# AM Transmitter question

berkeman said:
I googled "capacitor plate antenna" and got several hits. This one has a figure:

http://www.ece.msstate.edu/~donohoe/ece4990notes4.pdf [Broken]
Thats a great link...thanks for the information.

I just spoke to another antenna engineer and he set me straight.

My explanation wasnt completely off apparantly, but I was focusing too much on the level portions of the square wave.

During the transient from 0 - 10 volts we get current flow. If we kept the voltage at 10 volts after this we would have an equipotential surface (since the antenna is an open circuit) and the electrons would go back to electrostatic equilibrium and distribute themselves throughout the conductor due to the repulsive forces between them.

But if we continue to alternate the voltage from 0-10 volts and then back to 0 volts then this transient change in voltage will reverse their direction.

For some reason I was applying what happens during the transient portions of the square wave to the "level" portions of the square wave.

I think Im good now.

Thanks for the help berkeman.

Russ

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Averagesupernova
Gold Member
Russ the last part of your last post is pretty much what I tried to explain to you. The transition part of the waveform is not considered DC.

Averagesupernova said:
Russ the last part of your last post is pretty much what I tried to explain to you. The transition part of the waveform is not considered DC.
Right....the problem was that I was trying to picture where the electric field was coming from that causes the charges to move.

For charge to move via an electric field there has to be a potential difference, otherwise everything will just sit there as an equipotential surface (like shortly after we hook up DC to an antenna).

I was forgetting to take into account the time it takes for a change in voltage (or electric field) to "reach" the length of the wire.

Before we connect the antenna to anything, it is an equipotential surface sitting at 0 volts (Im assuming no E-fields being received by the antenna) and no charge is moving.

Then we connect one end to a source, say 10 volts, and during this transient change we have one end of the wire at 10 volts, while the far end is still at 0 volts.

Since changes in the electric field travel at the speed of light it takes (approximately) t= L/c (where L is the length of the antenna) seconds for the voltage/electric field change to reach the far end of the wire.

So for t<L/c we have essentially put a voltage across the length of the antenna and the electrons will move a bit. Then when we apply 0 volts we have the opposite situation and the electrons will move again, but in the opposite direction.

My error was in applying the voltage changes instantaneously over the entire antenna with a square wave input. I forgot to take into account the fact that it takes time for the changes in the electric field to propagate down the wire.

Stupid mistake.

Russ

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