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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.

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  • \$\begingroup\$ Do you have a measurement instrument that is accurate to 0.002 V on a range of (at least) 5V? \$\endgroup\$ – Wouter van Ooijen Feb 22 at 16:50
  • \$\begingroup\$ 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? \$\endgroup\$ – Elliot Alderson Feb 22 at 16:51
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    \$\begingroup\$ @CompuScie: What are you playing at with your question edits? \$\endgroup\$ – Transistor Feb 22 at 16:59
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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:

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

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

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

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