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General Vintage Technology Discussions For general discussions about vintage radio and other vintage electronics etc. |
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8th Apr 2019, 7:01 pm | #21 |
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Re: Maths help....Again...exp?
Your thinking's out. exp(-T/CR) is the reciprocal of (which is the same as saying 'one over') exp(T/CR). You used that fact in your calculation above when you divided exp(T/CR) into Vp-0.7 instead of mutiplying exp(-T/CR) by Vp-0.7, which is what the formula told you to do.
I fear that the appearance of exp(T/CR) has misled TonyDuell, but it's all right because of the above alteration you also made. That's why the answer came out right at 0.47V. Cheers, GJ (used to teach maths to physics undergrads, the poor souls)
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8th Apr 2019, 7:03 pm | #22 | ||
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Re: Maths help....Again...exp?
Quote:
Vr=(Vp-0.7v)(1-exp(- T/RC)) Lawrence. |
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8th Apr 2019, 7:08 pm | #23 |
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Re: Maths help....Again...exp?
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8th Apr 2019, 7:49 pm | #24 |
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Re: Maths help....Again...exp?
exp(-T/CR) = 1/exp(T/CR) = 1/(1.0338) from your previous post = 0.9672.
Cheers, GJ
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8th Apr 2019, 7:51 pm | #25 |
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Re: Maths help....Again...exp?
The notation "exp(x)" is another way of writing "e ** x", or e to the power of x. e is a fundamental mathematical constant, sometimes known as Euler's constant, about 2.71828. The function y = e ** x is also its own derivative; that means the slope of the graph at any value of x is also its height. So things tend to grow or decay exponentially, whenever the rate of growth is proportional to the existing amount or the rate of decay is proportional to the amount remaining.
e ** (-t / R*C) starts off at 1 (anything to the power zero is 1, because we always start from 1 when multiplying; if we have multiplied no times, then we must still have 1) and gets smaller by a bit less each time (since raising to a negative power is the same as inverting the same thing to a positive power), never quite reaching zero. But in practice, a capacitor is about as charged as it's ever going to get in 5 times R*C seconds. When you bring in complex numbers, e ** (j * x) is also equal to cos(x) + j * sin(x). So e ** (j * π) = -1. Possibly the most beautiful identity in all of mathematics .....
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If I have seen further than others, it is because I was standing on a pile of failed experiments. Last edited by julie_m; 8th Apr 2019 at 7:59 pm. Reason: Added missing minus sign and some text. |
8th Apr 2019, 8:16 pm | #26 | |
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Re: Maths help....Again...exp?
Quote:
So far as my thinking goes the back half of the formula is working out what the voltage has declined to in percentage terms after Tsecs from it's peak value, another way of computing Vr= Vmax-Vmin, but as mentioned earlier and as shown in the page link, T is taken as the peak to peak period not cutoff to conduction so it's an approximation in terms of that. Lawrence. Last edited by ms660; 8th Apr 2019 at 8:40 pm. |
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8th Apr 2019, 8:44 pm | #27 |
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Re: Maths help....Again...exp?
I never really got the maths of this sort of stuff. In both physics and chemistry e to the power of x (whatever x was made up from) featured in almost everything we studied. So, apart from not being very practiced with it, Emeritus pointing to the log tables was the revelation. A couple of days ago, a quick look in the natural tangents sorted out an angle (planning, but often in the workshop), but it ocurrs to me that nowhere in all that study starting over 60 years ago, did we ever refer to that page in the log tables. I even have an inkling that it may be on one of my slide rules.
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8th Apr 2019, 9:09 pm | #28 | |
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Re: Maths help....Again...exp?
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Lawrence. |
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8th Apr 2019, 9:32 pm | #29 | |
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Re: Maths help....Again...exp?
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Vout=Vin*(1-exp(-T/CR)) is the formula for the DC output voltage after time T has expired of an RC filter when a DC input is applied. This formula cannot be applied to the ripple on a PSU as that is a non-linear network. |
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8th Apr 2019, 10:17 pm | #30 | ||
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Re: Maths help....Again...exp?
