How do i analyze the below circuit? What is the starting point?
Hi,
It takes a little experience to understand this circuit because here are some shortcuts that you can take that eliminates some of the components and that makes the approximation easy. I'll run it by you and then supply some more details for an exact analysis.
First, this circuit does not need a negative supply to work, so I don't know why so many would be mentioning that. It's already biased to one-half of the positive supply, which also makes the output biased to about 1/2 of the power supply voltage also. That means if you have a supply of 10 volts then the output is at roughly 5v.
This works because of the two 220k Ohm resistors, which creates a voltage divider putting the junction (and voltage across the 0.5uf cap) at 1/2 of the supply voltage. With that we see 5v across the 0.5uf cap (10v power supply again). Since the 100k sees infinite impedance on the right for DC, that puts 5v at the noninverting input of the opamp also. The left side of the 30k resistor also sees infinite impedance because the input to the opamp is infinite and the 1uf cap has infinite impedance for DC. That means for DC, the 30k resistor makes the circuit look like a gain of 1 non inverting opamp circuit, which is where we get 5v at the output. Be aware again this is for DC which is for determining the bias voltages and/or currents. The AC operation will be different because the capacitors are not open circuits for AC only for DC.
Ok so we have established the DC operating point which for a +10v power supply means 5v at the noninverting input and 5v at the output, and the 30k is the only feedback for now so we have a gain of +1 circuit which is where we get the 5v at the output.
Now for the AC analysis.
For this the caps become non-zero impedances and this means there could be a small effect or a large effect, and for this circiut for most frequencies there will be a large effect, but only on the AC output result not the DC.
The input comes in through the 0.1uf cap which works with the 100k resistor, and that creates a voltage divider with the voltage across the 0.5uf cap. Since the input DC is biased at 1/2 of the supply voltage Vcc, the AC voltage will cause some AC voltage to ride on the 5v DC bias, which means the opamp effectly sees the AC voltage alone, and for a first approximation we consider the 0.5uf cap to be large enough so that the DC voltage across it does not change much, and this is reasonable for now.
Since the 0.1uf impedance will be fairly low compared to the 100k resistor for most AC frequencies, we can approximate the input at the noninverting input to be EQUAL to the input voltage itself.
Now before we saw the 1uf cap be open circuit for DC, but now we have AC so we have to consider it as having a significant effect. Since the 1uf and 2k shunt some of the current away from the inverting input, that means the gain will change. Before we had a gain of 1, now we have to change that because of the voltage divider effect of the 30k and the 2k in series with the 1uf cap.
The voltage divider effect has to take into account the impedance of the 1uf cap, there's no way around that except at very low frequencies. The impedance of the 1uf cap is:
Z=1/(2*pi*f*C)
and with C=1e-6 that makes the impedance:
Z=1/(2*pi*1e6*f)
which is about:
Z=1/(6.3e6*f)=f*1.6e-7
The series string with the 2k resistor makes that string:
Zs=Z+2000
This comes out to about :
Zs=2000.0-(159154.9430918953*i)/f
This will be close to 2000 for frequencies around 1kHz and above, so let's use that as a starting approximation.
Combined with the 30k resistor, the voltage divider effect is 1/16, so we get one-sixteenth of the output voltage at the inverting input. This means the output will be 16 times greater than the voltage at the inverting input, which means the AC again will be much higher than the DC gain.
Since this is a noninverting op amp circuit now, the gain is:
G=1+Ginv
where Ginv is the inverting opamp gain. Since Ginv=15 that puts the AC again at 16.
This of course means that if we input 0.1vac we will get about 1.6vac output at frequencies around 1kHz or above, up to the frequency where the opamp does not function well.
So that's the short cut approximation just so you can get an idea how this works without going too deep in the math yet.
If you want a full analysis, then you have to go to describing all the capacitors as complex impedances, combine those with the resistances, and then calculate the output. This analysis will allow calculating a result with frequencies lower than 1kHz as well as others.
I can show you how to do this if you would like to see how it is done. It is more complicated, but if you understand complex numbers it is basically easy.
Just to add a little here, opamps are not perfect devices so sometimes we have to take into account other parameters such as open loop frequency response and slew rate. For the analysis above we assume the opamp can follow any input perfectly, but if you need a higher frequency response then you have to go to a better opamp then a cheaper general purpose opamp. You can find details about this on the web.
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