OK, but empirical methods should have some backup theory to instill confidence in their application.
Ratch
It was my own discovery, so I have not seen it anywhere else, results verified by myself as both R&D and Test Engineer at the time.
Your case of 10% resulted in a choice of RC= 146ms for a half bridge at 20ms, so your result is RC=7.5T vs mine RC=10T
Charge interval defined above is for a full-wave rectifier. My example used half-wave rectifier.
How do you get the relation 10% ripple means RC=10 * charge interval?
Ratch
Hello,
This would come from a completely linear approximation of an RC filter:
dv/dt=1/RC
The deviation would be:
dv=dt/RC
so if you want only 10 percent droop in dt time you need RC to be 10 times dt.
dv=dt/RC
0.10=1/10
0.10=2/20
0.10=N/(10*N)
0.20=N/(5*N)
0.02=N/(50*N)
0.01=N/(100*N)
OK, I understand how that table is computed. But look at my example. You can see from the graph that a 0.020 sec time constant allows a dip to 0.499. According to the method you demonstrated, to go from a 0.50 to 0.60 to 0.70 to 0.80 to 0.90 dip requires a 4 times lengthening of the time constant to 0.080 sec. Yet, my computation is 0.147 sec. Can you discern where the difference lies?
Ratch
0.01 1667 1621
0.02 833 799
0.03 556 526
0.04 417 390
0.05 333 309
0.06 278 254
0.07 238 216
0.08 208 187
0.09 185 165
0.10 167 1470.10 167 147
0.20 83 67
0.30 55 40
0.40 41 27
0.50 33 20
0.60 27 14
0.70 23 11
0.80 20 8
0.90 18 50.005 3333 3272
0.010 1667 1621
0.015 1111 1072
0.020 833 799
0.025 667 635
0.030 556 526
0.035 476 448
0.040 417 390
0.045 370 345
0.050 333 309Hello again,
Very sorry, I am not sure what you mean by "go form 0.50 to 0.60...etc.".
The computation Tony is using is:
RC=dt/(percent_ripple/100)
where percent_ripple=10 for ten percent, and where dt is the time from the back end of the previous cycle to the front end of the next cycle, just as in your formula. Since your formula comes up with a good estimate of the discharge time, we use that:
(2*pi-acos(f))/(2*pi)=0.0155 seconds
so in the formula Tony used, we need a time constant RC of:
RC=10*0.0155
if we want to see a ripple of 10 percent for example. This leads to:
RC_Tony=0.155
so that means:
C_Tony=0.155/1000=155uf
In your complete formula since your time came out to 0.0155 we have:
0.0155=-ln((100-10)/100)*R*C
or:
0.0155=-ln(0.9)*R*C
which equals:
0.0155=0.10536*R*C
and dividing both sides by 0.10536 gives us:
0.147=R*C
and with R=1000 we have;
C_Ratchit=147uf
So it is 155uf vs 147uf, so the linear approximation isnt too far off as long as we use the right time period to begin with. If we instead estimate it to be from one peak to the other, we would have:
RC=0.0167*10
so we would come out with 167uf, which is a little larger than 155uf.
So all three formulas come out with a reasonable estimate.
For small percentages we have the following comparison table, arranged as percent ripple as a fraction, C_Tony, C_Ratch (all C in uf):
Code:0.01 1667 1621 0.02 833 799 0.03 556 526 0.04 417 390 0.05 333 309 0.06 278 254 0.07 238 216 0.08 208 187 0.09 185 165 0.10 167 147
So 5% ripple yields caps 333uf and 309uf.
As the percentage ripple increases we see more and more difference between the two:
Code:0.10 167 147 0.20 83 67 0.30 55 40 0.40 41 27 0.50 33 20 0.60 27 14 0.70 23 11 0.80 20 8 0.90 18 5
Also, notice how the second cap values in this next table scale almost linearly:
Code:0.005 3333 3272 0.010 1667 1621 0.015 1111 1072 0.020 833 799 0.025 667 635 0.030 556 526 0.035 476 448 0.040 417 390 0.045 370 345 0.050 333 309
In particular, note 309 is roughly 2*150, and 635 is roughly 2*2*150. Thus suggests a compromise strategy based on computing one value (like 150uf) with the more accurate formula, then using that to scale for other percent ripple.
