Prepared by PATEL DEEP J. ENROLL. NO PATEL JAINIL K. ENROLL.NO PATEL ASHISH. ENROLL.NO PATEL APURV ENROLL.NO

Kristin Ackerson, Virginia Tech EE Spring 2002 – VTech –Calvin Project For Prof. Ribeiro _ + Operational Amplifiers Tutorial Series

Table of Contents The Operational Amplifier______________________________slides 3-4 The Four Amplifier Types______________________________slide 5 VCVS(Voltage Amplifier) Summary: Noninverting Configuration____________slides 6-9 Inverting Configuration________________slides ICIC(Current Amplifier) Summary________________________slide 13 VCIS (Transconductance Amplifier) Summary_____________slides ICVS (Transresistance Amplifier) Summary_______________slides Power Bandwidth_____________________________________slide 19 Slew Rate____________________________________________slide 20 Slew Rate Output Distortion____________________________ slide 21 Noise Gain___________________________________________slide 22 Gain-Bandwidth Product_______________________________slide 23 Cascaded Amplifiers - Bandwidth________________________slide 24 Common Mode Rejection Ratio__________________________slides Power Supply Rejection Ratio___________________________slide 27 Sources_____________________________________________slide 28

The Operational Amplifier Usually Called Op Amps An amplifier is a device that accepts a varying input signal and produces a similar output signal with a larger amplitude. Usually connected so part of the output is fed back to the input. (Feedback Loop) Most Op Amps behave like voltage amplifiers. They take an input voltage and output a scaled version. They are the basic components used to build analog circuits. The name “operational amplifier” comes from the fact that they were originally used to perform mathematical operations such as integration and differentiation. Integrated circuit fabrication techniques have made high-performance operational amplifiers very inexpensive in comparison to older discrete devices.

i(+), i(-) : Currents into the amplifier on the inverting and noninverting lines respectively vid : The input voltage from inverting to non-inverting inputs +VS, -VS : DC source voltages, usually +15V and –15V Ri : The input resistance, ideally infinity A : The gain of the amplifier. Ideally very high, in the 1x1010 range. RO: The output resistance, ideally zero vO: The output voltage; vO = AOLvid where AOL is the open-loop voltage gain The Operational Amplifier +VS -VS vid Inverting Noninverting Output + _ i(-) i(+) vO = Advid RORO A Ri

The Four Amplifier Types Description Gain Symbol Transfer Function Voltage Amplifier or Voltage Controlled Voltage Source (VCVS) Avvo/vin Current Amplifier or Current Controlled Current Source (ICIS) Aiio/iin Transconductance Amplifier or Voltage Controlled Current Source (VCIS) gm (siemens) io/vin Transresistance Amplifier or Current Controlled Voltage Source (ICVS) rm (ohms) vo/iin

VCVS (Voltage Amplifier) Summary Noninverting Configuration + _ vin vO vid i(+) i(-) iO iF RFRL R1 i1 vid = vo/AOL Assuming AOL   vid =0 Also, with the assumption that Rin =  i(+) = i(-) = 0 _ vF + _ v1 + _ vL + _ iL Applying KVL the following equations can be found: v1 = vin vO = v1 + vF = vin+ iFRF This means that, iF = i1 Therefore: iF = vin/R1 Using the equation to the left the output voltage becomes: vo = vin + vinRF = vin RF + 1 R1 R1

VCVS (Voltage Amplifier) Summary Noninverting Configuration Continued The closed-loop voltage gain is symbolized by Av and is found to be: Av = vo = RF + 1 vin R1 The original closed loop gain equation is: Av = AF = AOL 1 + AOL  Ideally AOL  , Therefore Av = 1  Note: The actual value of AOL is given for the specific device and usually ranges from 50k  500k.  is the feedback factor and by assuming open-loop gain is infinite:  = R1 R1 + RF AF is the amplifier gain with feedback

VCVS (Voltage Amplifier) Summary Noninverting Configuration Continued Input and Output Resistance Ideally, the input resistance for this configuration is infinity, but the a closer prediction of the actual input resistance can be found with the following formula: RinF = Rin (1 +  AOL) Where Rin is given for the specified device. Usually Rin is in the M  range. Ideally, the output resistance is zero, but the formula below gives a more accurate value: RoF = Ro Where Ro is given for the  AOL + 1 specified device. Usually Ro is in the 10s of  s range.