Quote:
Vout=Vin*(1-exp(-T/CR)) ? that formula is a new one to me, I was referring to Vr not Vout. For average out, Vout(avg) = Vmax-Vr/2, but they used the time from peak to peak in their formula for Vr, not the time from peak to conduction, Vmax being the maximum output from the rectifier not the max input. Lawrence. |
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8th Apr 2019, 10:20 pm | #31 | |
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Re: Maths help....Again...exp?
Quote:
Getting back to this particular problem, if we look at the calculations more exactly, starting with T/CR=0.0333 then exp(-T/CR) = 0.967248 and exp (T/CR) = 1.033861 so 2-exp(T/CR) = 0.966139 You can see that the two numbers in bold are close to one another - different by only about one part in one thousand - but they're not exactly the same. If you make T/CR smaller and smaller the fractional difference between the two numbers will become less and less. But as PJL said, they won't become exactly equal until T/CR gets down to exactly zero. Cheers, GJ
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8th Apr 2019, 10:51 pm | #32 | |
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Re: Maths help....Again...exp?
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I'm hoping that someone can talk me through the procedure using the Window 10 scientific calculator I now have (thanks to GeoffK) As said before, I used an alternative method to that formula which I figured out using my limited powers of mathematical deduction but it would be nice to feed all the inputs and expressions as per the formula into the scientific calculator and see the correct result spewed out.... Lawrence. Last edited by ms660; 8th Apr 2019 at 10:55 pm. Reason: addition |
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8th Apr 2019, 11:49 pm | #33 | |
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Re: Maths help....Again...exp?
Quote:
The point I was making was that exp(-T/RC) will always be correct. 2-exp(T/RC) will always be wrong, unless T/RC is zero. So the trick is always to use the correct expression, exp(-T/RC), and not to worry that the wrong expression will only ever give 'approximately' the right answer. Cheers, GJ
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9th Apr 2019, 5:47 am | #34 | |
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Re: Maths help....Again...exp?
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(^ means raise to the power of in this reply) Unfortunately the way the equation is presented is not ideal as the (1-e^-(T/RC)) makes it look like an inverted exponential when it is not. The equation is one for exponential decay. e^-kt is a general form of exponential decay where k is some constant and t is time. The "k" or constant in your equation is merely 1/RC. Of note though e^0 is 1, which is the case if time = 0, the start of your measuring point. So at time 0 the capacitor is charged to Vp-0.7V. Later on at some time t the value of voltage on the capacitor is its starting voltage times the decay as its fallen to a lower value: (Vp-0.7)e^-t/RC..... t is>0 So to calculate the ripple (starting voltage minus finishing voltage): Ripple voltage = (Vp-0.7 ) - (Vp-0.7)e^-t/RC Which reduces to (Vp-0.7)(1-e^-t/RC), your equation. t is the time between peaks and the ripple voltage is a peak to peak amplitude from the max capacitor voltage to where is sags down to before it gets recharged. You will see the e^-kt in many places, often preceded by some peak amplitude A, For example exponential decay of a sine wave has the form: Asin(0)e^-kt. (For your equation I think the assumption is that the capacitor is fully charged to Vp-0.7V at each peak. The actual charging of the capacitor is an inverted exponential, but it charges from a much lower internal resistance of the transformer quickly and discharges more slowly via the load resistance R. Though a similar equation can be used to calculate the charging ripple if R is the internal resistance of the charging circuit and not the load and a shorter time is taken into account) Last edited by Argus25; 9th Apr 2019 at 6:14 am. Reason: typo |
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9th Apr 2019, 7:15 am | #35 |
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Re: Maths help....Again...exp?
Even the full and correct equation is only an approximation because it was constructed around some assumptions about the circuit... That the load discharging the capacitor between charging pulses is resistive where thd current taken is proportional to the voltage presented. This is what gives the inverse exponential decay of the capacitor voltage. The rate of drop of the voltage slows as the voltage drops.
