I have simulated the circuit shown in the image below using Multisim.

enter image description here

I understand that when the pulse is high, when t = 0, or a very small change in time, the voltage across the capacitor, Vc, is 0. I also understand that the Resistor current, Ir = 5V/100k = 5e-5A., Then after roughly 5 tau, the voltage across the cap is equal to 5V and the current shall approach 0.

I have attached the chart of the simulation below. enter image description here

I have circled a segment in the above chart where the current dips to a negative value. I don't understand why it is doing that. Can anybody help me understand that part?

NOTE: I have used a current clamp as shown in the simulation diagram. The current clamp converts a current in a wire to a voltage. In doing so, it allows the user to set a voltage to current ratio. The ratio I chose was 20V/mA to help see the signal better on the chart.

I am adding the chart for the discharge section for which I am trying to do the math on...

enter image description here

enter image description here

  • \$\begingroup\$ Hint: what's the relationship between voltage and current for a capacitor? \$\endgroup\$
    – The Photon
    Commented Oct 27, 2019 at 4:38
  • \$\begingroup\$ i = Cdv/dt? Right? \$\endgroup\$ Commented Oct 27, 2019 at 4:39
  • \$\begingroup\$ And what's happening to the capacitor voltage in the time period you circled? \$\endgroup\$
    – The Photon
    Commented Oct 27, 2019 at 4:46
  • \$\begingroup\$ discharging. so from the simulation, i can see that dv = -4.996, dt = 999ps.if i plug it in: i = (3e-6F)*(-4.9996V/999ps) = -14993.51. But the simulation is showing me -840.904mV. When i try to convert the -840.904mV back to current, i do the following: -840m*(20V/1mA) = -16800. My calculation is off... \$\endgroup\$ Commented Oct 27, 2019 at 5:00
  • 1
    \$\begingroup\$ @DeeTee, Channel A is the voltage across the voltage source. For the voltage across the capacitor (and how fast it's changing) you should be looking at channel B. \$\endgroup\$
    – The Photon
    Commented Oct 27, 2019 at 15:51

2 Answers 2


Where is the problem?

If the capacitor is charged from \$0V\$ (inital voltage) to \$5V\$ via \$100kΩ\$ resistor with the time constant \$T = RC = 0.3s \$. ,immediately after we connect the supply voltage a current will start to flow. And becouse all this voltage from the power supply appears across the resistor (empty capacitor \$V_C = 0V\$ and KVL \$V_{IN} = V_R + V_C\$) this inital current is equal to \$I = \frac{5V}{100kΩ} = 50 \mu A \$.

This is why we can say that the empty capacitor act just like a short circuit.

As the capacitor begins to charge, the voltage across the capacitor begins to increase. At the same time, the current flow begins to decrease. Why? Because the voltage across the capacitor begins to increase. And the voltage across resistor must decrease by the same amount (KVL in action). This is why charging current (I = (Vin-Vcap)/R) begins to decrease as the voltage across the capacitor rises.

So, after \$0.5s\$ the capacitor voltage will reach this value:

$$V_C = V_{IN}(1 - e^{\frac{-t}{RC}}) = 5V(1 - e^{\frac{-0.5s}{0.3s}})= 4.05562V $$

$$I_{(0.5s)} = \frac{5V - 4.05562V}{100kΩ} = 9.4 \mu A$$

At this time the \$V_{IN}\$ "switches" to \$0V\$ and the capacitor will start a discharge process. Now the capacitor "act" just like a voltage source. And we connect this voltage (the capacitor) directly across \$R\$ resistor. Hence the discharge current is flowing in the opposite direction than in the charging phase. And this is why during this phase the voltage across the resistor is negative. Because the current is now flowing in the opposite direction.

Initially this discharge current is large (\$I = 4.05562V/100kΩ = 40.55µA\$)

enter image description here

But as time goes on the capacitor continues to discharge process (via resistor ) and the voltage across the capacitor decreases exponentially with the time constant \$RC = 0.3s\$.

So, after 1s the capacitor voltage reaches this value:

$$V_C = V_{START}\: e^{\frac{-t}{RC}} = 4.05562Ve^{\frac{-0.5s}{0.3s}}=0.766V$$

And the current \$I_{(1s)} = \frac{0.766V}{100kΩ} = 7.66 \mu A\$

But now again the input voltage source switches back to \$5V\$.

And the capacitor will again start the charging phase (because Vin > Vc). At this very first moment of transient (when Vin raeches 5V) the charging current becomes is:

\$I_(1s) = \frac{5V - 0.766V}{100kΩ} = 42.3 \mu A \$

At the end of a charging prosess (t = 1.5s) the capacitor voltage will be equal to:

$$V_C = ({V_{\infty}} - V_{START} )(1 - e^{\frac{-t}{RC}})+V_{START} = (5V - 0.766V )(1 - e^{\frac{-0.5s}{0.3s}})+0.766V = 4.2V $$

and the current \$I_{(1.5s)} = \frac{5V-4.2V}{100kΩ} = 8 \mu A\$

Because again at \$t = 1.5s\$ the \$V_{IN}\$ switches again back to \$0V\$ and the discharge process begins (I_dis = 4.2V/100kΩ = 42µA).

And ends up at \$t = 2s\$ with the cap voltage: $$V_C = V_{START}\: e^{\frac{-t}{RC}} = 4.2Ve^{\frac{-0.5s}{0.3s}}=0.793V$$

And this is how the voltage across the cap will looks like:

enter image description here

And the current

enter image description here

And I hope that now you see what is going on in this simple circuit. AC Circuit Having Only Capacitor

  • \$\begingroup\$ Can you tell me what software you used for your diagrams and charts? \$\endgroup\$ Commented Oct 27, 2019 at 19:22
  • \$\begingroup\$ @DeeTee LTspcie for simulation and MSpaint + my personal lib for symbols. \$\endgroup\$
    – G36
    Commented Oct 27, 2019 at 19:34
  • \$\begingroup\$ thank you for sharing! \$\endgroup\$ Commented Oct 27, 2019 at 19:39

Charging the cap has to be done with positive current and discharging has to be done with a negative current.

So where is the Problem? What is it exactly that you don‘t understand?


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