Maths help....Again...exp?

[ms660]

Vr=(Vp-0.7v)(1-exp(- T/RC))

I don't get the above, however….

...Vr= ripple voltage
...Vp=AC peak into a rectifier
0.7v= rectifiers forward voltage drop

I get that bit....I also get the T/RC bit...Time over Resistance Capacitance....
….but what's 1-exp?

I tried using some exp calculators online but couldn't make much sense of them due to my ignorance but eventually I did manage to find one that I could use and used the result to compute Vr but not by using the above formula (as I see it) I tried other values against known results and my method agreed with them.

Here's the calculator I used:

https://www.medcalc.org/manual/exp_function.php

I entered the result of T/RC into the brackets and hit the button, with that result I divided it into Vp-0.7v I then subtracted that result from Vp-0.7v and got the correct value for Vr as would have been given by the formula in question

Excuse my maths dyslexia, I'd prefer a simple answer if it's possible.

Lawrence.

[Herald1360]

I think that particular "exp" is a version of "e" or ~ 2.7183.


..-t/cr
e..... is a common expression where time constants are involved.


Rather confusing sinc "exp" is commonly used as a workround when superscript isn't an option, like on here. (Bit of cunning formatting going on here 'cos you can't even paste in superscripts as far as I can tell- if you look hard, you can see the white ...s).


eg:

..-t/cr
e....... ≡ e exp(-t/CR)

[MrBungle]

If this is load ripple, it's easier to work in current and the capacitor and assume the transformer is ideal.

C=dV/dT is easier to work with. dV=ripple ... dt=10ms (for 50Hz). Once you have C you can work the voltages back from the load to the transformer adding dropout for regulator, ripple, two diode drops, load regulation of the transformer, mains voltage variation and a few % of headroom.

If I've got the wrong end of the stick, ignore me

[Herald1360]

Isn't dV/dt = i/C the expression needed? Or for constant current, V = it/C

[ms660]

Originally Posted by MrBungle

If this is load ripple, it's easier to work in current and the capacitor and assume the transformer is ideal.

C=dV/dT is easier to work with. dV=ripple ... dt=10ms (for 50Hz). Once you have C you can work the voltages back from the load to the transformer adding dropout for regulator, ripple, two diode drops, load regulation of the transformer, mains voltage variation and a few % of headroom.

If I've got the wrong end of the stick, ignore me

Where I got the formula from I think T is an approximation ie: their calcs used the time period from peak to peak.

I'm still no further forward with 1-exp.

Lawrence.

[Lucien Nunes]

The function you describe relates to ripple voltage on a capacitor loaded with a resistance, rather than a constant-current.

exp(x) here means e (the base of natural logs) raised to the power of x.

exp(-t) is a very general function that describes the exponential decay of a value as t increases. This occurs widely in the physical world, where the rate of decay is proportional to the instantaneous value. At t=0, exp(-t)=1, indicating that the decaying property is at its full initial value. As t increases, exp(-t) falls towards zero following the exponential decay curve.

exp(-t/RC) describes the decay of the voltage on a charged capacitor as it discharges through a resistance, as a fraction of the initial voltage.

Vp(exp(-t/RC)) gives the voltage at time t, as it falls from the initial peak charging voltage Vp.

If one assumes the ripple voltage to be sawtooth-like, where the capacitor charges in negligible time and then discharges into a constant resistance, the ripple voltage will be the difference between the peak voltage, and the voltage to which it has fallen by the beginning of the next charging impulse, which we are considering to be time t.

Hence, Vr = Vp(1-exp(-t/RC))

For small values of ripple, it is not necessary to compute the exponential, since the discharge current is approximately constant, as mentioned above.

[PJL]

That formula looks like the output voltage assuming a DC input. The ripple would be 0 without load current yet the load is not in your equation.

Try cheating http://www.duncanamps.com/psud2/

[GrimJosef]

Chris (Herald1360) is right that exp is the exponential function. It's like other functions in that you put a number in (in this case -T/RC) and get another number out. The output number in the case of exp varies from 1, when the input is 0, smoothly down to 0, when -T/RC is very large.

So 1-exp(-T/RC) varies the other way i.e. when -T/RC is very small (usually because you've got a big R and/or a big C, which makes a very effective ripple filter) the ripple voltage will be close to zero. If, on the other hand, you've got very small resistance and capacitance, so -T/RC is very large then exp(-T/RC) will be close to zero, 1-exp(-T/RC) will be close to one and the ripple will be the full AC input minus just the diode voltage drop.

Cheers,

GJ

[kalee20]

Vr=(Vp-0.7v)(1-exp(- T/RC))

How to calculate it? Do the brackets first.

So, work out T/RC and change sign to get -T/RC.

Then hit e^x which will give you exp(-T/RC). It's a number less than 1. Put it in memory.

