# What is Important in a Design, Voltage, Current or Power?

Before beginning a design, it is very important to think about the question, "What is important in this design, voltage, current, or power?" This is because it will determine how you design the electronics. It is such a basic question, that it is not often discussed, because it was determined long ago, and now is taken for granted. The answer to this question determines the best tools to use for analysis and design, and the answer is typically different for different types of electronic engineers. For example, typically. engineers dealing in the area of control systems are more likely to care about voltages, and currents. Typically these voltages or currents represent a controlled quantity (maybe position, velocity, etc), and they want to do some kind of signal processing operation to the voltages or currents in their controller. Typically it is not power that matters to them. On the other hand, RF engineers are often much more concerned with power than voltage, because they are trying to transfer power over transmission lines to and from antennas, and it is the power that is transmitted that determines how well the communications system they are working on works. The signal to noise power ratio is their primary concern. Sometimes a circuit has some parts where power matters most and other parts where voltage or current matter most.

The answer to this question is made more important, because the differences in what you really care about usually only lead to subtle differences in you design, and these subtle differences can be easily overlooked, but these subtle differences are often important.

#### Transfer Function Differences

To illustrate this let's design a simple low pass filter with a cutoff frequency of 300 MHz. Suppose this filter is driven with a 50 ohm signal generator and is connected to a 50 ohm load resistor. Suppose you decide that voltage is important. Assuming you can't get inside the signal generator to measure the source Thevenin voltage, the transfer function that is important to you is $V_o(\omega) / V_i (\omega)$. This is easily found to be: $\frac {V_o(\omega)} {V_i(\omega)} = \frac {R} {R + j\omega L} = \frac {1} {1 + j (\frac \omega {\omega_0})}$ where $\omega_0 = R/L$ and is the frequency where behavior changes from passing to rejecting. So it is natural to set $\omega_0 = 2 \pi 300 MHz$. For a resistor of 50 ohms, this gives us an L of 26.5 nH.

On the other hand, suppose you are an RF engineer who cares about power. In this case you might think, that is easy, we will just make the ratio of input to output power of the filter equal to 1/2 at the cutoff frequency, since the magnitude of the transfer function from above is $1/ \sqrt 2$. Unfortunately this is impossible, since the inductor dissipates no power, and so the ratio of output to input power is always one. So what does an RF engineer really want? Well, it is likely that they want the ratio of power dissipated in the load to what would be dissipated there if there was no filter. We call that $P_0 / P_{av}$ where $P_o$ is the output power with the filter in place, and $P_{av}$, or "available power" is the power that would be delivered to the load if there were just short wires between the source and the load. You can compute this ratio and you will come up with $\frac {P_o(\omega)} {P_{av}(\omega)} = \frac {4R^2} {|{4R + j\omega L}|^2}$

and by similar arguments to those used above you will get a value of L that is twice what you got the first time. Another way of saying this is that the cutoff frequency for the filter is different if you are talking about voltage and current, or power. This is illustrated in the analysis of the above in Qucs. ##### Decibel Preferences

Note that decibel scale for voltage was calculated following convention as $20 log_{10} (|V_o/V_i|)$ and for power $10 log_{10}(P_0 / P_{av})$ and that these are just not the same. The reason is that $P_{in} \neq P_{av}$. Even if we had a situation where the power ratio were from the right voltages, they would only be the same if the reference voltage, and measured voltage were both dissipated in the same value of resistance. Really the best we can say is that dB means different things to engineers that are concerned with voltage and current, than it does to engineers concerned with power. Also note that the -3dB points are different by a factor of two depending if you care about voltage and current, or about power.

#### Source and Load Impedance Preferences

Another difference is the ideal input resistance of a detector. If you care about voltage so you don't want to load the measured voltage down, you want the detector of that voltage to be high impedance; if it is current, you want the input impedance of the detector to be close to zero. If you care about power, you usually want the detector to receive the most power it can, and so you want to "match" the impedance of the detector to that of the source. Likewise, if you are constructing a source, if you care about the voltage, you want a source impedance near zero, and if you prefer current, you need a source impedance of as high as possible, however if you care about power, you want your impedance to be something standard, often 50 ohms, or some other common transmission line characteristic impedance.

#### Analytical Tool Preferences

Often engineers concerned with voltages and currents are dealing at lower frequencies, and lumped circuit analysis using the laws of Ohm, and Kirchoff work well. RF engineers who usually care about power more often work at high frequencies. This necessitates using transmission line theory along with the laws of Ohm and Kirchoff, because often parts of their circuits are too big to use lumped approximations. When dealing with waves, it is easier to consider forward and reverse going waves, of both current and voltage. When there is a voltage wave, $V_+$, there is a corresponding current wave, $I_+ = V_+/Z_0$, where $Z_0$ is the characteristic impedance of the transmission line. The power in a forward going wave is $|V_+|^2/Z_0$ and that is also the available power to a load of $Z_0$, which is the load that they would usually like to have. When a load of another impedance is encountered in a transmission line, a reflected wave, $V_-$ is generated. It is often nice to think of these waves, instead of the total voltage $V = V_+ + V_-$ and current, $I = I_+ + I_- = V+/Z_0 - V_-/Z_0$, because they relate to the available power better than either the voltage or current. The available power is the maximum power you can get from a source for any load. This happens when you have a matched $Z_0$ load, and with that load there is no reflection. This means that the voltage on the load is just due to $V_+$, so the available power is $P_{av} = |V_+|^2/{2Z_0}$. (The factor of two comes in because it is assumed that the phasor quantities are not RMS ones.)

##### Scattering Parameters

Since RF engineers are more concerned with power than voltage or current, and because they deal with high frequencies where wave models are necessary, they commonly use phasor wave quantities whose magnitude squared is power. The characteristic impedance is normalized into these quantities in a handy way. The column vector $\bold a$ is used to represent the incoming waves, and the column vector $\bold b$ are for waves reflected from the ports of an n port network. For example, a one port network would have $a = V_+/ \sqrt{Z_0}$, and $b = V_-/ \sqrt{Z_0} = S_{11} a$. For a two port network terminated in the correct characteristic impedance, the scattering matrix element $|S_{21}|^2$ is the ratio of actual output power to the available output power, which is just what an RF engineer would want. The waves $\bold a$ go in and the waves $\bold b$ splash back. For the two port below: $\begin{bmatrix} {b_1} \\ {b_2}\end{bmatrix} = \begin{bmatrix} S_{11} & S_{12} \\ S_{21} & S_{22} \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \end{bmatrix}$

or in short matrix notation $\bold b = \bold S \bold a$

which applies for an n port as well. If port 2, the "output" is terminated in $Z_0$ so that $a_2 = 0$, then the available power is $P_{av} =|a_1|^2$, the output power is $P_o = |b_2|^2$ and the ratio is $P_o/P_{av} = |S_{21}|^2$.