Unformatted text preview: red from opamp 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 opamp output in this example • Current is drawn into opamp output terminal
drawn
opamp
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 noninverting 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 OpAmps 02/14/2008 UCSD: Physics 121; 2008 UCSD: Physics 121; 2008 Differentiator (highpass) Lowpass 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 highpass 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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 Winter '08
 staff
 Power, Operational Amplifier, Cybernetics, Positive feedback

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