# How does this DC offset sinwave input work?

I have learned that this circuit will move the sin wave up to a DC points due to the voltage divider, that is:

V_out = V_in + Vcc*R2/(R1+R2)

I was using LTSpice to simulate this and it works, I try to prove this myself and I am stuck. Here is my attempt:

Using the first Kirchhoff’s first law (in Lapalace form): $$Cs(V_{in}-V_{out}) + \frac{Vcc/s - V_{out}}{R1} + \frac{0 - V_{out}}{R2} = 0$$ $$V_{out} = \frac{R1R2Cs}{R1 + R2 + R1R2Cs}V_{in} + \frac{R2}{R1 + R2 + R1R2Cs}(Vcc/s)$$ Now I need to manipulate above equation to: $$V_{out} =V_{in} + \frac{R2}{R1 + R2}(Vcc/s)$$ But I can't.

• May l know the name of the tool you are using to write the equations Commented Sep 8, 2021 at 2:35
• @V.V.T ok, I understand that, but in my equation, I consider Vcc is a fixed DC voltage that have Vcc (volt)
– Dat
Commented Sep 8, 2021 at 3:08
• @HARITO I use simple code, it is Mathjax, you can search for it on Stackexchange
– Dat
Commented Sep 8, 2021 at 3:13
• @James I concern about the V_out value, not currents
– Dat
Commented Sep 8, 2021 at 6:43
• Also note you're wasting 0.5W in those resistors : you may want to increase them.
– user16324
Commented Sep 8, 2021 at 12:56

You can simplify things a bit:

simulate this circuit – Schematic created using CircuitLab

(I'm sure you already know how to work out the Thevenin equivalent values for the right-hand side, above.)

From the perspective of $$\V_{_\text{TH}}\$$, this is a low-pass filter. From the perspective of $$\V_{_\text{IN}}\$$, this is a high-pass filter.

Using KVL, you should be able to achieve:

$$V_{_\text{OUT}}=V_{_\text{IN}}\overbrace{\left[\frac{s}{s+\omega_{_0}}\right]}^{\text{1st order high pass}\\\:\:\:\text{standard form}}+V_{_\text{TH}}\overbrace{\left[\frac{\omega_{_0}}{s+\omega_{_0}}\right]}^{\text{1st order low pass}\\\:\:\:\text{standard form}}$$

where $$\\omega_{_0}=\frac1{R_{_\text{TH}}\,C}\$$.

It's not difficult. The starting equation is just $$\V_{_\text{OUT}}=\frac{V_{_\text{IN}}\,R_{_\text{TH}}+V_{_\text{TH}}\,Z_{_\text{C}}}{R_{_\text{TH}}+Z_{_\text{C}}}\$$. The above is the result.

For very high frequencies, the second term goes away and only the first term remains. For very low frequencies, the first term goes away and only the second term remains. So, $$\V_{_\text{OUT}}=\overbrace{V_{_\text{TH}}}^{\text{DC}} + \overbrace{\vphantom{V_{_\text{TH}}}v_{_\text{IN}}}^{\text{AC}}\$$.

• I had feeling that this result is only true for high frequencies. I am stuck at final step because I don't think about filters. These are two independent inputs with two independent filters, simply plus them to get the result. Great answer!
– Dat
Commented Sep 8, 2021 at 8:05
• @user253751 That's the standard book-way of thinking about it. In fact, that's exactly what I expressed in the starting equation, shown in my answer. But it's the other ways that sometimes help more, I think. They break it down in a different way that can help "see" in ways that a "frequency-dependent" divider cannot achieve for many. Changing perspectives both simplifies things, as well as expands the mind, too.
– jonk
Commented Sep 8, 2021 at 8:41