Hi again,
tvtech:
Yeah it has been very interesting to do this and look at the solutions. We can also have a practical thing come out of all this soon, which should satisfy the non-theory lovers
Ratch:
Yes that looks good, and also nicely illustrates how straight that exponential really is when we do a practical circuit.
Electrician:
You reading my mind?
i was just about to do something like that, and also, it would be kind of cool to come up with a formula to express the ratio of the full wave case capacitor to the half wave case capacitor. This would tell us and others what benefit we can get from going to full wave as opposed to half wave. Then we could obtain an estimate for this ratio, that would be a really practical thing (since many people want and even need that kind of solution).
So who's up for it?
You reading my mind?
i was just about to do something like that, and also, it would be kind of cool to come up with a formula to express the ratio of the full wave case capacitor to the half wave case capacitor. This would tell us and others what benefit we can get from going to full wave as opposed to half wave. Then we could obtain an estimate for this ratio, that would be a really practical thing (since many people want and even need that kind of solution).
So who's up for it?
Hi,
BTW, here's something for you guys to think about
Since, and as Electrician noted, a repetitive waveform has voltages that repeat every cycle, regardless where we look, i chose to look at the peak of the first wave bump, and call that zero time. This makes the input sinusoid:
cos(w*t)
instead of sin(w*t), although it may not matter in the long run if we adjust things later, but i'll use the cosine form for now.
So we know the voltage across the resistor is:
cos(w*t)
and the voltage across the cap is:
cos(w*t)
so the current in the resistor is:
Ir=cos(w*t)/R
and the current in the capacitor is:
Ic=C*dv/dt=C*d(cos(w*t))/dt=-w*sin(w*t)
so you see where this is going...the sum of the two currents, which is the source current, is:
Is=Ir+Ic
which of course equals:
cos(w*t)/R-w*C*sin(w*t)
and if you havent figured this out already, we set this equal to zero to find the time delay t:
cos(w*t)/R-w*C*sin(w*t)=0
The solution is easy because sin/cos is tan, so we get the same (equivalent) expression for the time again:
t=atan(1/(w*R*C))/w
Only during the time the power source is energizing the capacitor. The rest of the time the capacitor is feeding the resistor current.
Yes of course, that is what we are looking for with this equation.Same answer as above. The power source stops feeding the resistor cos(wt) when its voltage drops below the cap voltage.
Only during the time the power source is feeding the resistor and capacitor.
There are periods of time when the source current is not supplying current to either the cap or resistor. Yet the resistor is constantly receiving current from either the capacitor or power supply.
I can't figure out what you are calculating.
We are calculating exact values here. But i could compare the results to your graphical results if you like?View attachment 93166
OK, the above graph shows what is happening. The blue line is the current present in the resistor and the orange line is the current supplied by the capacitor. The current of the resistor follows the voltage of the capacitor most of the time until the source voltage kicks in and raises the current to its max value. The current leaving the cap is dependent on the capacity value and the rate of voltage change. At about 7.15 msec, the power supply takes over and the current from the cap drops rapidly from a negative value to zero while the power supply takes over supplying current to the resistor and energizea the capacitor. So the power supply only supplies current from 7.15 msec to 8.25 msec as shown by the graph.
Ratch
Hi again Ratch,
Disagree totally. The waveform is periodic regardless what is conducting. That means if you look at the wave at 0.001 seconds, then again at 0.001+1/60 seconds, the voltage will be exactly the same. In fact, that is the definition of periodic: v(t)=v(t+Tp) always.
Yes of course, that is what we are looking for with this equation.
It is again the time from the peak of the cosine wave (or sine wave) to the start of the discharge of the cap. That is the small time delay (we're calling it small because it is usually smaller than the time for the discharge, but that's not a necessity). That isnt too hard to calculate as you can see, and the time from the peak to the end of discharge isnt too hard to calculate, so by subtracting we get the actual time the capacitor is discharging, and by using the time delay itself we can calculate the coefficient k for k*e^(-t/RC) which is the true discharge curve.
Didnt you notice that this comes out to the same time delay that Electrician obtained using Laplace Transforms?
There are also a couple other ways to prove that this result and Electrician's result are accurate. So we've done it both ways now: once using Laplace and once working strictly in the time domain, and in both cases we come up with the same result.
Note that we dont have to calculate the ENTIRE time that the cap is being controlled by the source, because the time between the end of discharge and the peak is uninteresting and that is because the cap is controlled only by the source so has to free will to change the waveshape until some time after the peak. If for some reason you do want to know that time though, simply subtract the entire time we found from a full cycle:
Tp-T=1/60-T
and that will tell you the time from the start of charge to the end of charge, for what it's worth.
