What is Important in a Design, Voltage, Current or Power?: Difference between revisions

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The answer to this question is made further important, because the differences in what you really care about only lead to subtle differences in you design, and so they can be overlooked, but these subtle differences are often important. 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.
The answer to this question is made further important, because the differences in what you really care about only lead to subtle differences in you design, and so they can be overlooked, but these subtle differences are often important. 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.
[[File:300MHz Lumped LPF.png]]
[[File:300MHz Lumped LPF.png]]

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 <math>V_o(\omega) / V_i (\omega)</math>. This is easily found to be:
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 <math>V_o(\omega) / V_i (\omega)</math>. This is easily found to be:



Revision as of 06:44, 17 May 2017

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 further important, because the differences in what you really care about only lead to subtle differences in you design, and so they can be overlooked, but these subtle differences are often important. 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. 300MHz Lumped LPF.png

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 . This is easily found to be:

where and is the frequency where behavior changes from passing to rejecting. So it is natural to set . 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 . 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 where is the output power with the filter in place, and , 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 Failed to parse (unknown function "\abs"): {\displaystyle \frac {P_o(\omega)} {P_av(\omega)} = \frac {4R ^2} \abs {{4R + j\omega L}}^2 } and by similar arguments to those used above you will get a value of L that is half 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.