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Input terminal by v in r2r 1 vout v in v in r2 r1

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Unformatted text preview: red from op-amp output) voltage drop across R2 is IfR2 = Vin ( R 2 /R1) so that output is higher than neg. input terminal by V in (R2/R 1 ) Vout = V in + V in ( R2 /R1 ) = V in (1 + R 2 /R1) – thus gain is (1 + R2 /R1 ), and is positive – negative sign earns title “inverting” amplifier • Current is sourced from op-amp output in this example • Current is drawn into op-amp output terminal drawn op-amp Winter 2008 9 Winter 2008 10 UCSD: Physics 121; 2008 UCSD: Physics 121; 2008 Differencing Amplifier Summing Amplifier R2 Rf R1 R1 V1 V R2 V2 + Vout + V+ Vout R1 R2 • Much like the inverting amplifier, but with two input voltages • The non-inverting input is a simple voltage divider: – inverting input still held at virtual ground – I1 and I2 are added together to run through Rf – so we get the (inverted) sum: V out = R f ( V1/R 1 + V 2/R2) – Vnode = V +R 2/(R1 + R 2) • So If = (V • if R2 = R1, we get a sum proportional to ( V 1 + V 2 ) • Can have any number of summing inputs – we’ll make our D/A converter this way Winter 2008 Lecture 9 Vnode)/R1 – Vout = V node IfR2 = V +(1 + R2/R 1)(R2/(R 1 + R2)) – so V out = (R2/R 1)(V + V ) – therefore we difference V + and V 11 Winter 2008 V (R2/R 1) 12 3 Op-Amps 02/14/2008 UCSD: Physics 121; 2008 UCSD: Physics 121; 2008 Differentiator (high-pass) Low-pass filter (integrator) R C C R Vin Vin + Vout Vout + • For a capacitor, Q = CV, so Icap = dQ/dt = C·dV/dt CV dQ C·dV • If = Vin/R, so C·dVcap/dt = Vin/R /R, C·dV – Thus Vout = IcapR = R C·dV/dt • So we have a differentiator, or high-pass filter – if signal is V0sin t, V out = V 0RC cos t – the -dependence means higher frequencies amplified more – and since left side of capacitor is at virtual ground: d Vout/dt = Vin/RC – so – and therefore we have an integrator (low pass) Winter 2008 13 Winter 2008 14 UCSD: Physics 121; 2008 UCSD: Physics 121; 2008 RTD Readout Scheme Notes on RTD readout • RTD has resistance R = 1000 + 3.85 T...
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