Superior Measurements with a Differential Amplifier
Posted by admin on July 25, 2012
PXI digitizers have limited input range resulting from space constraints. This paper explores the advantages that a differential input . . .
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billrussell42 said,
I know it's tricky.
How I look at it is: The opamp will drive the output to cause the voltage across the inputs to be zero, as long as the opamp is operating in it's linear region. Sounds simple, but it works every time.
For the inverting opamp, one input is grounded, so the other input is forced to be ground. Since the input draws zero current, the current in the input resistor equals the current in the feedback resistor.
Say those are 1k and 10k. Apply one volt to the input. The current in Ri, since the other end is at vritual ground, is 1v/1k = 1mA. That 1 mA has to also flow in the Rf of 10k. 1mA and 10k is 10 volts. Looking at the current flow, that has to be –10v at the opamp output.
Thus we have 1v input and –10v output, a gain of –10.
Rick said,
I know it's tricky.
How I look at it is: The opamp will drive the output to cause the voltage across the inputs to be zero, as long as the opamp is operating in it's linear region. Sounds simple, but it works every time.
For the inverting opamp, one input is grounded, so the other input is forced to be ground. Since the input draws zero current, the current in the input resistor equals the current in the feedback resistor.
Say those are 1k and 10k. Apply one volt to the input. The current in Ri, since the other end is at vritual ground, is 1v/1k = 1mA. That 1 mA has to also flow in the Rf of 10k. 1mA and 10k is 10 volts. Looking at the current flow, that has to be –10v at the opamp output.
Thus we have 1v input and –10v output, a gain of –10.
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Rick said,
I know it's tricky.
How I look at it is: The opamp will drive the output to cause the voltage across the inputs to be zero, as long as the opamp is operating in it's linear region. Sounds simple, but it works every time.
For the inverting opamp, one input is grounded, so the other input is forced to be ground. Since the input draws zero current, the current in the input resistor equals the current in the feedback resistor.
Say those are 1k and 10k. Apply one volt to the input. The current in Ri, since the other end is at vritual ground, is 1v/1k = 1mA. That 1 mA has to also flow in the Rf of 10k. 1mA and 10k is 10 volts. Looking at the current flow, that has to be –10v at the opamp output.
Thus we have 1v input and –10v output, a gain of –10.
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upgrade lah, space constraints? Despite having a bigger oven, I always bake one layer at a time. Cookies also. *cries*
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Ditto daniel's post, it's mediaeval (or medieval, depending on whose dictionary you read)]]>
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Rick said,
I can understand your confusion and frustration. Here's the key (critical) point. The gain of an ideal op amp is infinite. Therefore (so long as the output voltage is in the active, not saturated, range) the voltage across the inputs will be zero, for any output (because the gain is infinite). Now about virtual grounds; assume the + input is grounded (and the output is not saturated), then the – input will also be at ground (because of the infinite gain). So Vin/Rin=-Vout/Rf (the bias current is zero) and the gain is -Rf/Rin. You can do the same type of analysis for the noninverting configuration. Vin/R1=(Vout-Vin)/Rf, Vout/Vin=1+Rf/R1
Edit: You must understand Ohm's law in order to understand this. In the case of the inverting amp, the curretn through Rin is Iin=Vin/Rin. This same current must flow through Rf. Therefore, Vo=0-Iin*Rf. So, from Ohm's law If=Vo/Rf. Since If=-Iin, Vo=-Iin*rf. Again, the key is the infinite gain of the ideal op amp.
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It took ten minutes since he had to answer questions and use the chart to diagram time and space constraints.
said,
Ditto daniel's post, it's mediaeval (or medieval, depending on whose dictionary you read)]]>
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billrussell42 said,
I can understand your confusion and frustration. Here's the key (critical) point. The gain of an ideal op amp is infinite. Therefore (so long as the output voltage is in the active, not saturated, range) the voltage across the inputs will be zero, for any output (because the gain is infinite). Now about virtual grounds; assume the + input is grounded (and the output is not saturated), then the – input will also be at ground (because of the infinite gain). So Vin/Rin=-Vout/Rf (the bias current is zero) and the gain is -Rf/Rin. You can do the same type of analysis for the noninverting configuration. Vin/R1=(Vout-Vin)/Rf, Vout/Vin=1+Rf/R1
Edit: You must understand Ohm's law in order to understand this. In the case of the inverting amp, the curretn through Rin is Iin=Vin/Rin. This same current must flow through Rf. Therefore, Vo=0-Iin*Rf. So, from Ohm's law If=Vo/Rf. Since If=-Iin, Vo=-Iin*rf. Again, the key is the infinite gain of the ideal op amp.
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Emailed three articles to oped pages of the Dawn,reply was ‘can not be published due to space constraints’.Is someone here 4 help?
said,
Ditto daniel's post, it's mediaeval (or medieval, depending on whose dictionary you read)]]>
said,
Ditto daniel's post, it's mediaeval (or medieval, depending on whose dictionary you read)]]>
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