VCVS (Voltage Amplifier) Noninverting Configuration Example + _ vin vO vid i(+) i(-) iO iF RFRL R1 i1 _ vF + _ v1 + _ vL + _ iL Given: vin = 0.6V, RF = 200 k  R1 = 2 k , AOL = 400k Rin = 8 M , Ro = 60  Find: vo, iF, Av, , RinF and RoF Solution: vo = vin + vinRF = *2x105 = 60.6 V iF = vin = 0.6 = 0.3 mA R R Av = RF + 1 = 2x = 101  = 1 = 1 = 9.9x10-3 R AOL 101 RinF = Rin (1 +  AOL) = 8x106 ( x10-3*4x105) = x1010  RoF = Ro = 60 =   AOL x10-3*4x

VCVS (Voltage Amplifier) Summary Inverting Configuration + _ RL vO + - vin + _ R1 i1 RF iF The same assumptions used to find the equations for the noninverting configuration are also used for the inverting configuration. General Equations: i1 = vin/R1 iF = i1 vo = -iFRF = -vinRF/R1 Av = RF/R1  = R1/RF

Input and Output Resistance Ideally, the input resistance for this configuration is equivalent to R1. However, the actual value of the input resistance is given by the following formula: Rin = R1 + RF 1 + AOL Ideally, the output resistance is zero, but the formula below gives a more accurate value: RoF = Ro 1 +  AOL Note:  = R1 This is different from the equation used R1 + RF on the previous slide, which can be confusing. VCVS (Voltage Amplifier) Summary Inverting Configuration Continued

VCVS (Voltage Amplifier) Inverting Configuration Example + _ RL + - vin + _ R1 i1 RF iF Given: vin = 0.6 V, RF = 20 k  R1 = 2 k , AOL = 400k Rin = 8 M , Ro = 60  Find: vo, iF, Av, , RinF and RoF vO Solution: vo = -iFRF = -vinRF/R1 = -(0.6*20,000)/2000 = 12 V iF = i1 = vin/R1 = 1 / 2000 = 0.5 mA Av = RF/R1 = 20,000 / 2000 = 10  = R1/RF = 2000 / 20,000 = 0.1 Rin = R1 + RF = ,000 = 2,  1 + AOL ,000 RoF = Ro = 60 = 1.67 m  1 +  AOL *400,000 Note:  is 0.09 because using different formula than above

ICIS (Current Amplifier) Summary  Not commonly done using operational amplifiers + _ Load iin iL Similar to the voltage follower shown below: Both these amplifiers have unity gain: Av = Ai = 1 + _ iin = iL vin = vo vin + _+ - vO Voltage Follower 1 Possible ICIS Operational Amplifier Application

VCIS (Transconductance Amplifier) Summary Voltage to Current Converter + _ Load iL R1 i1 vin + _ OR + _ Load iL R1 i1 vin + _ + _ General Equations: iL = i1 = v1/R1 v1 = vin The transconductance, gm = io/vin = 1/R1 Therefore, iL = i1 = vin/R1 = gmvin The maximum load resistance is determined by: RL(max) = vo(max)/iL

VCIS (Transconductance Amplifier) Voltage to Current Converter Example + _ Load iL R1 i1 vin + _ Given: vin = 2 V, R1 = 2 k  vo(max) =  10 V Find: iL, gm and RL(max) Solution: iL = i1 = vin/R1 = 2 / 2000 = 1 mA gm = io/vin = 1/R1 = 1 / 2000 = 0.5 mS RL(max) = vo(max)/iL = 10 V / 1 mA = 10 k  Note: If RL < RL(max) the op amp will saturate The output current, iL is independent of the load resistance.

VCIS (Transresistance Amplifier) Summary Current to Voltage Converter General Equations: iF = iin vo = -iFRF rm = vo/iin = RF + _ iF iin RF vO + -

VCIS (Transresistance Amplifier) Summary Current to Voltage Converter Transresistance Amplifiers are used for low-power applications to produce an output voltage proportional to the input current. Photodiodes and Phototransistors, which are used in the production of solar power are commonly modeled as current sources. Current to Voltage Converters can be used to convert these current sources to more commonly used voltage sources.

