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# What is the digital circuit equivalent of a capacitor??

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#### SanjayKKK

##### Newbie level 4
In analog circuits we have capacitors, inductors etc..

What is the digital circuit equivalent of these elements which operate in digital values??

In digital, you have logic gates

kunu

### kunu

Points: 2
Hello Sanjay?,

sounds like an interesting philosophical question to me.
I deal a lot with mixed designs. Hence I could imagine a counter is a good equivalent to a cap or inductor.
Just think of adding/subtracting a charge with each increment/decrement. The counter value would simplyreflect the voltage.
For an inductor the increment/decrement would reflect the current caused by voltage jump. The very intersting point is how to emulate the sudden cut off of current and its negative voltage peak.

Hope that helps a bit

Rolf

kunu

### kunu

Points: 2
Schematically, You can think about it like system with following equations:
a) for capacitor
y[n] = x[n] + y[n-1]
b) for inductor
y[n] = x[n] - x[n-1]
It means You can simulate it with add and delay elements.

SanjayKKK

### SanjayKKK

Points: 2
For any MOS transistor, Gate to node A, Drain & Source to node B.
Node A & B give you the digital circuit equivalent of a capacitor.
This is a common practice in VLSI design.

In digital, you have logic gates

SanjayKKK

### SanjayKKK

Points: 2
In analog circuits we have capacitors, inductors etc..
What is the digital circuit equivalent of these elements which operate in digital values??

For my opinion, the question - in this general form -makes no sense.
Why do you expect that a "circuit equivalent of these elements" does exist in the digital domain?
I think the only possible question in this respect could be:
Which electrical property of a capacitor/inductor could possibly have something like an analogon in the digital domain? Possibly - as some answers imply - the charging process of a capacitor?
But I think there will be no equivalent as far as the voltage-current relationship in the time domain or the impedance characteristic in the frequency domain is concerned.

---------- Post added at 13:18 ---------- Previous post was at 11:42 ----------

A capacitor is a mechanical part with specific electrical properties; "analog" and "digital" are two special methods to process signals. A capacitor can be used in analog working circuits as well as in digital operating circuits, because it can act as a storage element for voltages. Thus, it really makes no sense to ask for a capacitor equivalent for digital applications.
By the way: For my opinion, there are no "digital parts" at all. These parts or circuits, which are believed to be "digital", are operated beween to distinct values only. Between these treshold values there is a (perhaps small) region which behaves "analog". On the other hand, an opamp can be used as a "digital" circuit if operated like a comparator between both extreme voltages.
Simple example: A CMOS inverter can be (and is) used as a linear amplifier if properly biased.

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SanjayKKK

### SanjayKKK

Points: 2
Perhaps you mean that as an analog circuit we can build an RC filter so that for each analog signal at its input there will be a well defined output.

Your question might be therefore: How can we build a digital system which would perform the same function of the previous RC filter? Obviously there will be an ADC at its input and a DAC at its output.

If this is the case, I leave its answer to those who work in digital filters :smile:

SanjayKKK

### SanjayKKK

Points: 2
If what you are asking is about functionality, i.e. integration, derivation, etc. (like an RLC filter) there is some equivalencies, and it depends on the transformation you use. You can transform any transfer function in the S domain (analog) to the Z domain (digital) using the bilinear transform (check Bilinear transform - Wikipedia, the free encyclopedia)

However, there are several transformations which use V/I or scattering matrix to implement a similar behavior in the digital world. These techniques are used to implement different digital filter topologies like LDI or wave digital filters.

Hi all, Thanks for all your replies..

Hey Max Planck, your answer is wat is given in my textbook
You said,

a) for capacitor
y[n] = x[n] + y[n-1]
b) for inductor
y[n] = x[n] - x[n-1]

Can we call the equation for capacitor as Differentiator and that of inductor as Accumulator???

I have problem understanding these equations viz
In mathematics diffention of a function will be like this
y = [f(x) - f(x+h)] / h
It would represent the rate of change or the slope..

How come in digital domain we represent diffrentiation like this
y[n] = x[n] + y[n-1]
How does this represent the rate of change??

Futher details:
I came across this when reading Delta Sigma Modulation.. where the integerator is used to pre-emphasize the signal..

I found this simple introduction. You may like to have a look on it, "Digital filter design for Analogue engineers.pdf":
www.jamesoakwood.co.uk/Digital filter design for Analogue engineers.pdf

How come in digital domain we represent diffrentiation like this
y[n] = x[n] + y[n-1]
How does this represent the rate of change??

I think you meant
y[n] = x[n] - x[n-1]
It seems the sampling rate here is the counterpart of your 'h'.

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SanjayKKK

### SanjayKKK

Points: 2
Hi SanjayKKK

Can we call the equation for capacitor as Differentiator and that of inductor as Accumulator???
I think that is inversely, ie. a capacitor works as an accumulator (integrator) and an inductor works as a differentiator (providing that You think about a voltage as an output signal).

In mathematics diffention of a function will be like this
y = [f(x) - f(x+h)] / h
It would represent the rate of change or the slope..
I would rather write:
y = [f(x + h) - f(x)]/h.
If You think about h as a unit increase previous equation will be: y = f(x + h) - f(x) or y = f(x) - f(x - h).

How come in digital domain we represent diffrentiation like this
y[n] = x[n] + y[n-1]
How does this represent the rate of change??

It represents rather accumulation not differentiation.
Rate of change represents rather following equation:
y[n] = x[n] - x[n-1]

SanjayKKK

Points: 2