Here in the Figure R5=50K. This resistor value is not in the equations to obtain output voltage, and in many tutorials non-inverting input is directly grounded without a resistor. What might be the reason to use it here?
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got trolled too much this week
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asked Dec 12, 2013 at 14:47
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\$\begingroup\$ The same question but for a non-inverting amp at electronics.stackexchange.com/questions/145897/… \$\endgroup\$– got trolled too much this weekCommented Jan 13, 2015 at 22:18
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2 Answers
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R5 equals R4 || R6 (resistors in parallel) calculated as $$R_5 = \frac{R_4 \cdot R_6}{R_4 + R_6}$$ and its purpose is to balance the two inputs of the opamp, it cancels the differences caused by the opamp input bias current
http://www.ecircuitcenter.com/Circuits/op_ibias/op_ibias.htm
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\$\begingroup\$ the amplifier in the circuit is: ti.com/product/tlv2404. Can I just use a 751 instead? The inout signal is DC coming from a potentiometer varying between 0V and 15V \$\endgroup\$ Commented Dec 12, 2013 at 15:11
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\$\begingroup\$ yes or the equivalent one. i mean i dont really understand why they use tlv2404 for this. \$\endgroup\$ Commented Dec 12, 2013 at 15:49
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\$\begingroup\$ LMV751 absolute max supply voltage is 5.5v, you can't use it for 15v \$\endgroup\$– alexan_eCommented Dec 12, 2013 at 17:45
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\$\begingroup\$ do you think any op-amp with 15V supply voltage could be used instead in this application? \$\endgroup\$ Commented Dec 13, 2013 at 11:56
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DC balance, aka input offset compensation.
Some opamps take a significant input current on each input pin - let's say 1 microamp. Work out the input impedance on the -ve input : 2 * 100k in parallel, giving 50K input impedance. Now that microamp input drops 50mv across that resistor, giving an output voltage of 50mv * the gain.
R5 compensates for this by dropping a matching 50mv on the other input, restoring the DC output to its correct value.
In practice there is also a voltage difference and a slight current imbalance between the inputs; called "input offset voltage" and "input offset current" which vary between amplifiers. This leaves an imbalance of (50k * offset current) + offset voltage at the input.
Some opamps draw tiny input currents; using these you can omit R5.
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\$\begingroup\$ the amplifier in the circuit is: ti.com/product/tlv2404. Can I just use a 751 instead? The inout signal is DC coming from a potentiometer varying between 0V and 15V. \$\endgroup\$ Commented Dec 12, 2013 at 15:10
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1\$\begingroup\$ It may be worth noting that on many op amps, the amount of current at the inputs may vary with temperature, phase of the moon, etc. but factors which affect one will tend to affect the other similarly. If the input current were known to always be exactly 1uA, connecting the non-inverting input directly to ground and sourcing or sinking 1uA from the inverting input some other way might be better than adding a resistor to the non-inverting input, but the resistor provides "automatic" compensation for any factors which affect both input currents equally. \$\endgroup\$– supercatCommented Dec 12, 2013 at 16:44
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\$\begingroup\$ Some algebra for calculating the value of the resistor at the positive input can be found here. \$\endgroup\$ Commented Dec 12, 2013 at 18:43