"Choose so that 500 µA flows through it when () is applied across it."Here is the schematic:
View attachment 213744
I am confused about how to calculate the current flowing back into the inverting input, beause the design requirement states, "Choose so that 500 µA flows through it when () is applied across it." I dont know how to interpret it because it says the current through Rf must be 500 microamps when a maximum voltage of 10V is applied, but surely if the supply is AC then the value won't be fixed at 500 microamps due to the changing reactance from the parallel capacitor, and if it is DC then the current wont be able to pass beyond the first capacitor C1?
Apologies, the orignal text said "Choose the feedback resistor so that 500 microamps flows through it when a voltage equivalent to the maximum output voltage (10V in this case) is applied across it. The answer you provided explains nicely to me how it works, but I am yet to find a method of calculating the correct resistance and capacitance values for a current of 500uA. I suppose I can work them out using a bode plot and some nodal analysis."Choose so that 500 µA flows through it when () is applied across it."
What is "it"? Are the two "it"s referring to the same thing, or two different things?
What does "when ( ) is applied" mean?
A common design approach for simple amplifiers is that the capacitors are sized so that, at the frequencies of interest, each either acts very much like a short or very much like an open. At DC, all capacitors look like open circuits and at very high frequencies all capacitors look like short circuits. So you really only have two possibilities here, either C1 transitions from an open to a short before Cf or the other way around. Sketch out the amplifier gain for both possibilities and see which one makes the most sense and go with that.
Resistor current does not change with frequency therefore a fixed voltage of any frequency causes the same AC current.Apologies, the orignal text said "Choose the feedback resistor so that 500 microamps flows through it when a voltage equivalent to the maximum output voltage (10V in this case) is applied across it. The answer you provided explains nicely to me how it works, but I am yet to find a method of calculating the correct resistance and capacitance values for a current of 500uA. I suppose I can work them out using a bode plot and some nodal analysis.
It is not a differentiator that has no input resistor. The input RC acts like a simple highpass filter.
Sorry AG, the double negative made this even more puzzling to me. At lost here.It is not a differentiator that has no input resistor. The input RC acts like a simple highpass filter.
This is not really a differentiator, it's a high pass filter so the degree to which it functionally acts as a differentiator depends on the component values and the spectral content of the signal. Regardless, amplifier is a pretty generic term that can be applied to just about any device that transforms one signal into another.@WBahn @Audioguru again
I am puzzled; why a differentiator is discussed as amplifier?
Could you enlight the audience?
So, what about what I found in Wikipedia?This circuit in this thread has an input resistor in series with a capacitor feeding the virtual ground input of the inverting opamp. Then it is a highpass filter that cuts low frequencies but passes high frequencies. The opamp has a determined gain of Rf/Ri at high frequencies. The cutoff frequency is determined by the input and its series capacitor values.
A differentiator has no input resistor so its input capacitor passes only extremely high frequencies, It almost has infinite gain at an infinite frequency.
A differentiator does often have an input resistor in series with the cap. That is to keep noise down somewhat.This circuit in this thread has an input resistor in series with a capacitor feeding the virtual ground input of the inverting opamp. Then it is a highpass filter that cuts low frequencies but passes high frequencies. The opamp has a determined gain of Rf/Ri at high frequencies. The cutoff frequency is determined by the input and its series capacitor values.
A differentiator has no input resistor so its input capacitor passes only extremely high frequencies, It almost has infinite gain at an infinite frequency.
Hi,So, what about what I found in Wikipedia?
Surprising claim (at least for me).Regardless, amplifier is a pretty generic term that can be applied to just about any device that transforms one signal into another.
I probably should have caveated it with saying that it almost certainly wouldn't be considered an amplifier if the output power doesn't come from something other than the input signal.Surprising claim (at least for me).