Basic impedance calculation with LM741 op-amp buffer circuit

Thread Starter

wlam13

Joined Apr 9, 2018
2
Hello all,

I have a simple impedance calculation question regarding the circuit below,

upload_2018-4-9_16-39-39.png
The circuit consists of a simple buffer circuit with a LM741 Op-Amp, the purpose of the resistors is to increase the overall circuit impedance (and other applications not related to this question). Also from the LM741 datasheet, the input resistance of this op-amp is typically 2MΩ.

I have replicated this particular part of the circuit in a lab and when I use my multimeter to measure impedance (from input to ground), the impedance measures around 4.2 MΩ.

How does the impedance calculate to this value?

What will the impedance read when I replace the LM741 op-amp with a OPA1662AIDRQ1 op-amp?


The differential resistance of the OPA1662AIDRQ1 op-amp is 170KΩ, the common mode resistance is 600KΩ .
 

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WBahn

Joined Mar 31, 2012
25,745
First off, the input impedance numbers apply to when the circuit is powered. If it is unpowered, then who knows what the internals of the opamp will result in. And you can't use a resistance meter to measure resistance in a powered circuit, so you need to get a little more elegant in your measurement techniques.

But let's assume that your measurement has yielded useful results. What would the input impedance of the opamp need to be in order to yield the 4.2 MΩ value? Is that in conflict with the data sheet (which, after all, only gives a typical or perhaps minimum value). Does it give a maximum value?
 

Thread Starter

wlam13

Joined Apr 9, 2018
2
When you say the circuit is powered, does that refer to the input voltage, the rail voltages to the op-amp, or a combination of the two?
When I measured the impedance I did not supply the input voltage, only the rail voltages into the op-amp.

For the input impedance of the Op-Amp I calculated for it to be about 6MΩ to yield the 4.2MΩ value.
(Please check my calculation)
There is no maximum value listed for this op-amp, only the minimum and typical,0.3 and 2.0MΩ respectively.

Can I conclude the input impedance is indeed 6MΩ (assuming my measurements were correct)?
 

danadak

Joined Mar 10, 2018
4,057
Whats the input Z of your multimeter ? If the circuit had ideal OpAmp
Zin = 10 M, because there would be no current flowing thru 1M, hence no
Vdrop across it.

Regards, Dana.
 

WBahn

Joined Mar 31, 2012
25,745
When you say the circuit is powered, does that refer to the input voltage, the rail voltages to the op-amp, or a combination of the two?
When I measured the impedance I did not supply the input voltage, only the rail voltages into the op-amp.
The opamp needs to be powered, which means that you can't just hook up an ohmmeter to the circuit to measure the resistance since the meter relies on the assumption that the circuit is unpowered. So what you need to do is get more fundamental and apply a voltage and measure the current and take the ratio. Depending on the circuit, you need to actually take two data points and determine the incremental resistance.

For the input impedance of the Op-Amp I calculated for it to be about 6MΩ to yield the 4.2MΩ value.
(Please check my calculation)
Yep. About 6.2 MΩ.

There is no maximum value listed for this op-amp, only the minimum and typical,0.3 and 2.0MΩ respectively.

Can I conclude the input impedance is indeed 6MΩ (assuming my measurements were correct)?
Assuming your measurements yielded a valid result, then at least at that input condition that would be correct. And it is not too unreasonable that a random part of the shelf might have around three times the typical resistance since the more the better and the specs cover a whole range of operating points, particularly over temperature and supply voltage, so the values right in the middle of all that is usually "better" than the value good over the entire range.

It is very likely that the effective input impedance changes as the input level changes and it probably has a reactive component to it as well, meaning that it will change with frequency.
 

MrAl

Joined Jun 17, 2014
7,588
Hello all,

I have a simple impedance calculation question regarding the circuit below,

View attachment 150155
The circuit consists of a simple buffer circuit with a LM741 Op-Amp, the purpose of the resistors is to increase the overall circuit impedance (and other applications not related to this question). Also from the LM741 datasheet, the input resistance of this op-amp is typically 2MΩ.

I have replicated this particular part of the circuit in a lab and when I use my multimeter to measure impedance (from input to ground), the impedance measures around 4.2 MΩ.

How does the impedance calculate to this value?

What will the impedance read when I replace the LM741 op-amp with a OPA1662AIDRQ1 op-amp?


The differential resistance of the OPA1662AIDRQ1 op-amp is 170KΩ, the common mode resistance is 600KΩ .
Hi,

What is the input bias current?

I ask because that probably affects this reading because the input impedance to a non inverting amplifier it pretty darn high. The reason for this is because the output is fed back to the inverting input, and it is in phase with the input voltage and that means the input impedance, theoretically, goes very high.
An example would be if you had a dependent voltage source with 2M across the sense inputs and internal gain of 200k and the output fed back to the negative sense lead, the input impedance at 1v would be something like 4e11 Ohms, which is much higher than the 2M input resistor. That's because as the input goes higher, the output also goes higher and that reduces the voltage rop across the 2M input resistor, hence the input current barely increases at all. It does increase slightly though because the output does nto follow the input prefectly as there is a tiny tiny difference, but it's so small it leads to a feedback voltatge that is very nearly the same as the input voltage, and hence the 2M resistor draws very little current and hence very high input impedance.

So if you measure something in the MegOhm range, it's probably because of bias current, but we could check that too.
 
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