Help understanding current draw from center tapped transformer... - Page 1
Forum section: electronics / beginners

author: Shredhead
author_rank: Regular Contributor
timestamp: July 08, 2016, 08:35:09 pm
content: 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?

author: IanB
author_rank: Super Contributor
timestamp: July 08, 2016, 09:05:03 pm
content: 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.)

author: Shredhead
author_rank: Regular Contributor
timestamp: July 08, 2016, 09:17:03 pm
content: Measure both secondaries in relation to the center tap and ignore the measured current flowing in the center tap then?

author: Ian.M
author_rank: Super Contributor
timestamp: July 08, 2016, 09:28:31 pm
content: 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.

author: MosherIV
author_rank: Super Contributor
timestamp: July 08, 2016, 09:40:24 pm
content: QuoteActually, 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.

author: bktemp
author_rank: Super Contributor
timestamp: July 09, 2016, 01:02:51 pm
content: Quote from: MosherIV on July 08, 2016, 09:40:24 pmIt 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.

author: Shredhead
author_rank: Regular Contributor
timestamp: July 09, 2016, 09:59:21 pm
content: 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 loadMains A B Positive rail Negative rail119.2VAC 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=1.0874W2.0276W 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 regulatorMains A B Positive rail Negative rail118.8VAC 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=1.3921W4.317W to 3.033W =70.3% efficient primary to secondary3.033W to 1.3921W =45.9% efficient secondary to regulated output4.317W to 1.3921W =32.2% efficient primary to regulated outputNote: each of the 2 LED's after the bridge rectifier draws 11.83mA @21.68VDC =.2565W

author: IanB
author_rank: Super Contributor
timestamp: July 09, 2016, 11:37:06 pm
content: 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?

author: Shredhead
author_rank: Regular Contributor
timestamp: July 10, 2016, 09:55:56 am
content: Wow, never thought of that. Thanks IanB.

author: bktemp
author_rank: Super Contributor
timestamp: July 10, 2016, 11:34:51 am
content: Quote from: Shredhead on July 09, 2016, 09:59:21 pmExactly, 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.

author: Ian.M
author_rank: Super Contributor
timestamp: July 10, 2016, 12:45:40 pm
content: See Fluke 87V specs in:http://media.fluke.com/documents/8xv_____cmeng0100.pdfQuote from: page 13Taking measurements at a higherCrest Factorthan 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.