[sdiy] Integrator frequency calculation

[David G. Dixon]

Justin,

The frequency of the triangle will be exactly the same as the frequency of
the incoming square wave.  The question is, what will the triangle
peak-to-peak amplitude be?

You must do a current balance around the integrator opamp.  If the voltage
at the - terminal (equal to the + terminal vias voltage) is V(-) and the
voltage of the square wave sourve is V(s), then the current into the
feedback loop from the square wave source:

I(in) = [V(s) - V(-)]/R(in)

Voltage rate of change at the integrator output:

dV/dt = -I(in)/C(cap) = -[V(s) - V(-)]/R(in)C(cap)

(The negative sign is because active integrators are inverting.)  Separating
terms of the differential equation:

dV/[V(s) - V(-)] = -dt/R(in)C(in)

SinceV(-), R(in) and C(in) are all constant, and V(s) is constant during
each half cycle of the square wave, the integration is very simple:

[V(out) - V(out,0)]/[V(s) - V(-)] = -t/R(in)C(in)

Where V(out) = V(out,0) at t = 0.  Rearranging to solve for V(out):

V(out) = V(out,0) - [V(s) - V(-)]t/R(in)C(in)

For the first case, for the higher square wave half-cycle, presuming the
bias voltage V(-) is the mean value of the square wave (which is required to
give a symmetrical triangle output):

V(-) = (5.29 + 6.25)/2 = 5.77V

R(in)C(in) = 28600R * 0.00000001F = 0.000286s = 286us

dV/dt = -[V(s) - V(-)]/R(in)C(in) = -(6.25 - 5.77)/0.000286 = -1678V/s

For a 200Hz square wave, the half-cycle time is 1/2*200 = 1/400 = 0.0025s.
Hence, the peak-to-peak amplitude of the resulting triangle is:

0.0025s * 1678V/s = 4.20V

That means that the triangle will vary over the range of 5.77 +/- 4.20/2, or
from 3.67V to 7.87V.  Of course, to ensure this, I think you must put a
large resistor (3.3M) across the capacitor to eliminate the integrator
output from climbing to the rail.



> Hello,
> 
> Since the answers I got from this list about my DC summing calculations
> were so much clearer than my interpretations of various textbooks - I
> thought I'd just come straight to the source this time.
> 
> I'm trying to understand how to calculate the frequency output of an op
> amp integrator. I understand that it's a relationship between the change
> in voltage over time and the current across the capacitor.
> 
> My integrator is being fed a square wave, the output is a triangle wave.
> 12V single supply with VCC/2 = 6V, 10nF cap and approx. 28K6 input
> resistor. Triangle isn't biased exactly on VCC/2 right now - but there is
> a reason for that.
> 
> Here's what I have so far:
> 
> Integrator outputs a triangle wave of approx. 200Hz from a square wave
> with a 'low' state of approx. 5.29V and a 'high' state of approx. 6.25V.
> That's a change of 0.96V every 0.005 seconds.
> 
> or..
> 
> Integrator outputs a triangle wave of approx. 574Hz from a square wave
> with a 'low' state of approx. 4.47V and a 'high' state of approx. 7.17V.
> That's a change of 2.7V every 0.0017etc. seconds.
> 
> That's what I know - the rest is speculation. So - how do I take V, I, R
> and C - and get F.
> 
> Thanks all,
> 
> Justin
> 
> 
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