Electrical – Capacitor in series with voltage divider

capacitorresistorsvoltage divider

I am trying to understand the dynamics in the following circuit:

I have a voltage source here.

Now, I measure current at C1 and voltage at C1/R2 and the result is this:

So, what I understand so far is that the capacitor "sucks in" all the voltage, so that the voltage that I measure is really low at the beginning. At the same time, the capacitor draws in a lot of current for loading until it is full and nothing more is drawn in.

But I do not get what happens when I change resistor values. According to voltage divider rule, the capacitor gets I_in * R2/(R2+R1) as voltage. So, I have 4/(4+1) = 0.8 in the circuit below. But when I change R2 to 1k, R2/(R2+R1) is 0.5, so the voltage provided to C1 should be lower. This means, it should load slower. But it doesn't the simulation will show the peak of the green curve earlier instead of later.

So, question is: What are the dynamics here? How can I calculate how fast the capacitor charges?

I tried to understand this circuit by using Ohm's law and the voltage divider rule but just nothing seems to work here…

Best Answer

The voltage in a capacitor in series with a resistor at time t can be determined from the following formula: $$V_c=V_s (1-e^{-t/RC})$$ ...where Vs is the voltage being applied to the capacitor, R is the resistance, C is the capacitance and t is the elapsed time.

However this only works for the characteristic circuit with a source, resistance and capacitance in series.

For your circuit we need to calculate the Thevenin source and resistance equivalent. Luckily for the prior we have a simple voltage divider, so the new Vs will be 4.5V. For the resistance we need to short circuit the voltage source and look at the resistance from the load terminals which will be R1 in parallel with R2. $$R_{Th}=\frac{1}{\frac{1}{R1} + \frac{1}{R2}}$$ $$=500$$ So we now have the values to calculate the new charging characteristics. A capacitor can be considered fully charged after 5 time constants: $$\tau =RC$$ $$V_c=4.5\times(1-e^{-5})$$ $$=4.47V$$ We can see from above that the only way to reduce the charging time is to reduce the R or C values. The C value remained the same, but the new R value was 500R instead of the original 800R and thus the new circuit will charge in 250ms instead of 400ms.

Related Solutions

Electronic – Voltage Divider w/ Varying Load Current

What do I take into consideration when designing the circuit?

Think it through.

The load voltage will be highest when the current is lowest
The load voltage will be lowest when the current is highest.
You're given the high and low load current specification as well as the high and low load voltage specification.

So, you have two points on the load line. This will allow you calculate the equivalent resistance (Thevenin resistance) of the voltage divider as well as the voltage division factor \$\frac{R_2}{R_1 + R_2}\$.

Can you take it from here?

I edited the question to remove the current fluctuation part. I was just warned to keep it in mind, it wasn't part of the actual question.

Since there are two resistor values to pick, you need two independent specifications. Certainly, one of those specifications is that the voltage must be within 20% of 6V when the load current is 500mA.

If you form the Thevenin equivalent circuit of the 15V source and voltage divider resistors, the voltage across the load resistor is seen to be given by

$$15V \frac{R_2}{R_1 + R_2} - I_L\cdot R_1||R2 $$

So, you need another specification. One approach would be to limit the open-circuit voltage to be 20% higher than the nominal voltage.

$$15V \frac{R_2}{R_1 + R_2} = 7.2V$$

Now you have two equations and two unknowns:

$$\frac{R_2}{R_1 + R_2} = \frac{7.2}{15} $$

$$R_1||R_2 = \frac{7.2 - 6}{0.5A}$$

This is just an example of choosing an additional reasonable constraint in order to uniquely specify the resistor values. You might reasonably choose a different additional constraint.

Electrical – RC voltage divider

It appears that with a square wave in, the output (OUT1, below) is:

$$V_o = \frac{C_2}{C_1 +C_2} (V_n)$$

The spikes From OUT2 do have a decay time, and if you want to find out what it is, the LTspice circuit list follows:

Version 4
SHEET 1 880 680
WIRE 144 -192 64 -192
WIRE 272 -192 208 -192
WIRE 352 -192 272 -192
WIRE 272 -160 272 -192
WIRE 272 -64 272 -96
WIRE 64 80 64 -192
WIRE 144 80 64 80
WIRE 272 80 208 80
WIRE 384 80 272 80
WIRE 464 80 384 80
WIRE 384 96 384 80
WIRE 64 112 64 80
WIRE 272 112 272 80
WIRE 64 208 64 192
WIRE 272 208 272 176
WIRE 272 208 64 208
WIRE 384 208 384 176
WIRE 384 208 272 208
WIRE 64 256 64 208
FLAG 64 256 0
FLAG 272 -64 0
FLAG 352 -192 OUT1
FLAG 464 80 OUT2
SYMBOL cap 256 112 R0
SYMATTR InstName C1
SYMATTR Value 1µ
SYMBOL cap 208 64 R90
WINDOW 0 0 32 VBottom 2
WINDOW 3 32 32 VTop 2
SYMATTR InstName C2
SYMATTR Value 1µ
SYMBOL voltage 64 96 R0
WINDOW 0 36 94 Left 2
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value PULSE(0 10 .1 1u 1u .1 .2)
SYMBOL res 368 80 R0
SYMATTR InstName R1
SYMATTR Value 10
SYMBOL cap 256 -160 R0
SYMATTR InstName C3
SYMATTR Value 1µ
SYMBOL cap 208 -208 R90
WINDOW 0 0 32 VBottom 2
WINDOW 3 32 32 VTop 2
SYMATTR InstName C4
SYMATTR Value 1µ
TEXT 30 280 Left 2 !.tran 1

Best Answer

Related Solutions

Electronic – Voltage Divider w/ Varying Load Current

Electrical – RC voltage divider

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