Impedance Matching and Power PDF Print E-mail
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Written by Bryce Ringwood   
Sunday, 19 May 2013 17:34
 

In this article, we investigate the transfer of electrcal power. This could apply to coupling a loudspeaker to a valve amplifier, couping a transmitter or receiver to an antenna, or the rather more prosaic business of transferring power to your stuff from a generator when ESKOM ditches you. If you are reading this, you will most probably be wanting to drive a loudspeaker from a valve amplifier, or maybe want to make an antenna tuner.

Before we look at transferring power, I'm going to do the difficult part first and attempt to explain power in AC circuits.

AC Power

I don't recommend you try this, but if you were to place a 1µF 600 volt AC working capacitor across the 230 volt mains - how much power would the capacitor draw ? Well, in theory, no power at all. As you know, the current leads the voltage and starts off at a high value, with the voltage gradually rising during the first quarter cycle of the AC waveform - the capacitor is charging. During the next quarter cycle, the capacitor discharges, because the current rises to a negative peak at the end of the first half cycle, whilst the voltage becomes zero at the end of the cycle. At this point, the power delivered during the first quarter cycle equals the power delivered back again during the next. The same holds true during the next half cycle, and so, for the entire AC cycle, no power is consumed by the capacitor.

 

No power is consumed in an AC circuitcontaining a capacitor

 

Unfortunately, my experiments to show this to be true using certain "unbranded" capacitors ended in fire and much smoke, but I have to say that good quality EPCOS AC capacitors survived with no temperature rise watt-so-ever. Please don't try this at home, as they say.

The same argument holds true for inductors. A mains transformer with no load on the secondary would be an example.

Now, if we put a resistor across the mains (a light bulb will do), we draw power in just the same way as a DC circuit. The only thing we need to remember is that the average voltage and current will be the so-called RMS value, because the AC voltage rises to a maximum, falls to zero, then to a negative maximum. A 100 Watt light bulb will draw a current of :

\displaystyle{{100 \over 230} = 0.436 Amps}

Suppose we now put an inductance in series with the lamp - what happens? The inductance causes the current to lag behind the voltage, so the power applied to the lamp is less, and it glows less brightly. The true power applied to the lamp will be less, but the current supplied to the circuit will still be 0.436 amps, so the apparent power will be 100Volt Amps, or VA.  The power supplied to the inductor will, of course be zero, but it will have reactive volt-amperes (VAR).

The ratio between the true power and the apparent power is the power factor. $\displaystyle{cos \phi ={ true \phantom ppower \over {apparent \phantom ppower}}}$

In an industrial setting, the utility charges for volts and amperes supplied - but for the factory, this is less than the true power received, because of the inductance of electric machinery, wiring, and so on. To even things out, capacitor banks are used to reduce the reactive power and increase the power factor. In a home setting, the problem is usually not significant, except if you are purchasing an inverter. Inverter manufacturers often specify the apparent power output - not the true power. This might (and usually does) mean that you purchase a 1000VA inverter that will only deliver 600Watts, because the manufacturers have sucked a power factor of 0.6 from their thumbs.

Example Calculations

(See section on complex numbers. Use complex Number Calculator, or skip to next section)

I am using the complex power formula for AC: $\displaystyle \bf{P=VI^*}$, which is why the reactive component points downwards. 

Electric kettle. Power rating 2200 Watts. Supply Voltage=230 Volts 50 Hz Mains frequency

Current:        $\displaystyle{I=2200 / 230 = 9.56 Amperes}$

Impedance: $\displaystyle{{\bf {Z = {V \over I}}} ={ {230+j0.0} \over {9.565217+j0.0}}= {24.05 }Ohms}$

Power:          $\displaystyle{\Re({\bf V \bf I^*}) = 230*9.56=2200 Watts}$ (Multiply real components of voltage and current ) 

Same Electric kettle in series with 10mH Inductance

$\displaystyle{{\bf X{_L}}=0 - {{2*\pi * 50* 10} \over 1000} =0-j3.142 }Ohms$

The current will lag behind the voltage and we calculate it from the AC flavour of Ohm's Law:

