Quote of the Day

The beginnings and endings of all human undertakings are untidy.

— John Galsworthy

## Introduction

This is a circuit designed by Stephen Woodward that I saw years ago in EDN. I originally was interested in the right-hand side of the circuit, which measures the real power usage of the load. I now have an interest in the left-hand side of the circuit, which measures the Volt-Ampere (*VA*) usage of the load. I will present an abbreviated analysis of the circuit operation along with some design equations.

I find a number of things about this circuit that are interesting:

- It measures both
*VA*(i.e. apparent power) and real power usage of the load. - It uses the idea of the differential resistance (another example), which is a nice illustration of the mathematical concept of a differential.
- It has an interesting output circuit for converting an AC waveform into a proportional DC level. I say interesting because it has aspects of a voltage doubler circuit.

My analysis will show that the output voltage is linearly related to the *VA* usage of the load by Equation 1.

Eq. 1 |

where

*P*is load’s_{VA}*VA*usage.*V*is the thermal voltage._{T}*η*is known as the diode ideality factor.*CTR*is optoisolator’s current transfer ratio.- The
*R*values are the values of the various circuit components.

The circuit and my analysis does have some weaknesses:

- There is one potentiometer required to calibrate the VA-portion of the circuit (two pots are required to calibrate the real power-portion of the circuit).
While I use potentiometers in my personal designs, I make every effort not to have them in production designs. Calibration using a mechanical potentiometer is just too much trouble.

- My analysis assumes the optocouplers and transistors are matched.
In real life, optocouplers and transistors vary widely. This is one reason why a potentiometer is needed to calibrate the circuit. While not discussed in this post, there is a potentiometer in the right-hand side of the circuit (i.e. real power measurement) that is used to calibrate-out the optocoupler differences for that function.

I have included links to my LTSpice circuit and Mathcad file in Appendix A.

## Background

### Why Use Volt-Amperes?

The *VA* usage of a load is simply the product of the load’s actual RMS voltage and current. The load’s power (*P*) and *VA* usage are related by formula , where *PF* is the power factor, *PF = cos(θ)*, and *θ* is the phase shift between the load’s voltage and current. While the power and volt-ampere usage of the load are equal for resistive loads, like heaters, the two numbers can be dramatically different for motors and certain lighting loads (example). Knowing both *VA* and *P* usage of a load allows one to compute the load’s *PF*.

### Basic Optocoupler Operation

The Wikipedia has an excellent description of an optocoupler (aka optisolator) and I will refer you there for more information on their operation. Figure 2 shows a basic optocoupler circuit. There are numerous uses for optocouplers – I often use them to provide galvanic isolation.

One item not covered is the current transfer ratio of the optocoupler, which is defined as follows (source).

- Current Transfer Ratio (
*CTR*) - Current Transfer Ratio (
*CTR*) is the ratio of the phototransistor’s collector current compared to the infrared emitting diode (IRED) forward current expressed as a percentage (%).

*CTR* is not a constant – it varies with the LED‘s forward current. Figure 3 shows an typical example.

For my analysis, I will be assuming that we can use a single, average *CTR* value. I will assume that a sine wave passed through the* CTR* function will remain approximately sinusoidal. While a variable *CTR* clearly makes this circuit nonlinear, the results are close enough that I consider the model useful.

## Analysis

For my simulations, I assumed a 11 Ω load. This is near the maximum power level that the circuit supports. I did not worry about setting the VA calibration potentiometer to an optimal point – I just picked an arbitrary value.

### Reference Circuit

Figure 4 shows a slightly marked-up version of Woodward’s original circuit. The red mark-ups show points in the circuit where I will derive formulas for determining component values.

### Functional Overview

The overall circuit function can be described as follows:

- The voltage at Point A is proportional to the AC voltage amplitude.
This a simple half-wave rectifier circuit and we can easily determine a formula for the voltage at Point A.

- The two optocoupler LEDs are biased slightly differently –
*D*_{3}is driven directly with the rectified voltage and*D*is driven with the rectified voltage minus the voltage drop across the 0.001 Ω sense resistor._{4}The voltage drop across the sense resistor is proportional to the current draw of the load. The voltage across the sense resistor is just a few millivolts. However, this small voltage is enough to generate a small current difference between the current sourced by

*Q*and sunk by_{3}*Q*. This current difference will be proportional to product of the the AC voltage and current values._{4} - The current difference output from
*Q*and_{3}will be amplified by*Q*_{4}*A*with a constant diode voltage drop (actually a base-emitter junction) added to the amplifier’s output voltage (Point C)._{2}The diode voltage drop will compensate for a diode drop in the output circuit.

- The output circuit (Point D) consists of a simple low-pass filter circuit that is driven by current from
*A*._{2}The current from

*A*is steered by the BE junctions of_{2}*Q*and_{7}*Q*. The current drawn through the BE junction of_{8}*Q*charges the 4.7 µF capacitor on the output of_{7}*A*. The discharge current through the 4.7 µF capacitor is steered to the output through the BE junction of_{2}*Q*._{7}

### Voltage at Point A

Figure 5 show my derivation for the voltage at Point A. I also compare my result with an LTSpice simulation. The formula accuracy is reasonable considering the accuracy of the modeling that I am using. I use the fact that the average value of a half-wave rectified sinusoid equals the peak of the signal divided by *π.*

### Current Through Capacitor at Point B

Figure 6 shows my expression for the current through the capacitor at Point B. Here is where I make of the differential resistance of the diodes *gm* = *I _{D }/(V_{T} · η)*.

In Figure 6, the variable *P*_{VA} refers to the measured VA value.

### Voltage at Point C

I derive an expression for the voltage from amplifier *A _{2}* (Point C). The agreement with the LTSpice simulation is reasonable.

### Voltage at Point D

Figure 8 show my final expression, which is for the voltage at Point D. This is the output voltage from this circuit. My expression is giving me good agreement with the LTSpice simulation.

The key result is the formula for the circuit’s output voltage (*V _{PointD}* in Figure 8). This formula shows that the output voltage is linearly related to the

*VA*usage of the load.

Note the level of ripple on the output.

## Conclusion

I have presented an analysis of the *VA* measurement portion of the circuit that provides sufficient detail for a designer to select component values tailored to their specific application.

## Appendix A: Files Used in Analyzing This Circuit

Figure 9 show my LTSpice Schematic.