[sdiy] TB303 Slide
Tom Wiltshire
tom at electricdruid.net
Fri Jun 8 22:30:50 CEST 2012
On 8 Jun 2012, at 18:03, Tim Stinchcombe wrote:
>> Good picture! Could you try delaying that red curve so that
>> it starts at time 18 instead of 0?
>
> Yes (but shift is 15ms):
>
> http://www.timstinchcombe.co.uk/synth/const_time_glide_shift.gif
>
> OK, point taken! I must be looking at it in some cock-eyed way... :-0
>
> Or maybe I *am* being fooled by the 'terminology' - confused now :-/
>
> Tim
The two *are* the same. You can check the maths below, but I think it's ok (plaintext is not ideal for maths). The equation for a cap charging is:
V(1-e^(-t/rc))
Where V is the voltage source, t is the time, and rc is the rc time constant.
Equation for the 0-2V curve:
2(1-e^(-t/rc)) = 2-2e^(-t/rc)
We can work out what the time is when that reaches 1V (e.g. when y=1):
1=2-2e^(-t/rc)
2e^(-t/rc) = 1 (rearrange)
e^(-t/rc) = 1/2 (divide by two)
-t/rc = ln(1/2) (natural log of both sides)
t = -rcln(1/2) = 0.015249 (the 15ms Tim found earlier)
Now we can start the 1-2V curve with that time offset. The equation for this is:
1(1-e^(-rcln(1/2)-t / rc)) + 1 (The additional +1 at the end is because we start from 1V not 0V)
2-e^(-rcln(1/2)-t / rc) (sort out a bit)
We can show that these equations are equal:
2-2e^(-t/rc) = 2-e^(-rcln(1/2)-t / rc) (assume that they are)
2e^(-t/rc) = e^(-rcln(1/2)-t / rc) (take two off both sides, multiply by -1)
ln(2) - (t/rc) = -rcln(1/2)-t / rc (natural log of both sides)
rcln(2) - t = -rcln(1/2) - t (multiply both sides by rc)
rcln(2) - t = rcln(2) - t logy(x) = -logy(1/x), so ln(2) = -ln(1/2)
These equations are identical and the curves must be the same.
Tom
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