# Amplitude Modulation Circuit Implementation I am trying to understand AM modulation at circuit level.

As per theoretical framework, we have to multiply $V_{G2}(t)$, $V_{G1}(t)$, so that frequency shifting happens at carrier frequency, i.e., convolution in frequency domain.

Now as per the above circuit, using superposition, I can calculate the $V_{out}$ as $V_{out} = R_1 \cdot V_{G2}(t) + R_2\cdot V_{G1}(t)$, where $R_1$, $R_2$ are real valued and they are function of resistors, capacitors, and transconductance.

I am confused on how the circuit is multiplying the two voltages, it looks like the circuit is adding the two voltages, whereas it should multiply as per AM modulation. I am not clear on what's the purpose of $C_3$ and $R_2$.

• The circuit you have chosen is not a good example of a modulator in that it produces an AM signal almost as a bi-product due to non-linearities. Multiplying carrier with baseband produces a DSB signal - to get AM you need to add back the carrier - see slideshare.net/sghunio/chapter04-am-modulators – JIm Dearden Oct 10 '16 at 12:43
• The nonlinear hFE is a function of emitter current , used here to multiply current gain of carrier signal to output. not a great modulator but it works – Tony Stewart Sunnyskyguy EE75 Oct 10 '16 at 13:24
• A better AM modulator can be realized using the classical transistorized differential amplifier (loing-tailed pair) if the BJT-current source in the common emitter path provides the carrier frequency. – LvW Oct 10 '16 at 13:36
• hFE has pretty much nothing to do with modulation in that circuit. Vaudio just changes bias point linearly (in affine fashion really) $I_\text{C,0}\approx(V_\text{B}-V_\text{BE}-V_\text{audio})/R_\text{E}$. And then you have transcodcutance $g_\text{m}\approx I_\text{C,0}/V_\text{T}$ and finally carrier gain which is $A_\text{v,carrier}\approx-g_\text{m}R_\text{C}$. Not such a bad modulator all in all – carloc Oct 10 '16 at 17:03

If $V_{G2}(t)$ is a carrier sine wave, then you can start the analysis by setting the audio voltage to zero. You'll find that the output is equal to $G\cdot V_{G2}(t)$, assuming operation in the linear region of the transistor.
In the calculation of the gain $G$, you'll find that it depends on the emitter current. Now let the audio voltage increase; this produces a reduction in the emitter current and a correponding reduction in the gain. If the audio voltage is negative, the emitter current increases, and the gain increases.