Hello there,
We've all seen what a half wave rectifier circuit looks like, but this one is ideal. By that i mean that every component in it is completely ideal:
Source: zero impedance, zero tolerance
Cap: zero ESR, zero ESL, zero tolerance
Resistor: Exact value, zero tolerace
Diode: Perfect short when 'on', perfect open when 'off.
The interesting question is:
What is the value of C required to obtain exactly 10 percent ripple, or put another way, the maximum discharge of the capacitor is 100*K in percent? (Use K=0.9 for 90 percent, as a numerical example).
The input is 1vac peak, the value of the resistor is 1000, the line frequency is 60Hz.
Since the input is 1vac peak, that means the capacitor is only allowed to discharge to 0.9 volts for K=0.9 and 0.8 for K=0.8, etc., so K is a variable but we can use 0.9 for now if you like. Actually R is a variable too, but we can use 1000 ohms for now if you like. The frequency is also a variable but we'll use 60Hz for now too if you like.
So either use variables K,R,F, or use those numerical values above, to calculate the exact value of C to say 6 decimal places or better.
To help show how precise we want this, if the cap discharges to 0.90002 when the next charge cycle starts that's too high, or if it discharges to 0.89998 that's too low, but acceptable might be 0.90001 or 0.89999, and i want to get this out of the way before we start thinking about capacitor tolerances and stuff like that, because in the problem description we dont allow any tolerances so it has to be exact to at least 6 decimal places, but better would be an equation to solve for this capacitor value.
I had solved this in my own way and am looking for other solutions to compare with. Note that this is a purely theoretical question, one that might be asked in a classroom, not a practical application question where there are tolerances in everything we deal with.
Should you decide to accept the challenge i want to wish you good luck. For such as simple circuit that we've all dealt with many times in the past, it's not quite a simple as it seems, maybe that is what makes it so interesting
Alternately, come up with a simulation that is precise.
We've all seen what a half wave rectifier circuit looks like, but this one is ideal. By that i mean that every component in it is completely ideal:
Source: zero impedance, zero tolerance
Cap: zero ESR, zero ESL, zero tolerance
Resistor: Exact value, zero tolerace
Diode: Perfect short when 'on', perfect open when 'off.
The interesting question is:
What is the value of C required to obtain exactly 10 percent ripple, or put another way, the maximum discharge of the capacitor is 100*K in percent? (Use K=0.9 for 90 percent, as a numerical example).
The input is 1vac peak, the value of the resistor is 1000, the line frequency is 60Hz.
Since the input is 1vac peak, that means the capacitor is only allowed to discharge to 0.9 volts for K=0.9 and 0.8 for K=0.8, etc., so K is a variable but we can use 0.9 for now if you like. Actually R is a variable too, but we can use 1000 ohms for now if you like. The frequency is also a variable but we'll use 60Hz for now too if you like.
So either use variables K,R,F, or use those numerical values above, to calculate the exact value of C to say 6 decimal places or better.
To help show how precise we want this, if the cap discharges to 0.90002 when the next charge cycle starts that's too high, or if it discharges to 0.89998 that's too low, but acceptable might be 0.90001 or 0.89999, and i want to get this out of the way before we start thinking about capacitor tolerances and stuff like that, because in the problem description we dont allow any tolerances so it has to be exact to at least 6 decimal places, but better would be an equation to solve for this capacitor value.
I had solved this in my own way and am looking for other solutions to compare with. Note that this is a purely theoretical question, one that might be asked in a classroom, not a practical application question where there are tolerances in everything we deal with.
Should you decide to accept the challenge i want to wish you good luck. For such as simple circuit that we've all dealt with many times in the past, it's not quite a simple as it seems, maybe that is what makes it so interesting
Alternately, come up with a simulation that is precise.