EECS 42, Spring 2005Week 5b1 Topic 1: Decibels – Logarithmic Measure for Power, Voltage, Current, Gain and Loss Topic 2: Operational Amplifiers

EECS 42, Spring 2005Week 5b2 Decibels – A Logarithmic Measure Curious units called “decibels” are used by EEs to measure electric power, voltage, current, the gain or loss of amplifiers, and the insertion loss of filters. The decibel (dB) always refers to the ratio of the value of a quantity to a reference amount of that quantity. The word decibel is a reference to powers of ten and to Alexander Graham Bell.

EECS 42, Spring 2005Week 5b3 Logarithmic Measure for Power To express a power, P, in terms of decibels, one starts by choosing a reference power, P reference, and writing Power P in decibels = 10log 10 (P/P reference ) Exercise: Express a power of 50 mW in decibels referred to 1 watt. Solution: P (dB) =10log 10 (50 x /1) = - 13 dB W. (The symbol dB W means “decibels referred to one watt”.)

EECS 42, Spring 2005Week 5b4 Aside About Resonant Circuits When dealing with resonant circuits it is convenient to refer to the frequency difference between points at which the power from the circuit is half that at the peak of resonance. Such frequencies are known as “half-power frequencies”, and the power output there referred to the peak power (at the resonant frequency) is 10log 10 (P half-power /P resonance ) = 10log 10 (1/2) = -3 dB.

EECS 42, Spring 2005Week 5b5 Logarithmic Measures for Voltage or Current From the expression for power ratios in decibels, we can readily derive the corresponding expressions for voltage or current ratios. Suppose that the voltage V (or current I) appears across (or flows in) a resistor whose resistance is R. The corresponding power dissipated, P, is V 2 /R (or I 2 R). We can similarly relate the reference voltage or current to the reference power, as P reference = (V reference ) 2 /R or P reference = (I reference ) 2 R. Hence, Voltage, V in decibels = 20log 10 (V/V reference ) Current, I, in decibels = 20log 10 (I/I reference )

EECS 42, Spring 2005Week 5b6 Note that the voltage and current expressions are just like the power expression except that they have 20 as the multiplier instead of 10 because power is proportional to the square of the voltage or current. Example: How many decibels larger is the voltage of a 9-volt transistor battery than that of a 1.5-volt AA battery? Let V reference = 1.5. The ratio in decibels is 20 log 10 (9/1.5) = 20 log 10 (6) = 16 dB.

EECS 42, Spring 2005Week 5b7 Gain or Loss Expressed in Decibels The gain produced by an amplifier or the loss of a filter is often specified in decibels. The input voltage (current, or power) is taken as the reference value of voltage (current, or power) in the decibel defining expression: Voltage gain in dB = 20 log 10 (V output /V input ) Current gain in dB = 20log 10 (I output /I input Power gain in dB = 10log 10 (P output /P input ) Example: The voltage gain of an amplifier whose input is 0.2 mV and whose output is 0.5 V is 20log 10 (0.5/0.2x10 -3 ) = 68 dB.

EECS 42, Spring 2005Week 5b8 Change of Voltage or Current with A Change of Frequency One may wish to specify the change of a quantity such as the output voltage of a filter when the frequency changes by a factor of 2 (an octave) or 10 (a decade). For example, a single-stage RC low-pass filter has at frequencies above  = 1/RC an output that changes at the rate -20dB per decade.

EECS 42, Spring 2005Week 5b9 Midterm 1 Midterm 1 – Thursday February 24 in class. Covers through text Sec. 4.3, topics of HW 4. GSIs will review material in discussion sections prior to the exam. No books at the exam, no cell phones, you may bring one 8-1/2 by 11 sheet of notes (both sides of page OK), you may bring a calculator, and you don’t need a blue book.

EECS 42, Spring 2005Week 5b10 Homework and Labs There will be no homework assignment for the next week, in view of the midterm. There will be a lab next week on RC Filters and the use of LabVIEW, a computer system for taking experimental data in a lab.

EECS 42, Spring 2005Week 5b11 OUTLINE The operational amplifier (“op amp”) Feedback Comparator circuits Ideal op amp Unity-gain voltage follower circuit Reading Begin Ch. 14

EECS 42, Spring 2005Week 5b12 The Operational Amplifier The operational amplifier (“op amp”) is a basic building block used in analog circuits. –Its behavior is modeled using a dependent source. –When combined with resistors, capacitors, and inductors, it can perform various useful functions: amplification/scaling of an input signal sign changing (inversion) of an input signal addition of multiple input signals subtraction of one input signal from another integration (over time) of an input signal differentiation (with respect to time) of an input signal analog filtering nonlinear functions like exponential, log, sqrt, etc

EECS 42, Spring 2005Week 5b13 Op Amp Circuit Symbol and Terminals +–+– V + V – non-inverting input inverting input positive power supply negative power supply output The output voltage can range from V – to V + (“rails”) The positive and negative power supply voltages do not have to be equal in magnitude.

