Operational Amplifiers I Lecture 7 Operational Amplifiers I

The Op Amp is a voltage – controlled device The perfect amplifier…almost. Vin = Vdiff → 0 Rin → ∞ Rout → 0 G → ∞

Non-Inverting Amplifier Iin Iin ~ 0 … why? Vin ~ V+ ~ V- … why? Vdiff ~ 0 … why? Iout = Vout / (R1 + R2) … why? Vin Iout Let Vin go positive…V+ input will draw no current! Vdiff = 0…therefore… Iout (supplied by op amp) all flows through R1 AND R2 Vin = V - = V + Vin = Iout R2 = R2 (Vout / (R1 + R2)) Gain = G = Vout / Vin = (R1 + R2) / R2 G = 1 + R1 / R2

The open-loop op amp gain A is infinite…A = ∞. The inputs draw no current…Rin is infinite…Rin = ∞. With negative feedback…the op amp does whatever it can to make Vdiff = 0.

Inverting Amplifier G = Vout / Vin = - R2 / R1 I I Assume Vin goes positive (the reverse will take place if it goes negative). Current I flows through R1 and R2 and is “sunk” by the op amp output. Vdiff = 0 Therefore… V - = V+ = 0 I = Vin / R1 Vout = - I R2 = - Vin (R2 / R1) G = Vout / Vin = - R2 / R1

Negative Feedback Vin Vdiff Vout β Vout Vdiff = V + - V - Vout = AOL Vdiff V - = β Vout Vdiff = Vin - β Vout Vout = AOL (Vin - β Vout) Vout (1 + β AOL) = Vin AOL G = Vout / Vin = AOL / (1 + β AOL) ~ 1 / β

How Negative Feedback Affects Zin & R0 Voltage Feedback A sample β of the output voltage is fed back… Vin = Vx - β Vout … Why? βVout is the (-) feedback voltage, which reduces the applied input voltage Vx to Vin … (Vin here was called Vdiff previously) Vx = Ix Rin + β Vout Vout = AV Vin = AV Ix Rin … (AV is the open-loop gain) Vx = Ix Rin + β (AV Ix Rin) Rin’ = Vx / Ix = Rin (1 + β AV) … (β AV is called the loop gain …) Rout’ = R0 / (1 + β AV ) … ( The output impedance is reduced ) A sample β of the output current is fed back… Rin’ = R0 / (1 + β AV ) … (The input impedance is reduced ! ) Rout’ = R0 (1 + β AV ) … (The output impedance is increased ! )

Non-Inverting Amplifier … again β = R2 / (R1 + R2) G = A / (1 + β A) ~ 1 / β G = (R1 + R2) / R2 G = 1 + R1 / R2

Inverting Amplifier … again A more ‘graphic’ way to think…imagine the resistance from input to output is continuous. What is voltage drop as you ‘move along’ this resistance? (What is V(R) ?) V Vin G = Vout / Vin = - R2 / R1 V = 0 R R1 R2 Vout

Difference Amplifier Gain of Difference Amplifier Exercise: Assume R1 = R2 = R Ground V2 … apply signal V1 … an inverting amp Vout = - V1 Ground V1 … apply signal V2 … non-inverting amp Vout = V2 Let V1 = V2 … Vout = V2 – V1 = 0 In general then, Vout = V2 – V1 (with all resistors = ). Formal Derivation… R1 ≠ R2 (i) Vy = V2 [ R2 / ( R1 + R2 ) ] = Vx (ii) I1 = ( V1 - Vx ) / R1 = (Vx – Vout ) / R2 Substitute Vx from (i) into (ii) and with a little algebra… Vout = (R2 / R1) ( V2 – V1 ) I1 x y Integrated op amp (INA105A) with uncommitted “sense” and “reference” pins. Resistors matched to +/- 0.01%. Classic dif amp…connect sense to output and ref to gnd. You can make ‘nifty circuits… Exercise: Show how to make (i) – (iii) Precision unity gain inverter Non-inverting gain of 2 Non-inverting gain of 0.5 Output to upper sense; in(+) to gnd Output to upper sense; in(-) to gnd Output to upper sense; ref to gnd

Are the Golden Rules Applicable Here? Probably a mistake…the designer wants a really, really high G amplifier…but no feedback and for reasons we’ll soon see, it will move into ‘saturation’. This is a goody… it works fine. It’s called a follower. This is a disaster… it’s feedback is positive…not negative. The Moral of the Story… Use feedback Make sure it’s negative Keep the op amp in the active region…not saturation.

Are the Golden Rules Applicable Here? These circuits work…conditionally. works… but only for (-) inputs…not (+) as shown. Why? works … but only for (+) inputs…not (-) as shown. Why?

