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Op-Amp Current and Voltage Offset

No Chip (or man) is an island.

In the last post on this subject, “Our Op-Amp Circuits Input Impedance“, I mostly talked about the the outer two squares on the diagram to the left.  Today we will talk about the two inner squares on the diagram.  This is some of the things I kind of “hand waved”  or did the “shuck and jive” to in that post.

First, before even talking about the op-Amp one of the other things I “hand waved” though last time was the input impedance of my voltmeter when I took some of the data readings.   I actually measured the impedance of the meter and it is something to learn on how I did that.  Imagine a known voltage source with a resistor connected to the positive lead of the source and then the device with unknown input impedance connected from the resistor to the negative lead of the source.  In other words the resistor and the device are in series.   If the voltage across the device, my meter in this case, is 1/2 the voltage of the source, then the resistor is exactly the same resistance as the input impedance of the device being tested.  (I am using impedance and resistance interchangeably. In DC circuits both are the same.)

The following is the results of some tests on my meter on the 20 V range.

Vin = 10.15 V,  Series resistor 10 MΩ,  Voltage across the meter 5.02 V.   So it is very close to 10 MΩ input impedance.
As a 2nd check.  Vin=10.15V, Series Resistor 20 MΩ, Voltage across the meter 3.33 V, So again close to 10 MΩ input impedance.

I was curious if the range setting on the meter made any difference in the input impedance so I used Vin = 5.12V.  The values were 1.69 V for a 20 MΩ resistor, and 2.51 V for a 10 MΩ resistor.  Again very close to 10 MΩ input impedance.

Test circuit 1:

Test circuit 1: – Voltage Offset effects.

Both op-amps on the chip were wired as inverting amplifiers.  The gain was set very high with an input voltage of zero.  The non-inverting input was also fed zero volts through  a 10 Ω input resistor.  With zero volts going to both inputs, the output should be zero.  With the 100K feedback resistor, I actually measured +2.28 and -1.16 Vout on my two op-amps.  100K/10 = a gain of 10,000.  This means the amplifiers are acting as if they are summing amplifiers with an additional input of -0.228 and 0.116 mV on each.  This is called voltage offset.   A second test was ran with a 10K feedback resistor giving a gain of 1000 and the Vout values were +0.22 and -0.11 V as expected.

Often this offset is not important in many circuits.  As the gain is set lower, the offset also is not amplified and could be an insignificant value of the total output.  If it is significant in actual circuits, voltage offset is dealt with by using a summing circuit and a adjustable resistor in a voltage divider providing one of the inputs. The adjustable resistor is used to null the output voltage with an input of zero to the input of the amplifier.

In most actual designs several op-amp circuits are used in series.  The first op-amp may only be used to provide a high input impedance and then the second one to amplify and modify the signal and then maybe a third one to reverse the polarity of the output.  The summing amplifier used to null the voltage offset could be part of any of the amplifiers in this amplifier string.

Test Circuit 2:

Test Circuit 2: Current offset effects.

In test circuit 2 I wired both amplifiers to be non-inverting amplifiers with a gain of 101. This configuration was used because in this case I was interested in the current flowing into the amplifier and wanted the a very high input impedance to develop a voltage due to this current.
In the first test I used a Rx value of 10MΩ and both amplifiers were saturated with an output of +9.47 V.  (Remember my +Vcc is only 10.15 V so the amplifier was as close to +Vcc as it can be.)

I replaced the 10MΩ Rx value with a 1MΩ and got measurable outputs of 2.42V on one amplifier and 1.86V on the other amplifier.  With our gain set at 101, this means we have an input of 24mV and 18 mv on the two amplifiers.   Since the resistor is grounded, the only way to get this voltages is a current flowing out of the non-inverting input.  This current is very small but because the resistance is so large the voltage becomes important.  24mv/10MΩ = 2.4 nA and 1.8 nA for the other amplifier.   These values are significantly less than the value on the data sheet, but are enough to cause us problems.

Current offset is only a problem if we are measuring across a high impedance source.   (There is another case where it can be a problem, but that has to do with AC and we have not got there yet.)   If there is a situation where current offset can be a problem the best answer is to purchase an op-amp that is compensated for the current offset.  Another compensation that can be done is to make sure both the inverting and the non-inverting inputs see the same impedance.   This is often done on an inverting amplifier by  putting a resistor from the non-inverting input to ground instead of just wiring the input to ground.  The size of the resistor is equal to Rg in parallel with Rf.

In the next post about Op-Amps I will summarize all of what we have talked about so far and quickly deal with the output voltage from the Op-Amp.   From there we will move on to using an Op-Amp in a temperature measuring circuit.  After that we will back up and start talking about energy storage devices, capacitors and inductors.  That will lead us eventually into AC and signals.   There is lots of stuff to learn and I will try to present practical uses throughout the process.

Gary


 

 

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