Current:                 $\displaystyle{{\bf{ {I ={ V \over Z} }}}= {230 \over{(24.045 - j3.142)}} ={9.404 + j1.228}} Amperes$

True Power =        $\displaystyle{\Re({\bf V \bf I^*}) = 230*9.404=2163} Watts$ (Fairly intuitivly)

Reactive Power = $\displaystyle{\Im({\bf{VI^*}})=230*1.228=282.4} VAR$ (Multiply reactive components)

Apparent power = $\displaystyle{230*{\sqrt({9.404^2}+{1.228^2})}=2181} VA$

Power factor = ?   One for you to work out.smiley

This shows that with an inductance in series, you are going to have to keep that kettle switched on for longer, and you will have to pay for that unwanted inductance. 

In a home / small office environment - power factor does not make much difference. I have this idea that ESKOM only measures true power for domestic use, but might be wrong. Anyone care to comment?

Impedance Matching

This "thought experiment" is going to use dry batteries, in case you get tempted to drag out the Honda genny, with dire consequences.

Power sources, such as batteries, generators and inverters all have their own source impedance. In the case of batteries, this would be their internal resistance. A load (kettle, light bulb, motor etc: ) has its own load impedance. The circuit looks like this:

Matching Supply to Load

Note that the power transferred to the load is $\displaystyle{V^2\over {R{_L}}}$, where V is the voltage across the load. =$\displaystyle{{{V_{batt}^2}} \times {R_L}\over{({R{_g}+R{_L})}^2} }$

I plotted this for an unlikely 10 Volt batery with a 50Ohm internal resistance, for loads varying from zero Ohms (Short circuit) to 300 Ohms.

Matching a 50 Ohm load to a 10V battery with 50 Ohm internal resistance

 

The maximum power transfer occurs when the source impedance is equal to the load impedance. NOT that this is great for your generator or inverter, since half the voltage would be dropped in the windings (or MOSFETS), and disaster would ensue. Hopefully the generator would just stall, but alas, your inverter would have to be returned to me for repair.

Audio Circuits

Speakers ("Drive Units") are usually manfactured with an impedance of 4, 8 and sometimes 15 Ohms. This allows  a fairly low mass speaker coil. I have to admit, Philips at one time made high resistance speakers for their Hi-Z radiograms. I dont know why, since they still had to use an output transfomer.

In conventional valve output circuits, a transformer is used to convert the high impedance of the output valve to the low impedance of the speaker so that power can efficiently transferred. Transistors are already low impedance devices, so output transfomers were used in low power transistor amplifiers. Nowadays, transistors are often coupled to the speaker via a large value capacitor. In an output stage, there is a standing current passing from the HT to the valve anode (or transistor collector). The ouput transormer isolates the speaker from this current, so that the speaker only responds to the AC part of the output waveform. If you put a Philips high resistance speaker directly in the anode circuit - the standing current through the output stage will pull the speaker cone one way, and no (or very distorted) sound will result.

Output transformers were at one time used in transistor output stages. The inter-stage matching transformer survived for longer, but nowadays a single IC is used containing transistors that can drive the speaker directly via an isolating capacitor.

For those who get their matching wrong, there is a price to pay. Valves are comparatively robust (electrically), and such sins as shorting the output transformer, or leaving it open circuit are usually forgiven. (But beware of using your AR88 for headphone-only listening - the punishment is a blown output transformer.) The valves may have a shortened life-span, glow blue, or have cherry red anodes, but will recover from temporary indignities. Not so transistors. Shorting the output or even placing a capacitor across the output can result in sudden death to the output transistors (You can drop transistors without breaking them, though). In my youth, I demolished several Sinclair IC10 amplifiers that I was trying to use as a transmitter modulator by placing a small capacitor across the output to smooth the somewhat harsh sounding modulation. 

I came across an explanation for this behaviour. Normally, I suppose you would design a 10 Watt output amplifier to dissipate 10 Watts of heat, on the basis of perfect impedance matching. You would then provide a 20 Watt power supply. If, however, you connect a capacitor or inductor across the output - NO power flows into the load and the amplifier now has to dissipate 20 Watts of power. (Or worse - the entire available power supply power, I guess).

