# How a switching mixer works in this example circuit?

I am reading about lock-in amplifiers and came across this example circuit of a mixer:

You have one sinusoidal input signal $$e_1$$ with a phase $$\phi_1$$ and the local oscillator has a square-wave signal $$e_2$$ with a phase $$\phi_2$$. On the positive half-cycles of the square wave the signals go through the upper transistor switch before outputting $$e_3$$. On the negative half-cycles, however, it goes through the lower switch and $$e_1$$ passes through an inverting amplifier with a factor -1, so it just reverses the signal.

Now, what is confusing me is the representation of the output signals on the bottom of the image. For $$\phi_1 - \phi_2 = 0$$, for example, whenever $$e_2$$ is negative $$e_1$$ becomes positive (because of the inverting amplifier) so the multiplication of both signals should still produce an output with positive and negative parts. The drawing of the output signal shows the result as if there was no inversing amplifier. A similar analysis goes for the case where $$\phi_1-\phi_2 = \pi$$. The drawing only works if there is no inversing amplifier at all.

So my question is, am I missing something or is the drawing wrong? And also, what is the purpose of this inversing amplifier at all?

• Would Electrical Engineering be a better home for this question? Commented Feb 6 at 11:34
• @Qmechanic Since this is an important experimental technique, it is a perfectly good physics question. Commented Feb 6 at 12:01

When $$e_2$$ is positive, input $$e_1$$ passes through to the output unchanged. When $$e_2$$ is negative, $$e_1$$ passes through to the output inverted. When $$\phi_1-\phi_2=0$$, the input $$e_1$$ is positive when $$e_2$$ is positive, so the output is positive, and the input $$e_1$$ is negative when $$e_2$$ is negative, so the output is positive (inverted).

If there was no inverting amplifier, the output would be the same as input $$e_1$$, because both paths controlled by $$e_2$$ would simply pass the signal through unchanged. $$e_2$$ would have no effect.

• Thank you very much for your answer. I apologise but I'm still not 100 percent clear on it. When e2 is negative so is e1. So their product is >0. In this sense, the output is inverted, but I thought the inversion was a separate operation, meaning that when e2 is negative the lower path will be conducting and e1 will pass through the inverting amplifier. Since the signals are in phase e1 would also be negative, but after passing through the amplifier it becomes positive. So the product of the both inputs would be <0. Commented Feb 6 at 12:40
• @NeonGabu "Since the signals are in phase e1 would also be negative, but after passing through the amplifier it becomes positive." True. "So the product of the both inputs would be <0." No, the switch passes the positive output of the inverter to the output. Are you imagining that the switch itself multiplies the signal by -1? It only switches. Commented Feb 6 at 12:59
• Ok, so the square-wave only controls which path will be open to e1, right? For some reason I thought it was not only controlling the switches but also being multiplied by e1, so I had an extra minus sign every time. It was a very basic question, but I appreciate you taking the time to explain it to me. Thank you very much. Commented Feb 6 at 13:59