If the load is active electronics, maybe a voltage regulator feeding constant voltage into whatever the end use is, or maybe some valves with cathode resistors setting their quiescent current in conjunction with their characteristic curves, then the load can look more like a constant current. If so, the capacitor will be linearly discharged and its voltage will drop on a straight line against time. The equation becomes simpler, losing the exponential function and replacing it with a simple linear ramp down. Real world electronics will be somewhere between these two extremes, and if your ripple voltage is not large, they both become close enough in their results to not make much difference. I usually use the constant current approach rather than the resistive approach. 1) the arithmetic is simpler! 2) it errs on the safe side for designing something 3) real world electronics tends to be this way. 4) when I designed the electronics being powered, I tried to make the bias currents stable. So I'd like to at least pretend I'd been partially successful David
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9th Apr 2019, 8:39 am | #36 | |
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Re: Maths help....Again...exp?
Quote:
Starting from Q = CV, then dQ/dt= current = I = CdV/dt. Solving for dV and assuming is linear Idt/C = the ripple. If dt is 10mSec (full wave ripple in a 50Hz system) and say the current I is 1 amp, and the capacitor 10,000uF then the ripple voltage is 1v. So knowing this you can nearly completely throw away the math because a 1 amp load on a 10,000uF cap will cause a 1V ripple in a 50Hz full wave rectifier system, that's all you ever need to know . So lets say the capacitor is 1000uF and the current is 0.1A the ripple is still one volt. Or it its half wave, the ripple is double. You can just scale it in your head for the capacitor value and current and the update interval to get the answer quickly. |
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9th Apr 2019, 8:43 am | #37 |
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Re: Maths help....Again...exp?
I usually use the constant current approach too because then I can do the sum in my head
Ripple voltage = (current drawn x time between charging pulses)/(reservoir capacitance) I rarely need to know the ripple voltage to an accuracy of better than, say, 10-20%. The capacitor is likely to be an electrolytic and its value will almost certainly be uncertain to that degree. The most critical application I routinely come across is working out whether the supply voltage will, at the trough of the ripple, drop below the dropout voltage of a regulator that follows the reservoir cap. I daren't work right up against the hairy edge of this happening so if I think there's any risk of it, especially taking into account mains fluctuations, I either find a transformer with a higher output voltage or I use rectifiers with a lower forward voltage drop or I find a more tolerant regulator or I put a bigger cap in. Cheers, GJ
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9th Apr 2019, 8:58 am | #38 |
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Re: Maths help....Again...exp?
It's important to understand the exponentials involved in resistor-capacitor circuits because they get used for timing things (intentionally and unintentionally) all over the place. So the effort in understanding the maths is still well-spent.
But often things aren't quite like the assumptions and get designed with a generous comfort factor, so simplified approaches work perfectly well. C times R is called the timeconstant. 63% of a step change of input voltage is on the output one timeconstant after the step hiit the input. 90% takes five timeconstants. 100% takes forever. 1/(2xPixRxC) gives you the frequency an RC lowpass has rolled off by 3dB and where it gives you a 45 degree phase shift (lag) Above that point a workable approximation is that the output drops 20dB for every 10-fold increase in frequency. See! this design malarky isn't anywhere as difficult as it appears. The art comes in knowing the short cuts and when they can reasonably be used, but you still need to be able to do the full trip round the houses when circumstances dictate. David
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9th Apr 2019, 9:57 am | #39 | |
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Re: Maths help....Again...exp?
Quote:
If the ripple is small, then a straight-line approximation is pretty good. For 50Hz operation, full-wave rectifier, drawing a load current of I amps into a reservoir capacitance of C μF, the ripple is Vp-p = 10,000 x I / C. So 100mA and 1,000μF will give 1V p-p ripple. What don't I agree with Dave about? If the load is a quick-responding switching regulator, then it will draw constant power so in this case, as voltage falls, current will increase so the ripple waveform, instead of ramping linearly down (constant current load) or curving with negative exponential (resistive load), it curves increasingly steeply downward according to a square-root law. And these days, this is becoming quite real-world too! |
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9th Apr 2019, 10:44 am | #40 |
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Re: Maths help....Again...exp?
I am not sure formula that are approximations are the best way of explaining the use of mathematics in electronics. You need to understand the basics before you can work out where to take short cuts or, in this case, if the proposed shortcut is appropriate.
Too much ripple and you simply increase the reservoir capacitor. Choose a transformer with insufficient regulation and the PSU will never work. The critical component does not even factor into this equation. |