Loudspeaker voice coils have a higher resistance than inductive reactance at audio frequencies - so the reactive component is not a worry. 

RF Circuits and Antenna Matching

The aim of a radio transmitter is to transfer as much power as possible from the power amplifier stage to the antenna. This involves matching the amplifier to the transmision line and the transmission line to the antenna. Getting it wrong means that (some of) the  power gets reflected back to the power amplifier causing valves to glow blue and transistors to die quicker than you can say "nanosecond". In my "ham" days, I was using a beam antenna with a characteristic impedance of 75 Ohms, cable of 75 Ohms impedance and a crude matching network to a transistor transmitter on 433MHz.  It wasn't my best effort and I blew many an RF transistor trying to get it to work. I eventually made a 4X150A (tetrode valve) transmitter with a HiFi valve modulator that "sounded like the BBC".

By someone's law of reciprocity, if you have an antenna system that's well matched to a 75 Ohm transmitter, then it will also be well matched to a 75 Ohm input receiver. Unfortunately, as short wave listeners, we generally use random bits of wire as antennae. These will have an unknown characteristic impedance. For this reason, we can employ an antenna tuning unit to match the antenna to the receiver input. (Just as we can use  a PI network to match a short wave radio transmitter power amplifer to an antenna system.)

An example of a PI network matching 400 Ohms to 50 Ohms was provided in the section on impedance calculations. Here is a picture of a short wave antenna tuner, which I don't use because it only makes all the interference stronger.

Antenna Tuning Unit

If you want to make a coil for an ATU, I have seen it done using ribbon cable wrapped around a 25mm former.  The ends were staggered and soldered. It was a commercial unit otherwise identical to the above effort.

(I will provide a calculator for PI network impedance matching in the software section, when I find time.)

Finally, in transistor radios, you will find that the collector of the IF transistor is tapped down the coil of the transformer and that the secondary is often an untuned circuit of low impedance coupled to theprimary. Transistors are low impedance devices and without the tapping would effectively load the IF transformer - providing low gain and wide bandwidth. The tap reduces the loading and raises the quality factor or "Q" of the tuned circuit. Valves, being high impedance devices don't require this strategy, and the IF transformers are usually 1:1 double tuned circuits. 

Conclusions

  • AC Circuits often have a "Wattless Component" of power called "vars".
  • The ratio of true power (Watts) to apparent power (VA) is called the power factor of an AC circuit.
  • You can think in terms of "complex power" - it may help in the long run.
  • Maximum power is transferred when load impedance = generator impedance BUT
  • Don't exceed the power ratings for your generator or inverter or you will pay.
  • Avoid very reactive loads on your amplifier, inverter or generator.
  • Try to match the output impedance of your  amplifier to your speaker for best quality audio
  • In some locations an antenna tuning unit can help reception. In others it just makes the interference from ADSL, PLT etc. much strongerangry.
  • Valve IF transformers don't work well with transistors and vice versa.

Appendix A

The Complex Power Formula

See section on complex numbers. Values in bold face are complex numbers. Values in italics are scalar quantities. Don't be confused by the terms "real" and "imaginary" - they are just ways of separating values. Believe me when I say reactive power is certainly not "imaginary",nor do I consider it to be a "fiction".

We saw when dealing with DC circuits that the formula for power is given by $P = I  \times V$. We also note that the power dissipated in a resistor $R$ Ohms is $\displaystyle{V^2\over R}$ or ${I^2} R$.

This leads us to the notion that we can express "complex power" as:
$\displaystyle{\bf {P=I V}}$, or more conveniently as $\displaystyle{\bf {I^*} V}$.

This is rather like the idea that we can express impedance as a single complex quantity, instead of expressing it in any other way, thus an impedance $\bf Z$ will give rise to a current $\displaystyle{\bf {I ={ V\over Z}}}$ Note that $\bf V$ is generally taken to have a zero phase shift, so we can simply call it the scalar quantity $V$.