Dont worry, when one ignores Cap ESR , diode drop and many other factors such as the change in Vdc from no load to full load in addition to ripple....`Thanks for the detailed explanation. It is good to see that both methods are in modest agreement. The only objection I have is the reference to my method being an "estimate". I believe my method calculates an exact value unless other factors act to change it.
Ratch
Thanks for the detailed explanation. It is good to see that both methods are in modest agreement. The only objection I have is the reference to my method being an "estimate". I believe my method calculates an exact value unless other factors act to change it.
Ratch
Hi,
You have to be careful what you call "exact". Tony sited a few practical considerations, but there are also other theoretical considerations assuming a perfect setup.
Your method is based on the capacitor discharge ending at the front of the next wave which is good, but it is also based on the capacitor starting to discharge at the EXACT PEAK of the previous wave, which is impossible except with infinite R, and then it doesnt really discharge anyway, so we might say that as R approaches infinity the cap discharge starts infinitesimally close to the peak.
One way to look at the start of the discharge is to consider that C starts to discharge into R when the line voltage gets below the cap voltage, but since the cap voltage is constantly decreasing at a certain rate, if the rate is faster than the rate that the cosine wave is decreasing then the cap voltage will follow the cosine voltage downward just as it did when it followed it upward on the charge. It's not governed by the input wave itself anymore, but by the wave AND the discharge of the cap. Since near the peak the cosine wave is decreasing very slowly, any discharge in the cap will tend to keep the cap voltage still following the cosine wave, for a short time. Thus, the cosine wave and the cap voltage fall at the same rate for a short time. In this view, once the cap voltage discharge rate becomes slower than the cosine wave decrease rate, the cap voltage wave 'leaves' the cosine wave and the diode stops conducting, and this is the typical e^-t/RC wave we see. So in mathematical form this view looks like this:
d[cos(w*t)]/dt=d[e^(-t/RC)]/dt
and solve this for 't'.
This means two things:
1. The time dt between start of discharge and start of recharge is less than the formula.
2. The LEVEL of the voltage at the start of the discharge is LESS than the peak.
Both of these change the droop somewhat, depending on the value of the capacitor. For lower values of C, the RC wave will follow the cosine wave longer.
Because of #2 the equation has to be improved to:
d[cos(w*t)]/dt=K*d[e^(-t/RC)]/dt
where K is also an unknown, and K<1.
Another view is to calculate the wave across the capacitor using a tiny resistance between the source and the cap, then write the equations and solve for the time that the cap discharge leaves the back end of the cosine wave. There are probably other ways to do this too.
Numerically for our example, the decrease in total dt for a 147uf cap comes out to approximately 48us, and for 70uf about 107us. Decreasing dt by these amounts would lead to a more exact computation, assuming K=1 for simplicity.
For extremely low values of C we'd see a big difference in dt though.
MrAl,
I found this interesting link https://scholarcommons.usf.edu/cgi/viewcontent.cgi?article=4856&context=ujmm which appears to somewhat follow my method. However, the author also disregards the fact that the de-energizing of the capacitor is dependent on the voltage differences between the power supply and the present voltage across the capacitor. I believe that applying the Laplace method by using the transform of a repetitive half-wave sinusoidal will give an exact solution. However, the inverse Laplace transform is very complicated involving several Ʃ's and other nasty terms. I will work on that when I get time.
Colin,
Yes, electronic pre-regulation or full electronic regulation is probably the best practical way to get more from less.
Ratch
MrAl,
I was able to compute a Laplace transform for a half-voltage source, but it did not help because the capacitor de-energizes through the voltage source when the source drops below the capacitor's energized value. A diode is needed, of course, but since a diode is a nonlinear device, and Laplace transforms are for linear circuits, I am stymied. I don't know how to model a diode using Laplace transforms. So I guess we have to use deterministic methods like Spice, tables, or rules of thumb to determine the rectifier parameters.
