NDSU Operational Amplifiers ECE 06 JSG Operational Amplifiers An operational amplifier is a input device with V o k(v V ) where k is a large number. For short, the following symbol is used for an differential amplifier: V V Operational Amplifier Characteristics Symbol for an operational amplifier (opamp) LM741 LM833 1.. V V 1.. V V V.. V V 1.. V V V Pin Layout for two common opamps: LM741 and LM833 LM741 LM833 Ideal Input Resistance M Ohms infinite Input Offset Current 0nA 5nA 0 Output Resistance 75 Ohms Output Short Circuit Current: 5mA 0 Input Offset ltage 1.0mV 0.3mV 0 Operating ltage / 1V.. / V /.5V.. / 15V any Diffential Mode Gain 00,000 100dB infinite Common Mode Rejectin Ratio 90dB 100dB common mode gain = 0 Slew Rate 0.5V/us 7V/us infinite Gain Bandwidth Product 1.5MHz 15MHz infinite Price (qty 100) $0.35 $0.5 1 July, 018
NDSU Operational Amplifiers ECE 06 JSG Input resistance / Input Offset Current: The input of the opamp does draw some current. If you keep the currents involved much larger (meaning at 1V, resistors are less than 50M Ohm), you can ignore the current into V and V. Input Offset ltage: If you have a lot of gain, there may be a slight DC offset in the ouput. You can model this as a 1mV (or 0.3mV) offset at the input, V or V. Operating ltage: A LM741 needs at least /1V to power it. Differential Mode Gain: The gain from (V V) to the output Common Mode Rejection Ratio: The gain from (V V) is this much less than the differential mode gain. Note that db = 0 log 10 (gain) Slew Rate: The ouput can't change from 10V to 10V in zero time. It can only ramp up this fast. Gain Bandwidth Product = 1.5MHz: If you want a gain of one, the bandwidth is 1.5MHz If you want a gain of 10, the bandwidth is 150kHz. etc. For a 741, for example, the output of an opamp is V o = k 1 (V V ) k (V V ) where k1 = 00,000 (the differential gain) and k = 6.35 (90dB smaller than the differential gain) Operational Amplifier Circuit Analysis Problem: Write the voltage node equations for the following circuit. Assume (a) a LM741 op amp. (b) an ideal opamp. 10k 1V 1k LM741 Figure : Find for this opamp circuit July, 018
NDSU Operational Amplifiers ECE 06 JSG (a) 741 Op Amp Analysis: First, replace the opamp with a model taking into account the input, output resistance and gains: 10k 1k Vm M 1V Vp M Vx 75 00,000(Vp Vm) 6.37(Vp Vm) Solution 1: Replace the opamp with its circuit model (LM741 used here) Since Vp = V = 0V, this simplifies a little: the gain of the opamp works out to 199,994 Vm. Now, write the voltage node equations: @Vm: V m 1V 1k Vm M Vm Vx 10k75 = 0 @Vx: V x = 199, 994V m Solving 1 1k 1 1199,994 M 10,075 V m = 50.4µV V m = 1V 1k V x = 199, 994V m = 10.07V V o = 75 10,00075 V o = 9.9994V V m 10,000 10,00075 V x 3 July, 018
NDSU Operational Amplifiers ECE 06 JSG (b) Ideal Op Amp: Note that many of the terms don't affect the output all that much: M Ohms in parallel with 1k is about 1k 1 199,994 is about 199,994. 50.4uV is about zero. If you approximate these terms, you're essentially using an idealop amp. The circuit simplifies to: 10k 1k Vm I=0 1V Vp = k(v V) Solution : Replace the opamp with an ideal opamp Now the voltage node equations are: @Vm: V m 1V 1k Vm 10k @: V o = k(v p V m) Note that: = 0 You can't write a voltage node equation at : the opamp supplies whatever current it takes to hold the output voltage. Since you don't know what that current is, you can't sum the currents to zero. For an ideal opamp, if the gain, k, is infinity and the output is finite, then V p = V m. Solving then results in V o = 10.00V which is very close to what you get for a 741 opamp. Notes: When analyzing an opamp circuit, you almost have to use voltage nodes. If assuming an ideal opamp, the voltage node equation at is V p = V m 4 July, 018
NDSU Operational Amplifiers ECE 06 JSG Example : Assume ideal opamps Write the votlage node equations for the following opamp circuit Find the voltages V 1V 0k V3 V4 V 3V V1 Example : Find the voltages There are four unkown voltage nodes. We need to write 4 equations to solve for 4 unknows. Start with the easy ones. For ideal opamps with negative feedback meaning V p = V m V 1 = V V 3 = V Now write two more equations. It's tempting, but you can't write the node equations at V or V4 Equation (1) and () are the node equations at the outputs you've already done that. You don't know the current from the opamp meaning you can't sum the currents to zero. Instead, find two mode nodes where you can sum the currents to zero: nodes V1 and V3. Solving V 1 3 V 3 V In Matlab: V 1 V = 0 V 3 1 0k V 3 V 4 = 0 1 0 0 0 0 0 1 0 1 0 0 0 1 7 1 V 1 V V 3 V 4 = 3 5 (1) () (3) * to clear the denominator (4) * to clear the denominator 5 July, 018
NDSU Operational Amplifiers ECE 06 JSG >> A = [1,0,0,0 ; 0,0,1,0 ;,1,0,0 ; 0,1,7,1] 1 0 0 0 0 0 1 0 1 0 0 0 1 7 1 >> B = [;;3;5] 3 5 >> V = inv(a)*b V1 V 1 V3 V4 8 This checks with the PartSim solution PartSim results for example : The votlages match our computations. 6 July, 018
NDSU Operational Amplifiers ECE 06 JSG Example 3: Assume ideal opamps. Find the node voltages. 1k V1 k 3k 4k V V3 V4 V 3V There are four unknown votlages, so we need to write 4 equations to solve for 4 unknowns. Start with the easy ones: at the output of each opamp, V = V V 1 = V 3 = 3 Sum the currents to zero at nodes 1 and 3 for the remaining two eqations V 1 1k V 3 V 3k In matrix form: Solving: V 1 V k = 0 V 3 V 4 4k = 0 1 0 0 0 0 0 1 0 1 1 1k k 1 k 0 0 0 1 3k 1 1 3k 4k 1 4k V 1 V V 3 V 4 (1) () (3) (4 = >> A = [1,0,0,0 ; 0,0,1,0 ; 1/10001/000, 1/000, 0, 0 ; 0,1/3000, 1/30001/4000,1/4000] 1.0000 0 0 0 0 0 1.0000 0 0.0015 0.0005 0 0 0 0.0003 0.0006 0.0003 >> B = [;3;0;0]; >> V = inv(a)*b V1.0000 V 6.0000 V3 3.0000 V4 1.0000 3 0 0 7 July, 018
NDSU Operational Amplifiers ECE 06 JSG Again, this matches with what PartSim gives PartSim Solution: The node votlages match our calculations. 8 July, 018