Inverting Amplifier The Follower We get this… Suppose we let R1 → 0 and R2 → ∞ Extremely useful circuit: Unity gain buffer Isolates input and output circuits High input, low output impedance

More Followers (a) (b) (c) (d) (a) Vout = Vin – VBE Rout = rE Rin = β RE (b) Vout = Vin Rout = 0 Rin = ∞ Iout limited (c) Vout = Vin Rout = 0 Rin = ∞ Iout up to ~ > 1 A But only ‘sources’ Iout (d) Vout = Vin Rout = 0 Rin = ∞ Iout up to ~ > 1 A ‘Sources’ and ‘Sinks’

Class A Audio Amplifier (Note: Goes with BJT lectures) 10 W loudspeaker amplifier… Output driven by single-ended emitter follower, Q2 Dissipates 165 W of quiescent power! … Why? A single-ended follower operating between split supplies can drive a grounded load up and down ONLY if a high quiescent current is used. Q1 included for 2 reasons: (i) reduces drive requirements. (ii) cancels Q2’s VBE offset, i.e., 0 V in = 0 V out 100 mA current source provides Q2’s large base current requirement (~ 50 mA) at the top of signal swing. Power Dissipation Q2 can swing ~ +/- 15 V (9 V rms) ILOAD ~ (9V / 8Ω ~ 1.1 A. PLOAD ~ (9V)(1A) ~ 10 W With 0V in, P8Ω = (V8Ω)2 / 8Ω = 113 W IQ2 ~ IE = 30V / 8Ω = 3.75 A, so P8Ω = 3.75A (15V) = 56 W

Push-Pull Output Q1 sources when signal + Q2 sinks when signal – No power dissipation with no signal in. But…Crossover distortion Vout lags Vin by VBE Therefore, Vout ~ 0 between +/- VBE Biasing remedy Place push-pull stage into slight conduction. Diodes do that. As signal “crosses over’ Q1 or Q2 is in conduction. R must insure enough base current when Q1 or Q2 on hard. 100 Ω resistor eliminates any “dead zone” during cross over

Cleaning Up Cross-over Distortion The “SILLY” connection generates a beautiful sinewave at the op amp output, but the output of the push-pull suffers cross-over distortion. The “SMART” connection uses the magic of feedback to modify the op amp output so that the beautiful sinewave appears at the push-pull output! Class AB Amplifier Audio speakers have inductance as well as resistance. The “snubber” circuit is used to prevent ringing.

Cleaning Up Cross-over Distortion SILLY Scope Traces for resistive load SMART SILLY Scope Traces for speaker load SMART

Summing Amplifier Otherwise Variant of Inverting Amplifier Point X is a virtual ground Input is V1 / R1 + V2 / R2 + V3 / R3 If R1 = R2 = R3 = R … Vout = - (V1 + V2 + V3) Inputs can be + or – If R’s are unequal…get a weighted sum. For example…suppose R1 = 2R2 = 4R3 = 8R4 … what do you get?

Transresistance Amplifier A resistor is a current to voltage converter… otherwise known as a transresistance amplifier. V Many devices generate currents, eg: Photodiodes Phototransistors Photomultiplier tubes, etc. Often the currents are small. They require amplification and conversion into a voltage. A very large resistor might work, but it might load the output of the current source. B. Phototransistor sources current. Phototransistor used as photodiode to sink current. Solution… Use an op amp to source (or sink) the output current through a large resistor. Careful tho…such circuits are prone to oscillate because of the output capacitance of the source.

Current Sources (a) (b) Circuit (a) is ~ ideal current source without a transistor’s VBE. ILOAD = Vin / R Disadvantage I…load “floats” … not tied to ground or a power supply. eg, Couldn’t generate a sawtooth wave wrt ground. Disadvantage II…All load current comes from op amp and its output current is almost always limited to small values, say +/- 25 mA. I Vin Vin I Circuit (b) much better… again… ILOAD = Vin / R but … (i) Uses transistor to source current, which is provided by power supply. (ii) Load doesn’t float…it’s tied to power supply. (iii) Transistor is within feedback loop…op amp feedback eliminates its annoying VBE. (iv) Op amp eliminates effect of VBE variations with temperature and current. (v) Op Amp makes transistor a “perfect” current source. Its collector resistance doesn’t matter.

PNP Current Source Iout = VCC (R1 / R) / (R1 + R2) Vin = VCC R2 / (R1 + R2) Iout ~ IE = (VCC – Vin) / R Iout = VCC (1 - R2 / (R1 + R2)) / R Iout = VCC (R1 / R) / (R1 + R2) Idiv IE Vin No VBE offsets No VBE variations with temperature No variations of VBE with IC No variations of VBE with VCE But IB might vary with VCE, so IC could change even though IE cannot. (IE is stabilized by the op amp…not IC!) Use a Darlington…or… Use a MOSFET. Vin is a fixed voltage referenced to VCC… Suppose you want a variable Vin referenced to gnd! This circuit does the trick!

Howland Current Source A “textbook” current source IF R4 / R1 = R3 / R2. Convert to non-inverting current source by switching Vin and ground. Resistors must be matched “perfectly”. Limited by CMRR. For large currents, R’s must be small; limits compliance. At hi frequencies, loop gain is low and output impedance can drop from ∞ to 100’s of Ω’s. I1 I4 Vx I2 I3 Solve for (V0 – Vx) Substitute into 2nd eqn above