VCIS (Transresistance Amplifier) Current to Voltage Converter Example + _ iF iin RF vO + - Given: iin = 10 mA RF = 200  Find: iF, vo and rm Solution: iF = iin = 10 mA vo = -iFRF = 10 mA * 200  = 2 V rm = vo/iin = RF = 200

Power Bandwidth The maximum frequency at which a sinusoidal output signal can be produced without causing distortion in the signal. The power bandwidth, BWp is determined using the desired output signal amplitude and the the slew rate (see next slide) specifications of the op amp. BWp = SR 2  Vo(max) SR = 2  fVo(max) where SR is the slew rate Example: Given: Vo(max) = 12 V and SR = 500 kV/s Find: BWp Solution: BWp = 500 kV/s = 6.63 kHz 2  * 12 V

Slew Rate A limitation of the maximum possible rate of change of the output of an operational amplifier. As seen on the previous slide, This is derived from: SR = 2  fVo(max) SR =  vo/  tmax Slew Rate is independent of the closed-loop gain of the op amp. Example: Given: SR = 500 kV/s and  vo = 12 V (Vo(max) = 12V) Find: The  t and f. Solution:  t =  vo / SR = (10 V) / (5x105 V/s) = 2x10-5 s f = SR / 2  Vo(max) = (5x105 V/s) / (2  * 12) = 6,630 Hz  f is the frequency in Hz

Slew Rate Distortion v t desired output waveform actual output because of slew rate limitation  t vv The picture above shows exactly what happens when the slew rate limitations are not met and the output of the operational amplifier is distorted. SR =  v/  t = m (slope)

Noise Gain The noise gain of an amplifier is independent of the amplifiers configuration (inverting or noninverting) The noise gain is given by the formula: AN = R1 + RF R1 Example 1: Given a noninverting amplifier with the resistance values, R1 = 2 k  and RF = 200 k  Find: The noise gain. AN = 2 k  k  = 101  Note: For the 2 k  noninverting amplifier AN = AV Example 2: Given an inverting amplifier with the resistance values, R1 = 2 k  and RF = 20 k  Find: The noise gain. AN = 2 k  + 20 k  = 12  Note: For the 2 k  inverting amplifier AN < AV

Gain-Bandwidth Product In most operational amplifiers, the open-loop gain begins dropping off at very low frequencies. Therefore, to make the op amp useful at higher frequencies, gain is traded for bandwidth. The Gain-Bandwidth Product (GBW) is given by: GBW = ANBW Example: For a 741 op amp, a noise gain of 10 k corresponds to a bandwidth of ~200 Hz Find: The GBW GBW = 10 k * 200 Hz = 2 MHz

Cascaded Amplifiers - Bandwidth Quite often, one amplifier does not increase the signal enough and amplifiers are cascaded so the output of one amplifier is the input to the next. The amplifiers are matched so: BWS = BW1 = BW2 = GBW where, BWS is the bandwidth of all AN the cascaded amplifiers and AN is the noise gain The Total Bandwidth of the Cascaded Amplifiers is: BWT = BWs(21/n – 1)1/2 where n is the number of amplifiers that are being cascaded Example: Cascading 3 Amplifiers with GBW = 1 MHz and AN = 15, Find: The Total Bandwidth, BWT BWS = 1 MHz / 15 = 66.7 kHz BWT = 66.7 kHz (21/3 – 1)1/2 = 34 kHz

Common-Mode Rejection Ratio The common-mode rejection ratio (CMRR) relates to the ability of the op amp to reject common-mode input voltage. This is very important because common-mode signals are frequently encountered in op amp applications. CMRR = 20 log|AN / Acm| Acm = AN log-1 (CMRR / 20) We solve for Acm because Op Amp data sheets list the CMRR value. The common-mode input voltage is an average of the voltages that are present at the non-inverting and inverting terminals of the amplifier. vicm = v(+) + v(-) 2

Common-Mode Rejection Ratio Example Given: A 741 op amp with CMRR = 90 dB and a noise gain, AN = 1 k Find: The common mode gain, Acm Acm = AN = 1000 log-1 (CMRR / 20) log-1 (90 / 20) = It is very desirable for the common-mode gain to be small.

Power Supply Rejection Ratio One of the reasons op amps are so useful, is that they can be operated from a wide variety of power supply voltages. The 741 op amp can be operated from bipolar supplies ranging from 5V to 18V with out too many changes to the parameters of the op amp. The power supply rejection ratio (SVRR) refers to the slight change in output voltage that occurs when the power supply of the op amp changes during operation. SVRR = 20 log (  Vs /  Vo) The SVRR value is given for a specified op amp. For the 741 op amp, SVRR = 96 dB over the range 5V to 18V.

Open-Loop Op Amp Characteristics Table DeviceLM741CLF351OP-07LH0003AD549K Technology BJTBiFETBJT Hybrid BJT BiFET AOL(typ)200 k100 k400 k40 k100 k Rin2 M  1012  8 M  100 k  1013  || 1 pF Ro50  30  60  50  ~100  SR0.5 V/  s13 V/  s0.3 V/  s70 V/  s3 V/  s CMRR90 dB100 dB110 dB90 dB