T=yourformula-smalltimedelayfoundabove
We are calculating exact values here. But i could compare the results to your graphical results if you like?
BTW, what value cap do you come up with for the full wave case?
I was hoping to find a relationship between the half wave case and the full wave case, as to the size of the capacitor required to get a certain ripple result.
Yes it is only for solving the time problem.You implied that voltage across the cap is cos(w*t). It is for a short time, but not during the whole period.
And especially not if the ripple is largeAs before, I am ignoring the insignificant time from the peak of the voltage source until the cap starts de-energizing. Compared to the time of the total period, it is small. Especially if the ripple is small.
Those values sound too low. Previously you got 146.8uf for the cap for half wave, using your own formula. What happened?I got 55 uf for my full-wave example and 120 uf for the half-wave example.
For which case, center-tapped transformer (two diodes) or non-center-tapped (4 diode bridge)? Furthermore, it is easily observed that the higher the input voltage, the less significant is the diode voltage drop.
Ratch
[LATER LATER]
I get 6.72418986e-5 for C with the full wave.
Yes it is only for solving the time problem.
And especially not if the ripple is large
For example, if C=1uf the delay period is around 3.2ms, which is almost the entire quarter period of 4.16667ms. So we'd see a large part of the sinusoid before we saw the discharge curve start.
Those values sound too low. Previously you got 146.8uf for the cap for half wave, using your own formula. What happened?
It is hard to ignore diode voltage drop when the source voltage is low. In my last example 0.7 volts is a significant part of 3.0 volts.We are still using ideal diode(s). If you want to use diodes with a voltage drop then that's an entirely different paradigm. We might as well consider ESR too then, and some line resistance, or diode spice model.
This is exactly correct; I forgot to apply the amplitude correction in my previous result. Here's the computation:
The ratio of the approximate half wave formula to the approximate full wave formula is a constant of about 2.1676, independent of the value of K.
I don't know, I am falling in lust with graphs. I like the BIG picture.
Ratch
Why don't you analyze on SPICE and FALSTAD and compare results to get a bigger picture and a second opinion. Assume 200Vp ac 50 Hz 1Kohm 100uF
then plot the error for each method for Vmin / Vmax which may be more important for SMPS or LDO's etc.
This is exactly correct; I forgot to apply the amplitude correction in my previous result. Here's the computation:
The ratio of the approximate half wave formula to the approximate full wave formula is a constant of about 2.1676, independent of the value of K.
But the inputs can have real values for caps, like ESR and Vf @1A for diodes . Without these, your false assumptions will ignore I square R rise in caps when the current rises 20x Avg during peak charge for 5% ripple. Voltage. Each component can be scope for VI to confirm calculations. Even true RMS power calculations or VAR. and coupling factor and turns ratio for xfmrs..... Etc etc. Falstad even has programmableThat would be like performing a calculation on two or more different calculators to see if they all give the same answer. Same with graphing programs. Of course they are going to give the same answer if the identical inputs are made. The uncertainly is whether the inputs are correct, and no amount of multiple runs are going to determine that.
Ratch
I do have to differ a little on the constant however, if we allow K to go down to say 0.5, which should be allowed. I think it varies a little too much to call it an actual constant.
But the inputs can have real values for caps, like ESR and Vf @1A for diodes . Without these, your false assumptions will ignore I square R rise in caps when the current rises 20x Avg during peak charge for 5% ripple. Voltage. Each component can be scope for VI to confirm calculations. Even true RMS power calculations or VAR. and coupling factor and turns ratio for xfmrs..... Etc etc. Falstad even has programmable
LED characteristics and colour.
Your model is too simple to be accurate.
But electronics is a small part of his list. https://www.falstad.com
What I said in post #88 is: "The ratio of the approximate half wave formula to the approximate full wave formula is a constant of about 2.1676, independent of the value of K." I was not referring to the results from "exact" computations where the delay and amplitude adjustment are taken into account.
The approximate formulas I'm referring to are the ones I've given earlier:
The ratio of results from those two formulas is a constant for all K.
Here is a plot of the actual ratio from our more intense calculations. This shows the ratio C(FullWave)/C(HalfWave) for K ranging from 0.2 to 0.95, and a 3d view for f(K,R) would show a 3d curved sheet with that plotted function as the generating function, swept horizontally along R with zero slope. So this plot should be good for any R. Might want to check it over though just in case i made a mistake.
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?