# Voltage in an RC Circuit

I made an RC curcuit on a breadboard gave it a of 25V measured the voltage across it with an oscilloscope. During my experiment it has come to my attention that the peak voltage level was getting over 25V by a very small factor during the pulse switches ie. in 5T time below the graph. The value was 25.002V. Why would this occur, the maximum voltage i can supply to my capacitor is 25V? It doesnt make sense to me.

• Do you have a measurement instrument that is accurate to 0.002 V on a range of (at least) 5V? – Wouter van Ooijen Feb 22 '19 at 16:50
• What makes you think your square wave was switching between 0.0000V and 5.0000V? What makes you think your oscilloscope is really accurate to 2mV at the scale you were using? – Elliot Alderson Feb 22 '19 at 16:51
• @CompuScie: What are you playing at with your question edits? – Transistor Feb 22 '19 at 16:59

Well, when we have a series RC-circuit we can use Laplace transform to analyse it in detail. Using Faraday's law we can write:

$$\text{v}_\text{s}\left(t\right)=\text{v}_\text{R}\left(t\right)+\text{v}_\text{C}\left(t\right)\tag1$$

Using the relations of the voltage and current in a resitor and a capacitor we can rewrite equation $$\(1)\$$ as follows:

$$\text{v}_\text{s}'\left(t\right)=\text{i}_\text{R}'\left(t\right)\cdot\text{R}+\text{i}_\text{C}\left(t\right)\cdot\frac{1}{\text{C}}\tag2$$

Because it is a series circuit we know that the input current, $$\\text{i}_\text{in}\left(t\right)\$$, is the same as the current trough the resistor and the capacitor so we can write:

$$\text{v}_\text{s}'\left(t\right)=\text{i}_\text{in}'\left(t\right)\cdot\text{R}+\text{i}_\text{in}\left(t\right)\cdot\frac{1}{\text{C}}\tag3$$

Using the Laplace transform and assuming that the intial conditons are equal to $$\0\$$ we can write for equation $$\(3)\$$:

$$\text{s}\cdot\text{V}_\text{s}\left(\text{s}\right)=\text{s}\cdot\text{I}_\text{in}\left(\text{s}\right)\cdot\text{R}+\text{I}_\text{in}\left(\text{s}\right)\cdot\frac{1}{\text{C}}\space\Longleftrightarrow\space\text{I}_\text{in}\left(\text{s}\right)=\frac{\text{s}\cdot\text{V}_\text{s}\left(\text{s}\right)}{\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\tag4$$

Writing the supply voltage in the s-domain we get:

$$\text{V}_\text{s}\left(\text{s}\right)=\frac{1}{1-\exp\left(-10\text{T}\text{s}\right)}\cdot\int_0^{5\text{T}}\hat{\text{u}}\cdot\exp\left(-\text{s}t\right)\space\text{d}t=\frac{1}{\text{s}}\cdot\frac{\hat{\text{u}}\exp\left(5\text{s}\text{T}\right)}{1+\exp\left(5\text{s}\text{T}\right)}\tag5$$

So, for the input current we get:

$$\text{I}_\text{in}\left(\text{s}\right)=\frac{\text{s}}{\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\cdot\frac{1}{\text{s}}\cdot\frac{\hat{\text{u}}\exp\left(5\text{s}\text{T}\right)}{1+\exp\left(5\text{s}\text{T}\right)}=\frac{1}{\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\cdot\frac{\hat{\text{u}}\exp\left(5\text{s}\text{T}\right)}{1+\exp\left(5\text{s}\text{T}\right)}\tag6$$

So, the voltage across the capacitor is given by:

$$\text{V}_\text{c}\left(\text{s}\right)=\frac{1}{\text{s}\cdot\text{C}}\cdot\frac{1}{\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\cdot\frac{\hat{\text{u}}\exp\left(5\text{s}\text{T}\right)}{1+\exp\left(5\text{s}\text{T}\right)}\tag7$$

• Really informative. – Compu Scie Feb 25 '19 at 12:17
• @CompuScie You're welcome, I'm glad that I could help! – Jan Feb 25 '19 at 12:53

It doesn't make sense to anyone, I suspect, unless you have a measurement error.

• Check that your scope is on DC setting.
• Check that if you are using 10:1 probes that the trim capacitor adjustment is correct.
• Put both probes on the squarewave and confirm that they are identical.

In the end you are expecting about 4.965 V and seeing 5.002 V. That's an error of 0.7%. How does that compare with the scope specification?

5V versus 5.002V level is almost for sure to be measurement error or calibration differences between your scope probes or scope input channels.