer is bifilar wound two matched secondaries are possible, but they will usually be brought out to four terminals so they can be connected in series or parallel.

Another possible cause of imbalance is differences in diode Vf in the bridge rectifier.

Smf