EECS 42, Spring 2005Week 5b14 Op Amp Terminal Voltages and Currents +–+– V + V – All voltages are referenced to a common node. Current reference directions are into the op amp. – V cc + V cc – +vn–+vn– +vp–+vp– common node (external to the op amp) ipip inin ioio +vo–+vo– i c+ i c-

EECS 42, Spring 2005Week 5b15 Op Amp Voltage Transfer Characteristic  In the linear region, v o = A (v p – v n ) = A v id where A is the open-loop gain  Typically, V cc  20 V and A > 10 4  linear range: -2 mV  v id = (v p – v n)  2 mV Thus, for an op amp to operate in the linear region, v p  v n (i.e., there is a “virtual short” between the input terminals.) vovo v id = v p – v n The op amp is a differentiating amplifier: “linear” “positive saturation” “negative saturation” V cc -V cc Regions of operation: slope = A >>1

EECS 42, Spring 2005Week 5b16 Achieving a “Virtual Short” Recall the voltage transfer characteristic of an op amp: Q: How does a circuit maintain a virtual short at the input of an op amp, to ensure operation in the linear region? A: By using negative feedback. A signal is fed back from the output to the inverting input terminal, effecting a stable circuit connection. Operation in the linear region enforces the virtual short circuit. vovo vp–vnvp–vn V cc -V cc slope = A >>1 ~1 mV ~10 V vovo vp–vnvp–vn V cc -V cc slope = A >>1 Plotted using different scales for v o and v p –v n ~10 V Plotted using similar scales for v o and v p –v n

EECS 42, Spring 2005Week 5b17 Familiar examples of negative feedback: Thermostat controlling room temperature Driver controlling direction of automobile Pupil diameter adjustment to light intensity Familiar examples of positive feedback: Microphone “squawk” in sound system Mechanical bi-stability in light switches Negative vs. Positive Feedback Fundamentally pushes toward stability Fundamentally pushes toward instability or bi-stability

EECS 42, Spring 2005Week 5b18 Op Amp Operation w/o Negative Feedback (Comparator Circuits for Analog-to-Digital Signal Conversion) 1. Simple comparator with 1 Volt threshold:  V – is set to 0 Volts (logic “0”)  V + is set to 2 Volts (logic “1”)  A = Simple inverter with 1 Volt threshold:  V – is set to 0 Volts (logic “0”)  V + is set to 2 Volts (logic “1”)  A = 100 If V IN > 1.01 V, V 0 = 2V = Logic “1” If V IN < 0.99 V, V 0 = 0V = Logic “0” V0V0 V IN  V0V0 +  1V V IN V0V0 If V IN > 1.01 V, V 0 = 0V = Logic “0” If V IN < 0.99 V, V 0 = 2V = Logic “1” +  V0V0 V IN +  1V

EECS 42, Spring 2005Week 5b19 Op Amp Circuits with Negative Feedback Q: How do we know whether an op amp is operating in the linear region? A: We don’t, a priori. Assume that the op amp is operating in the linear region and solve for v o in the op-amp circuit. –If the calculated value of v o is within the range from -V cc to +V cc, then the assumption of linear operation might be valid. We also need stability – usually assumed for negative feedback. –If the calculated value of v o is greater than V cc, then the assumption of linear operation was invalid, and the op amp output voltage is saturated at V cc. –If the calculated value of v o is less than -V cc, then the assumption of linear operation was invalid, and the op amp output voltage is saturated at -V cc.

EECS 42, Spring 2005Week 5b20 Op Amp Circuit Model (Linear Region) vpvp vnvn RiRi +–+– RoRo vovo A(vp–vn)A(vp–vn) R i is the equivalent resistance “seen” at the input terminals, typically very large (>1M  ), so that the input current is usually very small: i p = – i n  0 Note that significant output current (i o ) can flow when i p and i n are negligible! ipip inin ioio

EECS 42, Spring 2005Week 5b21 Ideal Op Amp Assumptions: –R i is large (  10 5  ) –A is large (  10 4 ) –R o is small (<100  ) Simplified circuit symbol: –power-supply terminals and dc power supplies not shown +–+– +vn–+vn– +vp–+vp– ipip inin ioio +vo–+vo– i p = – i n = 0 v p = v n Note: The resistances used in an op-amp circuit must be much larger than R o and much smaller than R i in order for the ideal op amp equations to be accurate.

EECS 42, Spring 2005Week 5b22 Unity-Gain Voltage Follower Circuit V0V0 +  V IN V 0 (V) V IN (V) Note that the analysis of this simple (but important) circuit required only one of the ideal op-amp rules. Q: Why is this circuit important (i.e. what is it good for)? A: A “weak” source can drive a “heavy” load; in other words, the source V IN only needs to supply a little power (since I IN = 0), whereas the output can drive a power- hungry load (with the op-amp providing the power). v p = v n  V 0 = V IN ( valid as long as V –  V 0  V + ) vnvn vpvp I IN