Mastering RC Voltage Calculation for Chopped/Full-Wave Rectified AC Waveforms

In summary: Yes, the simplified cases are applicable. I have been measuring the time it takes to charge the capacitor up to a certain voltage, and I have found that it takes approximately 1.5-2.0 cycles to reach the desired voltage. My results so far indicate that the simplified cases are accurate for this type of situation.
  • #1
dsa
11
0
I am trying to figure out how to come up with an equation to calculate the voltage
across a capacitor for an RC circuit that is supplied by a chopped / full-wave rectified ac waveform shown below.

The shaded portion is the actual input signal to the RC circuit.


anybody have any ideas ?
 

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  • #2
Is this a homework/coursework question? If so, I can move this thread to the Homework Help forums.

Welcome to the PF, dsa. If the V(t) waveform is meant to represent the output of a full-wave rectified version of the chopped AC mains, then you need to include the bridge diodes in the schematic, because they will keep the lower portions of the input waveform from discharging the capacitor. If there is no load connected to the output wire that you show, then the voltage on the capacitor will be a peak hold. That is, the voltage will be the max of the input waveform, minus the two series diode drops of the input bridge in conduction. If there is a load connected to the cap, then the cap voltage will likely charge up to the peak voltage, and then droop some between half-cycles based on the load impedance.
 
  • #3
berkeman said:
Is this a homework/coursework question? If so, I can move this thread to the Homework Help forums.

Welcome to the PF, dsa. If the V(t) waveform is meant to represent the output of a full-wave rectified version of the chopped AC mains, then you need to include the bridge diodes in the schematic, because they will keep the lower portions of the input waveform from discharging the capacitor. If there is no load connected to the output wire that you show, then the voltage on the capacitor will be a peak hold. That is, the voltage will be the max of the input waveform, minus the two series diode drops of the input bridge in conduction. If there is a load connected to the cap, then the cap voltage will likely charge up to the peak voltage, and then droop some between half-cycles based on the load impedance.

Berkeman, this is not a homework question, but part of a design for a circuit I am doing at work.

I understand all of the above mentioned, I was tryin to isolate my problem area.

I am trying to come up with an equation to determine how long it will take to charge the capacitor up to a certain voltage, not the peak. The output wire has voltage detection circuitry with very high impedance so the load can be ignored.
 
  • #4
Okay, then you will need to include the source impedance of the voltage source and the input full wave bridge diodes in your model, in order to get an accurate calculation. Then set up the equations for the input current as a function of time into the capacitor, and integrate that current over time to calculate the capacitor voltage. Does that make sense?
 
  • #5
Somewhat...
I do not really care about super exact calculations, I know roughly what the times are from experimentations. I am just trying to come up with a general formula for V(t)
across the cap.
 
  • #6
dsa said:
Somewhat...
I do not really care about super exact calculations, I know roughly what the times are from experimentations. I am just trying to come up with a general formula for V(t)
across the cap.

For the general equation (including the source impedance and bridge diode drops), you need to use integrals and the differential equation for the current into the capacitor versus the capacitor voltage. Are you comfortable setting up those integral and differential equations?

For simplified cases (-1- Cap charge-up takes less than one quarter cycle, or -2- Cap charge-up takes many cycles), simple algebraic equations can be derived. Are either of the simplifying cases applicable? What are your experimental results so far?
 

1. How do I calculate the voltage drop in an RC circuit?

The voltage drop in an RC circuit can be calculated using the formula V = V0e-t/RC, where V is the voltage at time t, V0 is the initial voltage, R is the resistance, and C is the capacitance.

2. What is the difference between voltage drop and voltage across a capacitor in an RC circuit?

Voltage drop refers to the decrease in voltage across a resistor, while voltage across a capacitor refers to the voltage stored on the capacitor. In an RC circuit, the voltage drop across the resistor is equal to the voltage across the capacitor at any given time.

3. How do I choose the correct resistor and capacitor values for an RC circuit?

The values of the resistor and capacitor in an RC circuit depend on the desired time constant and the input voltage. The time constant, RC, determines the rate at which the capacitor will charge or discharge. To choose the correct values, you can use the formula RC = T, where T is the desired time constant. The input voltage will also affect the charging and discharging rates, so it is important to consider this when selecting values.

4. Can I use the same formula to calculate voltage drop in a series and parallel RC circuit?

No, the formula for voltage drop in a series RC circuit is different from that of a parallel RC circuit. In a series circuit, the voltage drop across each component is equal to the total voltage of the circuit. In a parallel circuit, the voltage drop across each component is equal to the input voltage. Therefore, the formulas for voltage drop in these circuits are different.

5. How does the frequency of the input voltage affect the voltage drop in an RC circuit?

The frequency of the input voltage has a direct effect on the voltage drop in an RC circuit. As the frequency increases, the voltage drop across the capacitor decreases, and the voltage drop across the resistor increases. This is because at higher frequencies, the capacitor has less time to charge and discharge, resulting in a smaller voltage drop. This relationship can be seen in the impedance equation of an RC circuit, Z = √(R² + (1/ωC)²), where ω is the angular frequency.

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