# Measurements on a simple power supply

*Last Modification: September 23, 2013*

The purpose of this measurement example is to show that the kind of parameter (mean, RMS) must be chosen carefully and depends on the goal of the measurement.

## Circuit

The measurement is done on a simple power supply, existing of a mains transformer, bridge rectifier, smoothing capacitor and a load resistor. Figure 1 shows the measurement circuit with the locations where the measurements are done. The measurements are carried out with a digital oscilloscope and to measure the current a DC-current clamp is used that has a very small introduced impedance. All channels are DC-coupled. For all the scope screen images in this article applies that the horizontal centerline is the zero reference.

With all measurements the following settings are valid:

- Channel 1: (green) voltage, 5 V/div
- Channel 4: (blue) current, 0,5 A/div
- Trace A: (orange) power, 10 W/div

From each: the voltage, current and power the mean value as well as the RMS value are determinate. The "c" in "cmean" and "crms" stands for cycle, which means that the parameter is calculated over a whole number of periods.

The article Theory and definitions will explained how the RMS-power is by definition an erroneous term. To show this on the basis of real measurements, this parameter is also included.

## The measurement

In multiple sections of the circuit the voltage and current is measured. With these data the dissipated power can be calculated.### AC input

The first measurement is carried out on the alternating voltage input (fig. 2). V_{AC} and I_{AC} are measured. The mean voltage and current are approximately zero. These values are correct, but they give no indication that there is actually a voltage is present or that a current is flowing. The RMS-values of the voltage and current does give this indication, but they can't be used for determining the input power. The multiplying of V_{RMS} (11,37 V) and I_{RMS} (899 mA) would assume that the consumed power will be 10,22 W. The real consumed power is the average power and amounts 8,57 W.

On the other hand the RMS-current is usable to check the load of the transformer. The heat development in the copper windings of the transformer has a direct correlation with this parameter.

### Rectifier

Figure 3 shows the voltage (V_{diode}) and current (I_{diode}) of one of the diodes of the bridge rectifier. The dissipated power of a single diode is 0,26 W. Because a bridge rectifier consists of four diodes, the total rectifier losses are four times the measured value: 1,04 W. This is with the assumption that the four diodes have equal properties.

### After rectifying

The measurement of the current coming from the rectifier (I_{DC}) and the voltage across the electrolytic capacitor (V_{C}) is shown in figure 4.

The RMS-current is equal to I_{AC} of figure 2. This makes sense because the same current with the same shape flows in both sections. (For the RMS calculation it's irrelevant that the signal becomes negative.)

When I_{DC} is compared with the current of a single diode (I_{diode}) the following is noticeable: The diode is conducting only a half period. The current must than also be halved. However, the RMS-value I_{DC} is a factor 1,42 higher than I_{diode}. This is because both current shapes are different. The mean current however is accurate. The diode current (I_{diode} = 302,9 mA) is exactly the half of the current after rectifying (I_{DC} = 607,1 mA). This is because the mean current measurement measures the real electron current.

### Electrolytic capacitor

Figure 5 shows the voltage across (V_{elco}) and current to (I_{elco}) the electrolytic capacitor. This capacitor is charged at the moment the voltage reaches it's maximum and discharges when the transformer voltage becomes to low. When the electrolytic capacitor doesn't leak there are moving an equal amount of electrons to the capacitor too charge as from the capacitor too discharge. The average current will therefore be 0 A, exactly that is measured (-4,4 mA read is an offset error).

The RMS-current (661,3 mA) refers always to the dissipation in a resistive load and therefore also to the ESR of the electrolytic capacitor. The power loss in the capacitor can be calculated with this RMS-current if the parasitic series resistance is known. In this example the ESR of the used electrolytic capacitor is 22 mΩ. The dissipation will be 0,661^{2} / 0,022 = 9,6 mW. This power loss (-10 mW) is too small to measure in this configuration.

The oscilloscope calculates also the RMS-power (7,92 W). Such a large value for a component that doesn't warm up noticeably indicates that this is a nonsensical number.

### Load

The last scope screen image (figure 6) is from the measurement on the load resistor R_{l} (approximately 20 Ω). The shape of the voltage (V_{elco}) and the current (I_{Rl}) are reasonably flat. The difference between the mean and RMS value is therefore small.

Because the average current to the electrolytic capacitor is zero the load current must be equal as the current coming from the rectifier (I_{AC}). This is consistent with the measurements.

The RMS load current does not match the RMS-current after the rectifier, even if the RMS capacitor current is used in the calculation. Use therefore always an instrument that measures the mean value with DC applications.

## Adding measured power

The power delivered by the transformer must be equal to the dissipated power in the load resistor plus all the losses in the components. In this case the components are the diodes and the capacitor. The losses in the transformer are ignored because the first measurement is done after the transformer.

The checksum is: P_{Rb} + P_{elco} + 4 * P_{diode} = P_{totaal}, And this must be equal to the measured AC-power (P_{AC}.

Calculation with the average power values: 7,26 W + -0,01 W + 4 * 0,26 W = 8,29 W, (measured 8,57 W)

And with the RMS power values: 7,33 W + 7,92 W + 4 * 0,55 W = 17,45 W, (measured 12,56 W)

These calculations show how large errors arise when calculated RMS-powers are used for adding individual powers. The power is always an average value. The adding of the mean powers correspond with the total power, taken the measurement errors into account. These deviations are partially due to fluctuations of the mains voltage while the different measurements are carried out.