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Input Attenuator

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New Member
Hello, I'm a beginner in electronics. I've been tasked with a practical that I've attached on here. To my understanding this is just a simple voltage divider. As it states, I need to modify the circuit such that the output is half the input AC signal. The audio generator that I have has an output of 600 ohm. Would I just include that in the voltage divider rule V_out = (R2/R1+R2) x V_in? And then change either R1 or R2 until I get the half the input signal?


Dan Soze

There is a lot to unpack from your question. The attachment of the assignment you posted does not mention the output impedance of the V1 signal source but you said your audio generator has a 600 ohm output. In a typical audio generator where the output metering is shown in aduio decibels (dBA) this needs to be connected to a specific impedance, for audio the standard is 600 ohms. The output impedance of your audio signal generator does not appear to be an important part of your assignment.

The circuit in your attachment attenuates the input by 1/3. Your assignment is to modify the circuit to attenuate the input by 1/2 and explain what is to be changed and why, then test the modified circuit.

So you have an error in you post. The "voltage divider rule V_out = (R2/R1+R2) x V_in" contains a typo. The correct expression is: "V_out = (R2/(R1+R2)) x V_in". You are missing the parentheses around the R1+R2 term.

To make a voltage divider that divides the Vin by 2 then R1 must equal R2. The simple change to the attached circuit would be to change the value of R2 from 1K to 2K ohms. But there is a side affect that total resistance of R1+R2 changes from 3K to 4K ohms.

A second way to get the same attenuation is to keep the total resistance of R1+R2 as 3K ohms and change both R1 and R2 to 1.5K ohm resistors.

A third choice is to add the idea that this is an audio signal that requires a 600 ohm impedance for this application so we should make R1 and R2 both 300 ohms each so the total resistance will be 600 ohms.


Well-Known Member
Assuming there are no trick questions hidden here, Dan has it right, that the source impedance is irrelevant. The only goal is that v_out be one half of v_in, so, by the voltage divider rule the simplest change to realize this is to modify R2 to be the same resistance as R1.

As an aside, the symbol used for the voltage source appears to be that of an ideal voltage source and an ideal voltage source, by definition, has zero output impedance. But, as Dan teaches, this is irrelevant to the task. You could use any voltage source with any output impedance and as long as R1=R2, then v_out will be one half of v_in.

As for how else it could be done, you could modify the value of both resistors to any other resistance, as long as they remain equal.
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