# Negative power measurement of transformer short-circuit test. What happened?

I have a dissectible transformer, made up of two 120-turn coils. I connect a function generator to the primary side, set the AC voltage to be constantly 10V, and short circuit the secondary side. And as I vary the input frequency, and measure the power reading on the primary side, I have the following graph of power against frequency: So what is happening?

• Time to get a new power meter. Seriously though - there will be a phase shift between voltage and current that increases with frequency; if the power measurement doesn't account for this properly, anything could happen. What happens to the curve when you change the voltage - is the negative offset constant? – Floris Feb 26 '15 at 1:26
• actually I didnt use a power meter to measure because our school lab doesnt have it. instead I use a data logger to measure current and voltage over time. then I calculate average power by averaging instantaneous power obtained from voltage and current readings. Yes I think I should try to change the input voltage – Norm Do Feb 26 '15 at 2:58
• Please label the axes on your plot and include a circuit diagram. – DanielSank Aug 20 '15 at 1:32

To get a negative number for power, the time averaged product of voltage and current must have been negative. Now if you are properly sampling voltage and current at the same time, for the transformer you are measuring (where power is obviously being dissipated in the transformer, not generated by it), you would expect at low frequencies to have the voltage and current in phase; and as the frequency increases, for the current to lag behind (because of the inductance). At sufficiently high frequencies, the voltage and current would be in quadrature and no power is dissipated - but that will not happen as there will be some "real" component of the reactance that keeps the current from lagging a full 90° behind the voltage.

Some simple plots (not to scale) of what happens (horizontal axis = time in arbitrary units; vertical = voltage and current in arbitrary units; frequency increases from top to bottom. Blue = voltage, green = current, red = instantaneous power.) As you can see, for an inductive load like the shorted transformer, you expect power to flow in and out depending on the phase (depending on whether you are storing energy in the magnetic field to increase it, or whether the magnetic field is decreasing and giving energy back). The average of the red curve becomes lower as the frequency goes up - because the phase difference becomes larger. It is conceivable that your method for sampling the voltage and current is catching the negative values more often for the high frequency. But you will find the transformer getting warm - a sign that some power is being dissipated.

UPDATE

We can model the circuit (with the series resistor) as an inductor in series with a resistor: For which the phase response with frequency looks like this: The simplest way to analyze this (apart from using an online circuit analysis program - as can be found for example in the editor of our sister site electronics.stackexchange.com) is to use complex arithmetic.

The impedance of an inductor with internal resistance $R$ and inductance $L$ can be written as a function of frequency $\omega$ as

$$Z = R + j\omega L$$

At low frequencies, this will tend to $R$ and the response is real; at higher frequencies, the response tends to infinity and there will be a $\pi/2$ phase shift (90°) between current and driving voltage.

Adding another resistor $R_s$ to the circuit, you can draw yourself a simple voltage divider and repeat the same calculation.

The power dissipation in the inductor is given by the real part of the current only - the imaginary part, as I tried to explain above, is "giving and taking" energy over the course of a complete cycle and therefore doesn't contribute. But just to make life a little bit more interesting, it is not simply the internal resistance $R$ that contributes to the losses - for an inductor with a core (like a transformer), there is an additional loss term due to eddy currents (which result in another "real" component of the impedance). Also, the internal resistance is a function of frequency because of the skin effect - at higher frequencies, current tends to flow in the surface of a conductor only, and not throughout the core. This effect is strongest at high frequencies and for larger conductors - it is already a concern in high voltage power transmission lines, even though they operate at 50 or 60 Hz - but usually we think of this as a problem at RF frequencies.

Ignoring for a moment these subtleties, we can compute the power in two steps. First, we find the amplitude of the current as a function of frequency (for the circuit with just internal resistance in the transformer, and no additional external resistor):

$$|I| = \left|\frac{V}{Z}\right| = \frac{V}{|R + j\omega L|} = \frac{V}{\sqrt{R^2 + (\omega L)^2}}$$

You can see that the current will decrease with frequency - this will become significant once $\omega L > R$. Then the power dissipated is the real part of the current multiplied by the voltage (and a scale factor for integrating over the full cycle - just like the conversion from amplitude to RMS), or

$$P \propto V I = V^2 \frac{R}{R^2 + (\omega L)^2}$$

At low frequencies this tends to the familiar result $P \propto \frac{V^2}{R}$, and at high frequencies it tends to zero. Power should drop by a factor $\sqrt{2}$ when $R = \omega L$. You should be able to measure the series resistance of the transformer and deduce the inductance by looking at your power curve to see where that point occurs.

Incidentally, you can do some good measurements with an analog oscilloscope. For example, if you put the scope in XY mode and put the drive voltage on one axis and the voltage across the transformer along the other axis, you should be able to see the shape of the XY plot change from a straight line (at very low frequencies) to a more open shape as the frequency increases. Using the cross hairs (which your scope presumably has) you can determine the value of current and voltage intercept as the curve crosses the X, Y axis (make sure the signal is properly centered), and you might be able to estimate how much larger the signal is at the peak. Using math similar to that shown above, you can learn a lot about the voltage and current from this. See for example this picture which shows what you would expect for a very small phase shift (0.1), medium shift (0.5) and a full $\pi/2$ • just now i did a couple of more experiments and i get these results: 1. when i vary the voltage and keep frequency to be 3khz, I get the following graph: imgur.com/69x7jYD (note that here i flipped the sign of the y values.actually all the 5 power values are negative) 2. I do experiment with constant voltage 10V and varying frequency. this time i do up to 10khz, then I get: imgur.com/DBkZyI0 so now the problem is whether the power will approach 0 or some positive values?what type of curve is this.and what actually happen when frequency goes beyond 1500hz? – Norm Do Feb 26 '15 at 6:48
• Could you show the individual values of the data logger that you obtain? I think there's a problem with your device - possibly an offset that is a function of frequency. This could be a result of using it above its design limits (what is the limit on the input frequency). And how do you measure the current? I would want to put a resistor in series, then measure the voltage across it and the driving voltage, and show the two on an oscilloscope. – Floris Feb 26 '15 at 15:05
• The idea of putting a resistor in series then observe the voltage on an oscilloscope is a good one. I didnt think of it. however it has some problems: 1. including the resistor in the circuit will mess up the phase diagram and complicate the calculation. 2. but the bigger problem is that our school oscilloscope is an analog one which cannot give precise result and cannot measure phase difference. – Norm Do Feb 27 '15 at 9:44
• Next i also check the possibility of the data logger and i did do zero calibration and repeat the experiment for a few times and the result is still negative. even when I correct the offset the result does not change significantly. It is not the problem of frequency either. By the way do you know the function of this curve? – Norm Do Feb 27 '15 at 9:58
• Analog scope is fine for this purpose - you are confusing "accurate" and "precise". How do you currently measure the voltage and current "at the same time"? Is this a true simultaneous measurement or sequential sampling by the same ADC (which give rise to a constant offset in time). You could test this by switching the voltage and current inputs to your logger - do you get the same result? I will draw up some equations for you... Give me a few hours – Floris Feb 27 '15 at 12:41