intergrator waveform measurements and symmetry

Thread Starter

relicmarks

Joined Oct 13, 2006
355
My function generator is set to a squarewaveform at 1Khz at 5 volts

My intergator RC time is 100us

The voltage will RISE to 5 volts in the time of 100us?

Slope = time/voltage

100us/5volts = slope time of a intergator?

Linear intergation measurements is measuring the rise time and fall time

Exponential intergation measurements look like a shark's fen or it will tilt the "top" of the squarewaveform, the top either leading or lagging slanting

Slope = time/voltage

Symmetry
1.) the rise time and fall time ratio


Duty cycle is ON time / period X 100%

Symmetry formula is Rise time/ rise time + fall time X 100%

example of Symmetry :

1.) function generator set to a squarewaveform
2.) Intergator RC time is 100milliSec.
3.) Output would be a triangle waveform
4.) Symmetry rise time is 20mS and fall time is 80mS = 100mS

I haven't found a intergator circuit that has a variable rise time and fall time symmetry adjustment parameter, but its the RATIO of the rise time and fall time

A intergator that has a seperate rise time value and a seperate fall time value

I think the correct formula is this one , but i'm not to sure

Symmetry formula is Rise time/ rise time + fall time X 100%

Another Main point about intergators is the Charge and Discharge time

Rise time and fall time are "linear"

Charge time and discharge time is Exponential

Exponential is not a slope it more like a curve for the rise and a scoop or skirt for the fall, i guess thats what you call them, do you guys know the names ?

The rise of a charge time looks ilke a sharks Fen and the Discharge time looks like a skirt or scoop

Rise time , fall time = linear slope

Slope = time/voltage

Charge time , discharge time= exponential curve

Curve = time/voltage
 

beenthere

Joined Apr 20, 2004
15,819
When you are charging a capacitor through a resistor, the formula is T = RC. But the capacitor will not be at full charge after period T. It doesn't get to 97% (I think) until 5 periods of T. That is where the exponential part comes in.

If you are integrating a square wave in order to produce a symmetrical triangle wave then you don't want the capacitor to fully charge anyhow. Try some different values to see how it affects the triangle waveshape.
 

Thread Starter

relicmarks

Joined Oct 13, 2006
355
The RC ratio values are going to output a linear slope or a exponential CURVE?

I think a passive RC network can't output a linear ramp/slope voltage

I think only a op-amp RC intergator can output a linear slope/ramp

How would you guys mostly measure intergation waveforms? slopes? ramps? and exponential curves?
 

Thread Starter

relicmarks

Joined Oct 13, 2006
355
RC circuit charge curve is exponential

Yes, but how do you measure the time constants or curves on the oscilloscope? or curve tracer?
 

Thread Starter

relicmarks

Joined Oct 13, 2006
355
Exponential "growth" time interval
1.) The exponential rise or exponetial charge time its called " Growth time" not rise time
Exponential decay ( time interval )
1.) Is that of the discharge of a capacitor through a resistor.

RC time measurements:
1.) Measure the time constant of the exponential decay
2.) Measure the Slope ( rate of change )
3.) curve fitting
4.) To measure an RC time constant and to observe the "shape" of the charging and discharging curves.
5.) Oscilloscope trace used to "measure half-life"

Function generator Settings for RC time constants:
1.) The period of the signal from the signal generator T = 1/f should be "several times"
the time constant
2.) Adjust the DC OFFSET of the generator so that the generator output alternates between a positive
voltage and zero voltage
 

Thread Starter

relicmarks

Joined Oct 13, 2006
355
Vc at 1/3
Vc at 2/3

T1/2=RCln2=0.693RC

rising shape 1-e-t/RC and the falling shape e-t/RC.

T = RC ln[(V0 – V1)/(V0 – V2)]

decaying and growing exponentials as a DC square wave voltage source is applied.


drawing a line tangent to the curve at time x. The difference between the time at which this tangent crosses the horizontal axis and time x is the time constant.


the charging/discharging time of the capacitor increases as the capacitance increases, resulting in flatter waves as the capacitance increases.

see, the charging/discharging time of the capacitor increases as the resistance in the RC circuit increases. This is because as the resistance increases, the voltage applied to the capacitor becomes smaller and smaller.

frequency increases, the charging/discharging time increases. This is because at lower frequencies, the capacitor has longer to stabilize its voltage before the source voltage changes again.


measurement of the phase difference between VR and VC
Transient Response of an RC Network
decay= 36%= 0.368 volts

measure the HALF TIME Of a discharing cap

1 RC time constant, the voltage is 37 % of the original for a discharging capacitor, and 63 % for a charging capacitor.
After RC seconds the voltage is 37 % of the original. To increase the time taken for a discharge we can:
Increase the resistance.
Increase the capacitance.
The half-life is 69 % of the time constant
exponential rise in charge and voltage. We get an exponential fall in the current
Time constant: RC = 2000 ohm × 5000uf = 10 s.
Half-life of the decay: t1/2 = 0.693 × RC = 0.693 × 10 = 6.93 s.
The time period taken for the capacitor to reach this 4T point is known as the transient period. After a time of 5τ the capacitor is now fully charged and the voltage accross the capacitor (Vc) = the supply voltage (Vs), so no more current is flowing in the circuit. The time period after this point is known as the steady state period.

The time constant τ is found using the formula τ = R x C in seconds.

Therefore the time constant τ is:

τ = R x C = 47k x 1000uF = 47 Secs

a) What will the voltage be across the capacitor at 0.7 time constants?
At 0.7 time constants (0.7τ) Vc = 0.5Vs. Therefore, Vc = 0.5 x 5V = 2.5V
b) What will the voltage be across the capacitor at 1 time constant?
At 1 time constant (1τ) Vc = 0.63Vs. Therefore, Vc = 0.63 x 5V = 3.15V
c) How long will it take to fully charge the capacitor?
The capacitor will be fully charged at 5 time constants.

1 time constant (1τ) = 47 seconds. Therefore, 5τ = 5 x 47 = 235 secs
 

Thread Starter

relicmarks

Joined Oct 13, 2006
355
On the oscilloscope time constant measurments:

Discharging or Decay time constant
1.) Measure the time interval between V peak to 2/3 of V peak
2.) Discharge down to 100%-63% = 37% of the full voltage.

Charge or Growth time constnat
1.) Measure the time interval between V peak to 1/3 of V peak
2.) charge time is 63%
 
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