# In an RC circuit, how do you explain the fast rise of voltage at a point close to source? (Weak pull up)

In the above figure, the voltage at the intermediate point rises quickly whereas the voltage after R2 is increases slowly ( as governed by the capacitor voltage.)

I am not getting the intuitive understanding on why/how the 'intermediate' voltage can rise quickly even though there is a capacitor in series to prevent the fast rise in voltage.

I can imagine the extreme cases => R1=0 => intermediate voltage = source volatge and the node rises quickly to source voltage, but I attribute it to the effect of source to pull the node voltage quickly. When there is a resistor in between source and capacitor, how do we know which effect predominates? Pull up of source or slow rise of capacitor voltage?

I understand that V(vc)'s slow rise is clear to you, as capacitor C1 is charged via (R1 + R2).

R1 + R2 can be seen as a simple resistive voltage divider for the voltages between the source V(12v5_pri) and the capacitor V(vc), thus, V(intermediate) is linearly between those two:

• V(intermediate) = [R2 * V(12v5_pri) + R1 * V(vc)] / (R1 + R2);

If R2 is dominating (>>R1, like in your example) it's nearer to V(12v5_pri); if R1 is dominating, it's nearer to V(vc).

Since you want to intuitively understand the circuit, I will suggest a few intuitive approaches.

## "Pulling" resistors

When there is a resistor in between source and capacitor, how do we know which effect predominates? Pull up of source or slow rise of capacitor voltage?

In fact, you came up with the idea yourself by imagining the circuit as two voltage sources (V1 and VC) "pulling" the common midpoint through two resistors - the "pull-up" R1 and "pull-down" R2. Each of the voltage sources tries to set its voltage at midpoint. Since VC does not "move" when V1 "jumps", the midpoint also "jumps" but weaker.

## "Shifted" voltage divider

As other said, R1 and R2 constitute a voltage divider. This is an ordinary voltage divider only in the first moment when V1 "jumps". After that, this is a "shifted" voltage divider since the lower end of R2 is not grounded but is slowly "moving".

## Voltage summer

Actually, the network of two resistors in series act as a voltage summer with weighted inputs (it can be easily analyzed by the superposition principle). So the V1 "jump" appears at the summer's output (the midpoint).

## Voltage diagram

But the most interesting representation would be to "open" the resistors to "see" the voltage at each point of the resistive film. I started doing it (real and mental, using my imagination as a "simulator") from 1990 until now. I have attached below my authentic recordings from that time (unfortunately in Bulgarian but I have translated them below the pictures).

Step 1. The idea was simple - to represent each local voltage by a segment of proportional length; thus I got a "voltage diagram" as a set of bars.

Fig. 1. A resistor visualized by a voltage diagram.

Step 2. For simplicity, we can depict only the envelope of the diagram (as a triangle).

Fig. 2. What is voltage diagram? * voltage distribution along the resistive film; * spatial diagram; * voltage as a function of position relative to the origin - V = f(l)

Step 3. It is interesting that, 30 years ago, I asked myself the same question about the RC integrating circuit and visualized it using such a voltage diagram. So, when the input voltage V1 (E in the picture) "jumps", all local voltages along the resistive film instantly and proportionally rise from right to left.

Fig. 3. RC integrating circuit. Voltage diagram of the resistor when V varies: We change V and Vc strives to "catch up" with it, to be equal to it, to obtain an equipotential voltage diagram.

Step 4. Then, when VC (Uc) slowly rises, all local voltages slowly and proportionally rise from left to right.

Fig. 4. Voltage diagram of a charging RC integrating circuit. The voltage at the left end of the resistor "rises" sharply. At the first instant, the voltage at the right end is zero because C is uncharged. The current I is maximum. Then the voltage Vc starts to rise and the voltage diagram rises (rotates around point A) until it takes an upper horizontal position (Vc = V). The current flows from left to right.

Step 5. Finally, the capacitor is charged and all local voltages along the resistive film are equal to V1 (E).

Fig. 5. Voltage diagram of an RC circuit at the end of the process. The capacitor C is fully charged to the source voltage V. The voltages at both ends of the resistor are the same. The pressure diagram is an equipotential line. No current flows through the circuit.

Step 6. When V1 (E) becomes zero again, the capacitor becomes a "voltage source" and begins discharging through V1 to ground.

Fig. 6. Voltage diagram of an RC circuit under discharge: At the beginning, the capacitor voltage Vc is equal to the input voltage V. The input voltage suddenly drops to zero but the voltage at the right end of the resistor remains equal to V. A current begins to flow in the opposite direction - at the first moment I = V/R but then it decreases and the capacitor discharges. The voltage diagram begins to descend from position AB and reaches its lowest zero position AC. At the end of the process, the capacitor discharges completely and the current stops.

Step 7. A more special case is when V1 (E) alternatively varies and the capacitor stays uncharged.

Fig. 7. AC virtual ground of an uncharged capacitor. When we "wiggle" the inpout voltage V above and below ground, the capacitor stays uncharged (Vc only slightly "viggles"). Point A has practically zero voltage (ground)... something like "AC virtual ground".

The voltages after R1 and R2 rise exactly with the same time constant (the time to reduce the remaining discrepancy by a factor of $$\e\$$). It's just that the voltage after R1 has a lot less distance to cover. But once you remove the DC offset and scale the voltages to match, they are identical. The voltage does not rise any faster after R1, rather it jumps immediately most of the distance when switching on the voltage. For time point of the jump, you can consider the capacitor a short (since the voltage across it had no opportunity to rise yet) and calculate the voltages right after the jump with the capacitor being shorted. Afterwards, voltage over the capacitor rises with a time constant $$\(R_1+R_2)\,C_1\$$ and the voltage over the resistors is reduced accordingly, split across the resistors according to their value.

The time constant of the change is the same everywhere, just the DC offset and the constant factor with which you see it may differ according to the resistance network.

I am not getting the intuitive understanding on why/how the 'intermediate' voltage can rise quickly even when there is a capacitor in series to prevent the fast rise in voltage.

Imagine that capacitor had infinite capacitance. It would behave like a short circuit. Do you agree? If it behaves like a short circuit then all that remains is a resistive potential divider like this: -

And, I'm sure you can agree that a resistive divider (as shown) will be able to replicate the input voltage (including input dynamics) at nearly the same amplitude (75/76.5) at the node you have named "intermediate".

So, why should rise times be any different with a smaller value of capacitance?