[help] Designing am transmitter

Brownout

Joined Jan 10, 2012
2,390
The analog input changes the gain of Q3, so the output is a gain-modulated version of the oscillator input. There is some math that shows this, but I don't have time to post it. Just remember, the transistor has an exponential gain, so the higher the input, the more gain you get out. That's why the analog input changes the gain.
 

Thread Starter

cck

Joined Jul 11, 2014
9
ok i wil try to find it. can you tell me that which equation i use?

Can i get it from Vdd*10k/20k - 10k//10k* Ib - Vbe - Ie* 1k = 0
 

Brownout

Joined Jan 10, 2012
2,390
You get it from expanding the exponential equation for a transistor using Taylor series

e^x = ... + x^2 + (higher order terms): This isn't exactly it, look it up.

Use the x^2 term:

(sina + sinb)^2 = sine^2 + 2sinasinb + sinb^2 now use the middle term

2sinasinb = 2sinω(a)T*sinω(b)T, where ω(a) is the carrier frequency and ω(b) is the audio frequency.

So, by using the exponential transfer function, cleaver biasing and filtering of unwanted signal products, you have effected a multiplication of the carrier and audio signals.

This is the basic idea. There is a lot more to it, but maybe this gets you started.
 

Brownout

Joined Jan 10, 2012
2,390
i will search taylor series and i think i can handle with it but which equation you talk about?

Ic = Is * ( e^Vbe/Vt - 1) from it?
yup. .The site you linked had a better development than what I wrote. You should read and understand that, but you can use my simplified math to maybe get the general idea.
 

Brownout

Joined Jan 10, 2012
2,390
Sure. A crystal is electrically an LC pi network. Extracting the parameters might be a challenge though.
 
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