In practical terms,then, we use the formula:


$\displaystyle{{\bf P} = I V e^{-j \phi}}$,
where $\phi$ is the phase shift caused by the impedance $\bf Z$ and the current $I$ is given by $\displaystyle{V \over {\bf |Z|}}$.

Up to now we have used a symbol  "$V$" to represent voltage. In Ohm's law for AC it didn't matter too much, since voltage and current are proportional to each other and if we use a peak value for voltage, then we use a peak value for current. In any case, everyone refers to 230 Volts AC and use that value without much further thought. Here we take a look at the relationship between peak and RMS volts. 

Voltage - Peak and RMS values

Suppose we have no reactive components at all - just a "pure" resistance R. We now apply a varying AC voltage in our complex power formula. There is only a "real" component. 

 

$V(\theta)=V_{peak} \cos(\theta)$ to the resistor, 
where $\displaystyle{\theta=2 \pi f t}$, with a frequency $f$ over time $t$.


In a small portion of the AC cycle $\delta \theta$, we apply:


$\displaystyle{{{V_{peak}}^2 \over R} \cos^2 (\theta)\delta \theta}$ Watts.


Clearly on average, the power supplied over a full AC cycle is:


$\displaystyle{{I_{peak}V_{peak} \over {2 \pi }}}\int^{2 \pi }_0{\cos^2 (\theta)d\theta}$

(You can go to the wolframalpha web site and let it evaluate the expression for you. It is simply the area under the power curve divided by the total time. ) 

This evaluates to:


$\displaystyle{{I_{peak}V_{peak}} \over 2}$ or $ \displaystyle{ {I_{peak} \over \sqrt 2} \times {V_{peak} \over \sqrt 2}}$

We normally rewrite this as $V_{rms} \times I_{rms}$ or simply use the symbols "$V$" and "$I$" to mean the rms or root-mean_square values of voltage and current. They are the AC voltages you measure on a multimeter, and the voltage we refer to when we say "the mains voltage is 230 volts." The complex power equation is an extension of the "AC power through a resistor formula" and clearly has to work with RMS values.

The foregoing is not rigorous - it outlines where the idea of complex power comes from. The part represented on the real axis is the true power absorbed to do useful work and is measured in Watts. The part represented on the imaginary axis is the reactive power that  transfers back and forth between the inductance (usually) of the circuit and the generator. The apparent power is simply the hypotenuse of the power triangle - Volts times Amperes or VA used in the ratings of motors and some inverters.

Please note that this discussion applies to sinusoidal waveforms for voltage and current. Loads such as those presented by switchmode power supplies  fall outside this definition.

As this seems to be an emotional subject, I will now sit back and wait for the howls of protest at my treating it this way.

Appendix B

Audio transformer design notes

There are companies in Johannesburg who will design and make audio transformers for you. Proper design is a little complicated and you might need a bit of trial-and-error (mainly error) before you get your design right.

Transformers for push-pull audio stages have standing current going to the anode of each valve, but the currents cancel each other out in the transformer windings, so the core of the transformer is not magnetized by the anode currents. On the other hand, transformers for single-ended output stages will have some magnetization from the anode current of the output valve, which could lead to distortion if the transformer core becomes saturated (unable to take any further magnetization on positive peaks of output). To prevent this problem, the core of the transformer may be given an air-gap (or plastic-gap), or simply be overdesigned. If you are winding your own, then clamp all the E laminations together, clamp all the I laminations together and place a THIN piece of plastic at the join of the two. You will have to put a metal frame round the assembly to stop it all falling apart.

Considering the power to the input of the transformer = power to the output, then it is easy to see:
$\displaystyle{{{Z_p}\over{Z_s}}= ({{n_p}\over {n_s}})^2}$ , where the $Z_p$,$Z_s$ are primary and secondary impedances and $N_p$,$N_s$ are the number of turns on the primary and secondary. If you are dealing with a transformer manufacturer, you should be able to simply tell him you need an output transfome for an EL84 or whatever, and he will know what to do.

Last Updated on Saturday, 24 August 2013 19:07
 
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