I was bored the other day and wanted measure the input capacitance of one the opamp that I was working with. The opamp in question is ADA4522 from Analog Devices. It's a great zero-drift opamp that has a rather large common mode input capacitance, which is about 35pF. I want to measure that!
The first thing that comes to mind is to insert a large series resistor with the opamp's non-inverting input to form a Low-pass filter at the input and measure the bandwidth at the output.
The experiment setup is depicted in figure.1. I chose the value of Rin to be 100kΩ, but it can be practically anything, we just have to know its value before we start the experiment. If the opamp's bandwidth is sufficiently larger than the low pass filter connected to the non-inverting input we can measure fc_in (cut-off frequency of the low-pass filter) at the output of the opamp. When fc_in is obtained Cin can be calculated using the formula shown in figure.1.
Simple right? However, even if you try simulating this circuit you don't get the results you expect. Let me show you.
Before starting and using the formula for Cin in figure.1 we expect to get about 45kHz of bandwidth, but if you look at figure.2 you can clearly see that the bandwidth is 5kHz lower than that. The input resistance and capacitance are modeled the same both in the datasheet and SPICE model, so this is not an incorrect model. Based on the measurement technique that we use here, the input capacitance is more that 39pF (39.34pF to be exact) compared 35pF, what is specified in the datasheet and modeled in the SPICE model of the opamp. What's going on here? What 's causing this extra input capacitance?
Well as it turns in some cases you can't just model the input opamp as a single capacitance to ground. It's kind of more complicated than that. Don't worry I will show you how to model exactly in this way with a little bit of tweak. Here are the SPICEfiles for you to play with.
The cause of that extra input capacitance is the differential resistance, not capacitance it's resistance. If you don't believe me go to ADA4522.cir (SPICE model) and delete the line that models the differential input capacitance "Cdm 1 2 7.0E-12", and you should get the same result. If you are still not convinced delete the line modeling the differential resistance "Rdm 1 2 30.0E+03" and rerun the simulation, in that case you measure about 35pF of input capacitance, which was the expected value. Opamp's finite and frequency dependent open loop gain also plays important role into explaining why this happens.
I'm not going to go into the extensive calculation to explain that. You can refer to Sergio Franco's Design with Operational Amplifiers and Analog Integrated Circuits book Section 6.3 for comprehensive explanation. Instead I'm going to give intuition into why differential resistance causes this issue. Scribd link to the book
Let's start with something simple. Consider a 1MΩ resistor connected between 10V and 9V voltage source like shown below in figure.3.
Even though, the resistor value is 1MΩ when it's "viewed" from 10V voltage source it looks a like a 10MΩ resistor to ground. It's because voltage difference across the resistor is small this makes the current coming out of the voltage source small, making the effective resistance (referred to ground) larger than the actual value.
How is this related to any of this? You are completely off topic, you might have yelled to your screen.
This is exactly what happens at the opamp's inputs. Imagine an ideal opamp which has infinite open loop gain, in that case the voltages at the inputs of the opamp are the same, hence the voltage difference is zero. This means that the differential resistance or impedance is irrelevant because there is no voltage across it, making it effective resistance infinite.
However, real opamps have finite open loop gain. Even at very low frequencies close to DC, there is going to be a very little voltage difference between the inputs of the opamps. Larger the open loop gain is, smaller this voltage gets. Since this difference is very small when you try to refer the differential input resistance to ground it's going to look much much larger, as it was shown in figure.3. No problem so far, but what happens when you go up in the frequency? As you might know, as frequency increases open loop gain decreases in real opamps. Lower open loop gain causes larger differential voltage across the differential resistance. This makes the effective resistance smaller compared to the effective resistance at very low frequencies.
Because of these reasons effective resistance "seen" at the non-inverting input to ground decreases as the frequency increases. Can you name anything else whose impedance also decreases with frequency? A capacitor. That's right. Even though the differential resistance is purely resistive because of the frequency-dependent property of the open loop gain it acts like a capacitor.
In figure.4 a more complete input model is shown including the effect of the differential resistance. I came across this model in Sergio's book mentioned earlier. It's explained very well in the book. I am going to briefly explain each component in the model, but I am not going to go into how they are exactly derived.
Obvisouly, Rin_cm and Cin_cm models the common-mode resistance and capacitance of the opamp. No problem there. Rin_eff is the ground referred version of the differential resistance at DC. As you can see in the figure in formula (1) it is affected by opamp's open loop gain and feedback factor. it's proportional to open loop gain and inversely proportional to closed loop gain.
Series connected Rin_dm and Cin_eff models aforementioned decrease in effective impedance. Rin_dm is the same as the differential resistance. Cin_eff affected by -3 dB point of opamp's open loop gain and Rin_eff that was calculated before.
This model seems a little complicated and useless, but we can simplify it as shown in figure.4. Rin_dm can be ignored if it's not too large compared to source resistance .You can even ignore Rin if it's too large (most of the time it is too large) and be left with only Cin which is the sum of Cin_cm and Cin_eff.
Returning back to the problem that we faced in the beginning of the post. Can you now see where the extra capacitance come from? It comes from Cin_eff.
To calculate Cin with Cin_eff and Cin_cm included we have to know opamp's differential resistance, open loop gain at DC and -3dB point of open loop gain. These are obtained from opamp's SPICE model. The calculation are shown below in figure.5.
Cin_eff is calculated to be 4.15pF. With the common mode capacitance 35pF total input capacitance, Cin, is about 39.15pF, which is very close to what we measured in figure.1.
These fancy calculations look good and reasonably accurate on paper and computer screen. However, there is a very little point trying to calculate effective input capacitance accurately. It is due to the fact that it depends on poorly controlled parameters like open loop gain and differential resistance. For example, open loop gain can vary as much as 20dB (100 times) from its specified value and you can't even find any information about differential resistance sometimes, but there are some useful conclusions we can draw some conclusions from what we discussed here. I'm going to discuss these conclusions in part 2.
This is the end of the first part. In the next part I'm going to compare different opamps with the same setup in SPICE and make some measurements.
Comments