Ratch
Hello again,
Well actually Laplace Transforms can be used for some non linear circuits also, just with a little more difficulty. You usually just have to handle it as more than one circuit in a circuit that switches (such as a buck regulator).
The idea is to break up the circuit into it's different sequential modes of operation, then compute each piece individually. This means solving for the times that the circuit switches, and then using the final values from the previous mode with the Laplace for the very next mode.
So we can use Laplace but we have to do it for more than one circuit in most cases. In this case though we only need one because after the diode switches we have what we want which is the time it switches. So we only have to use the transform to solve for the time that the diode switches from on to off, and then if we really wanted to continue we'd have to use those final values as the initial values of the next mode, which for this circuit would be the mode where the cap discharges.
We dont have to rely on spice or anything like that, this is computable and analytic up to the last step where we bring in a numerical equation solver to solve for the cap value. The time value however can be computed from the circuit equation(s) without the need for a numerical calculation in the end, at least in the solution i found earlier.
Does this help, or at least make sense?
Hi,
Well to solve for the cap value you only need to solve for the time delay, you dont need the whole waveform, especially since this time delay simply subtracts from your previous formula to get the corrected time from the back end of the previous wave to the front end of the next wave. You dont have to compute any Laplace for the diode either because it is an ideal diode.
To make a continuous function that describes the whole period would just mean solving for the individual periods anyway, then finding a way to combine them cleverly, which would be novel but just a waste of time (see "Spock Curve"). Once you have the periods solved, you dont really have to combine them.
It's a little harder than most simple circuits like this, but it's very interesting to find out how it really works. Maybe you could show me what you have so far and i can see if it makes sense to me.
Hi again,
I am pretty sure that is impossible, because the curve has to be smooth for that to work, and this curve in it's ideal form cant be smooth because the place where the two modes meet form a vertex, which cant be smooth. It's almost like trying to make the circumference of a square a smooth curve. If you round the corners for the sake of smoothness, then you void the true solution, so the best we can do is piece all the sides together using another function or functions.
However, the curve is continuous, by the definition of continuous being a curve that you can draw with a pencil without ever having to lift up the pencil point from the paper, it just wont be smooth where there are derivatives everywhere, and it shouldnt be because that's not how it looks in theory.
To make it continuous as above you could modulate the parts with a two phase rectangular wave. That would mean that a curve drawing program could plot the entire curve without having to know where one curve ended and the next started. That's really how it works anyway when we have an ideal diode, which we have here for the purpose of exploring that avenue of analysis.
It takes a little thought that's all
I had made a few attempts before i was able to settle on a solution that i believe is exact. However, since your approach sounded different than mine i was hoping to see yours too and compare to mine. I checked it with a simulation program (which was a little difficult to set up in itself for some reason) and it showed a solution of 0.9 out to five decimal places with the capacitor value calculated using that formula for C, so i assumed that was good enough.
If you feel like setting up LT Spice for this problem we could take a look at that too. I used a different simulator for now.
Your method is based on the capacitor discharge ending at the front of the next wave which is good, but it is also based on the capacitor starting to discharge at the EXACT PEAK of the previous wave, which is impossible except with infinite R, and then it doesnt really discharge anyway, so we might say that as R approaches infinity the cap discharge starts infinitesimally close to the peak.
One way to look at the start of the discharge is to consider that C starts to discharge into R when the line voltage gets below the cap voltage, but since the cap voltage is constantly decreasing at a certain rate, if the rate is faster than the rate that the cosine wave is decreasing then the cap voltage will follow the cosine voltage downward just as it did when it followed it upward on the charge. It's not governed by the input wave itself anymore, but by the wave AND the discharge of the cap. Since near the peak the cosine wave is decreasing very slowly, any discharge in the cap will tend to keep the cap voltage still following the cosine wave, for a short time. Thus, the cosine wave and the cap voltage fall at the same rate for a short time. In this view, once the cap voltage discharge rate becomes slower than the cosine wave decrease rate, the cap voltage wave 'leaves' the cosine wave and the diode stops conducting, ...
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