Clear the calculator, enter 1 - (memory) and hit equals, you now have (1 - exp(-t/RC)) which is also less than 1. Put it in memory, overwriting previous result. Clear the calculator.

Work out Vp - 0.7v (easy) and hit equals. Then multiply by (memory) giving you what you want!

[ms660]

Originally Posted by kalee20

Vr=(Vp-0.7v)(1-exp(- T/RC))

How to calculate it? Do the brackets first.

So, work out T/RC and change sign to get -T/RC.

Then hit e^x which will give you exp(-T/RC). It's a number less than 1. Put it in memory.

Clear the calculator, enter 1 - (memory) and hit equals, you now have (1 - exp(-t/RC)) which is also less than 1. Put it in memory, overwriting previous result. Clear the calculator.

Work out Vp - 0.7v (easy) and hit equals. Then multiply by (memory) giving you what you want!

There's no arrows on my calculator, no e either.

Lawrence.

[emeritus]

Here's a table of exponential functions from my old School log tables! Go down the left hand values of x to the nearest value of T/RC and read off the value in the second column [e to the minus x]. It should give adequate results, unless you are using close tolerance components.

[Al (astral highway)]

Hi Lawrence,

No problem...

Here's one of many online scientific calculators.

You can transcribe the calculation as it unfolds, with pen and paper, if that helps keep track of it.

Cheers!

[John KC0G]

exp(x) = 1 + x + (x^2)/2 + (X^3)/6 + .. successive terms are (x^n)/n! where n! = n(n-1)(n-2)....3*2*1

If T/RC is much less than 1, then exp(x) is approximately 1+x. This will be the case when (x^2)/2 is much less than x
Your formula then becomes
Vr=(Vp-0.7v)(T/RC))

HTH and 73

John

[ms660]

Here's the example numbers for the formula I posted:

Vp = 15 volts
T = 0.01667 secs.
R = 5,000 ohms.
C = 100uF

The answer given according to that formula is Vr (ripple voltage) = 0.47 volts but I can't get that formula to work as it's written ('cos I don't fully understand it) so here's what I did:

Work out T/RC, easy for me.....0.01667/0.5 which = 0.0333

Bung that into the exp calculator I used, that returns an answer of 1.0338

Divide that into Vp-0.7 (14.3) that = 13.83

Subtract that from Vp-0.7 (14.3) = 0.47 volts ripple

That's the same answer as was given by the formula that I don't yet fully understand.

Here's the site I got the formula and example from:

http://tuttle.merc.iastate.edu/ee201...rectifiers.pdf

I've used the other examples given for R and using the above method everything more or less tallies up with the given answers.

Lawrence.

[TonyDuell]

I think you've forgotten to change the sign of the (-T/CR) term.

The original formula works for me :

-T/CR = 0.01667/(5000*100*10^-6) = -0.0334 (I think that's where you got the sign wrong).

exp(-T/CR) = .9672...
1-exp(-T/CR)=.03279...
Vp-0.7=14.3V

So (Vp-0.7)*(1-exp(-T/CR)) = 0.47V or thereabouts.

There is no point in me giving the key sequence I used, I am using an HP RPL calculator...

I get lazy and chuck this sort of thing at ltspice https://www.analog.com/en/design-cen...SAAEgLmbfD_BwE warning... you can spend hours experimenting and looking at waveforms.

[ms660]

Originally Posted by TonyDuell

I think you've forgotten to change the sign of the (-T/CR) term.

The original formula works for me :

-T/CR = 0.01667/(5000*100*10^-6) = -0.0334 (I think that's where you got the sign wrong).

exp(-T/CR) = .9672...
1-exp(-T/CR)=.03279...
Vp-0.7=14.3V

So (Vp-0.7)*(1-exp(-T/CR)) = 0.47V or thereabouts.

There is no point in me giving the key sequence I used, I am using an HP RPL calculator...

Is 2-exp(T/CR) the same as exp(-T/CR) or is my thinking out?

Lawrence.

[GeoffK]

There's no arrows on my calculator, no e either.

Lawrence.

The computer has a built in scientific calculator on the start button, change the view to select type.

[PJL]

Lawrence, what is the circuit you are trying to model? Is it a PSU design and you are trying to determine the ripple voltage on the HT? If so, your model is far too simple and you need to look at the PSU designer link I gave you. Choose your PSU design, enter some values, and the PSU designer will tell you all you need to know.

[ms660]

Originally Posted by PJL

Lawrence, what is the circuit you are trying to model? Is it a PSU design and you are trying to determine the ripple voltage on the HT? If so, your model is far too simple and you need to look at the PSU designer link I gave you. Choose your PSU design, enter some values, and the PSU designer will tell you all you need to know.

I'm not trying to model anything, just trying to figure out the maths or at least how to do the calcs on a calculator.

Lawrence.