One of the possible reasonsI know is that capacitors are available in wide range of values and can be made more accurate than inductors. What other factor leads to the use of capacitors over inductors?
Electronic – Why is it that we use capacitors in Differentiator & Integrator circuits (comprising Op Amps) & not inductors
analogoperational-amplifier
Related Solutions
You need to amplify your -1 to +1V signal to 0-10V, so a gain of 5 is required (with a input shift of +1V for single rail)
An opamp set up in non-inverting mode is a simple and accurate way of doing this. YOu also need to shift the +1V to -1V AC waveform so it's all positive (if you are using a single rail supply) Here is a circuit similar to the one you linked to, but setup for 10V output. L1 and L2 represent the transformer, V2 supplies 30A at 100V for simulation. R5 is included in your CT, so all you need is R1, R4 and everything to the right of them:
Simulation - here are the waveforms, you can see the 30A input, the 1VAC output (shifted to 0V-2V by the 1V output from the R1 and R4 divider) and the opamp output from 0-10V.
The opamp can be any rail to rail input and output part (unless you are using dual rail) with a supply range of over 10V (Opamp positive supply needs to be >+10V)
The lower circuit, RC
only works when:
- The output impedance of Vin source is 0Ω
and at the same time:
- Vin is much larger than VC . Ideally the capacitor voltage is 0V, but that defeats the whole function as an integrator. The voltage across the capacitor influences the current through the resistor.
and at the same time:
- The load of the circuit (VC) side is ∞
These requirements are not easily to met. The latter requirement is caused by the fact that any load on the capacitor C or resistor R will change the frequency responce of the integrator.
The opamp circuit relaxes these requirements, at least for the infinite load and for the ratio between input and output voltage.
Because the output is driven by an opamp, the output impedance of the circuit is resonably low (in practice maybe couple tens of ohms) and therefore it is sufficient if the output load is about 100 times higher than that; say > 1000Ω A requirement that is a lot easier to meet.
Also because the opamp will try to keep its inputs at the same voltage (V- is virtual ground), the voltage across the capacitor is no longer influencing the input current of the integrator. The input impedance of the opamp circuit is entirely specified by the resistor, hence an ideal integrator.
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Best Answer
A lot of circuits require medium/high impedance values for even moderately low frequencies such as audio and a capacitor of say 10 nF at 1 kHz has an impedance of 15.9 kOhm. An inductor having this impedance at 1 kHz would have a value of 2.53 henries.
Now that isn't a small value to fit in a space for a surface mount component. That's one reason and the next is cost - try finding a 2.5 henry coil in farnell, digikey or mouser and see how much change you have out of a dollar or a pound. The 10nF will cost you about a couple of pennies maximum.
So, it's not small and it's not cheap and a side effect of it not being small is that it will possess significant parasitic capacitance (several pico farads if not more) and this makes it less than perfect. I'm certainly not saying caps are perfect but they are a couple of orders of magnitude more perfect than an inductor.
Also it will have a significant DC resistance as it is made as small as it can be. It won't be good at handling currents like capacitor is and the core will probably saturate.
Bad idea - use capacitors.