differentiator

[jit_singh_tara]

hi there,
can anyone tell me wat does differentiator do(an rc circuit.)wat are the out puts of diff signals applied to input of a differentiator

[audioguru]

Hi Jit,
Welcome to our forum.
A differentiator is just a highpass filter. Its series capacitor has a high capacitive reactance to block low frequencies but is like a dead short to pass high frequencies.
It can be used to pass a spike at the edges of a square wave.

[Kevin Weddle]

Just for fun, the differentiator performs the derivative function mathematically. It turns a certain rate of change into a voltage level. The capacitor is in the gain circuit of the opamp and the amplitude of the input will be determined by the capacitive reactance.

The integrator is identically different. It turns your voltage level into a rate of change. The slope is determined by the fact that the RC time constant produces the time of the ramp and the voltage level attained within that time determines the rate of change. A greater voltage will produce a greater ouput voltage that the slope must rise to in the time of RC.

What is confusing is how the capacitor will charge to any voltage in the same amount of time. But on paper, you can see how a greater voltage in the same amount of time leads to a higher rate of change. And this is what actually occurs. The RC produces a constant time and the voltage you charge to determines the rate of change.

There. That is two explainations of the same concept.

[audioguru]

Kevin,
An opamp has an extremely high gain. When it is used with negative feedback in a differentiator or integrator circuit, the capacitor is charged linearly with a constant current, not just an RC time-constant curve.
Therefore the rate-of-change doesn't depend on the input voltage.

[Kevin Weddle]

Okay I agree. A constant current. But remember too that because the input is DC the output is DC. Once you get the DC input, you should get the DC output but the capacitor has to charge and it charges following the capacitor charging curve just like when you apply a DC to a series RC.

[audioguru]

No Kevin,
The voltage of a capacitor that is charged by a constant current is an absolutely straight linear line. Its voltage rise has a constant rate of change with time increments, not following the curve when charging with a simple resistor. When a capacitor is charged by a resistor, the charging current decreases as the capacitor's voltage builds up.  ::)
A triangle-wave generator uses an opamp to produce a constant current for the capacitor's charging and discharging. That's why its slopes are straight, not curved.

[Kevin Weddle]

How else do you get the rise with time. Something about the DC applied to a capacitor produces a ramp. But what you are saying is constant current. I don't care what the current does, I know it has a curve too. I care what the voltage does. And the voltage rise is not as linear as it approaches final charge. That picture of the capacitor is the voltage with respect to time.

[audioguru]

Kevin,
You can have a simple integrator where a capacitor is charged and discharged by only a resistor. The slope of the capacitor's voltage is curved as it charges or discharges because as it charges, the voltage across the resistor becomes less, resulting in less charging current so the capacitor charges slower and slower until it is fully charged. Capacitor discharging by a simple resistor is the same.

If you use a constant current circuit to charge or discharge a capacitor, the voltage rise or fall of the capacitor is completely linear and shows an absolutely straight line on a 'scope.
If the constant current circuit works correctly, the voltage rise is linear up until the current source becomes saturated.

[Kevin Weddle]

You know the current is C dvdt. If you have a current you have a dvdt. The dvdt is the rate of change of the voltage. And what you are saying is that an opamp integrator uses this concept of a 0 dvdt to charge the capacitor. But of course the current is not zero and the capacitor's voltage is that of it's curve. But you miss the point I think when it comes to what the capacitors voltage does.

The input to the system is a signal with a dvdt. The capacitor uses the RC time constant to compare the time of the signal with the time of the RC. Remember the good old days when life was simple. Anyways, the capacitor will use up so many RC time constants between consecutive voltages. So what you have is a small amount of distortion between voltage levels. For example, if the signal time is 5 seconds and the 5RC is 25 seconds then you will use one time contant between consecutive voltage levels. If the signal changes to 10 seconds then you will have two time constants between consecutive voltage levels.

This is the harsh reality of the life of a capacitor.

[audioguru]

Kevin,
An opamp integrator doesn't have an RC time constant. It has the constant current times the reciprocal of the capacitance.
An opamp integrator also doesn't have dvdt. The rate doesn't change because the charging current is constant.

[prateeksikka]

hi all!
i just now had a look at this topic.
well i have confirmed this thing now from a book by "MILLMAN TAUB "
that displacement current through a capacitor is a 100% fictitious current.actually the engineers are forced to suppose this in support of Maxwell's equations .If it is not supposed then the Maxwell's equation wont hold true.Actually,there is no current through the capacitor!!!!it just charges and stops!!.the same phenomena makes us say that capacitor is open for pure dc and short for pure ac.because as audioguru said.pure dc without "R" in circuit
will charge the capacitor in 0 time.
:)prateek 

[audioguru]

You can pass an AC current through a capacitor just like it is a piece of wire.  ;D

[prateeksikka]

exactly!
why like a piece of wire ?
it is a piece of wire in its analogy with inductor also which is definitely a piece of wire for dc and similar is case with cap and ac.
prateek