I'm an electrical apprentice, just studying electrotechnology. I know this doesn't really have anything to do with my trade and that I don't need to understand, but I'm curious as to how a change in frequency will affect power, or if it affects power at all, or if I'm on the wrong track completely. What does a change in frequency do to current, etc.?
Electronic – Relationship between frequency and power
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Voltage required is affect by significantly more than clock speed, but you are correct, for higher speeds you will need higher voltages in general.
Why does power consumption increase?
This is a lot messier than a simple circuit, but you can think about it being similar to an RC circuit.
RC circuit equivilent
At DC an RC circuit consumes no power. At a frequency of infinity, which is not attainable, but you can always solve this theoretically, the capacitor acts as a short and you are left with a resistor. This means you have a simple load. As frequency decreases the capacitor stores and discharges power causing a smaller amount of power dissipated overall.
What is a microcontroller?
Inside it is made up of many many MOSFETs in a configuration we call CMOS.
If you try to change the value of the gate of a MOSFET you are just charging or discharging a capacitor. This is a concept I have a hard time explaining to students. The transistor does a lot, but to us it just looks like a capacitor from the gate. This means in a model the CMOS will always have a load of a capacitance.
Wikipedia has an image of a CMOS inverter I will reference.
The CMOS inverter has an output labeled Q. Inside a microcontroller your output will be driving other CMOS logic gates. When your input A changes from high to low the capacitance on Q must be discharged through the transistor on bottom. Every time you charge a capacitor you see power use. You can see this on wikipedia under power switching and leakage.
Why does voltage have to go up?
As you voltage increases it makes it easier to drive the capacitance to the threshold of your logic. I know this seems like a simplistic answer, but it is that simple.
When I say it is easier to drive the the capacitance I mean that it will be driven between the thresholds faster, as mazurnification put it:
With increased supply drive capability of the MOS transistor also increases (bigger Vgs). That means that actual R from RC decreases and that is why gate is faster.
In relation to power consumption, due to how small transistors are there is a large leakage through the gate capacitance, Mark had a bit to add about this:
higher voltage results in higher leakage current. In high transistor count devices like a modern desktop CPU leakage current can account for the majority of power dissipation. as process size gets smaller and transistor counts rise, leakage current becomes more and more the critical power usage statistic.
22 awg wire has a resistance of about 16.14 ohms per 1000 meters and has a max amperage of 0.92A according to this source.
Lets say you have a purely resistive load and maximize your current with 0 wire length, your load would be 9V/.92A=9.7826 Ohms. Using this as a voltage divider you would have:
250 ft - > 2.7 V drop
1000 ft -> 5.6 V drop
5000 ft -> 8.0 V drop
This are very rough numbers, but it is what I would use as an estimate.
Low AC frequency usually acts a lot like DC.
Please note though that you probably have a ground wire that has to go the same distance, so treat my lengths as "round trip".
I don't know if my method is 100% correct, but this is what I do.
Best Answer
To keep it very simple, if you have a 1000 watt heater intended to operate on 120 V 60 Hz power, it will produce the same heat at 50 Hz or 400 Hz, provided the voltage remains at 120 V. When you get into motors, transformers, and other non-resistive loads, the power may vary with frequency.