Combination Notch and Bandpass Filter Clever filter design for graphic equalizer can perform both notch and bandpass functions Gain or attenuation is controlled by a potentiometer for specific frequency bands 1 v in -r r V o L 2 C Lectur 7-1
Combination Notch and Bandpass Filter 1 v in -r r V o L 2 C Lectur 7-2
Combination Notch and Bandpass Filter 1 v in -r r V o L 2 C Lectur 7-3
Combination Notch and Bandpass Filter 4.7k v in 1k 9k V o 1H 4.7k 0.1uF Lectur 7-4
Combination Notch and Bandpass Filter 4.7k v in 9k 1k V o 1H 4.7k 0.1uF Lectur 7-5
Integrator via Negative Impedance Converter Presents a negative resistance at the input terminals Best analyzed by applying a test voltage and measuring the input current 2 1 V o i in v in If it is behaving like a linear circuit, we can calculate the Thevenin equivalent If it s passive, we can simply calculate its impedance (resistance in this case) Lectur 7-6
Thevenin/Norton Equivalents By definition, a linear circuit has a straight-line i-v characteristic i V oc v slope=1/ i 1 2 v linear circuit I sc v = i V oc Which can be represented by v i 1 or i = v v -- I sc i I sc 1 V oc 2 2 Lectur 7-7
Thevenin/Norton Equivalents If the line passes through the origin, then it is a passive linear circuit --- a single impedance Only one (v x,i x ) point is needed to determine the slope i 1 2 v linear circuit i (v x, i x ) slope=1/ v =v x /i x A negative resistance is recognized by a negative slope (with directions shown) i 1 2 v linear circuit (v x, i x ) i v slope=1/ Lectur 7-8
Thevenin/Norton Equivalents Note that the same Thevenin/Norton conversion steps --- applying test voltages and measuring test currents --- works for complex impedances too Z = I(s) I s ( s) 1 Z C = ----- sc Z L = sl V(s) We just can t draw them as two dimensional i-v characteristics Lectur 7-9
Calculate v in /i in 2 Negative Impedance Converter 1 Vo i in v in Lectur 7-10
Negative Impedance Converter 2 1 Vo i in v in Lectur 7-11
Voltage-to-Current Converter The negative impedance converter can be used to create a voltage-to-current converter where the output load current is independent of the load impedance 2 1 Vo V S Z L I L Lectur 7-12
Voltage-to-Current Converter 2 1 Vo V S Z L I L We could also write out all of the current equations and get the same result Lectur 7-13
2 Integrator Use the voltage-to-current converter to design an integrator 1 Vo V S Z L I L Lectur 7-14
Integrator Need a low output impedance for this circuit -- why? If the output impedance is not low enough, what is another design option? 2 1 Vo V S sc I L Lectur 7-15
More Nonidealities Along with the frequency dependence of the gain, and the finite output/input impedances of the devices, there are other nonidealities associated with opamps that can cause distortion Saturation: the output is really limited to a voltage that is 1 to 3 volts less than VCC Slew ate: limited gain of transconductance input amplifier can cause severe distortion in the output CM: the signal component that is common to both differential inputs is amplified somewhat, and the CM specifies the quality with which this phenomenon is rejected dc Offset Voltage: the input differential voltage required to set the output to zero when no other signals are applied Finite Input/Output Impedances: the input resistance/impedance of the inputs and the limited current sourcing capability of the output dc Input Bias Current: small currents required to bias the transistors at the input stage of the opamp Lectur 7-16
Slew ate Limitations We know that an opamp behaves like a low pass filter due to the frequency dependence of the gain A unity gain amplifier has a bandwidth of ω t 200dB 741 Open Loop Characteristics e-1 e0 e1 e2 e3 e4 e5 e6 e7 As ( ) = A o --------------- s 1 --- ω b 100 db 0dB v in -100dB DB(VMOUT/VMIN) frequency Lectur 7-17
Slew ate Limitations So we can write an expression for the closed-loop gain as: V i V o ----- V i = 1 -------------- s 1 ---- ω t Which is like a STC with a time constant of In the time domain we d expect a step response of the form: v o () t = V 1 e Which has a maximum possible change in output voltage of dv o () t -------------- dt = V -- τ 1 ---- ω t t - τ Lectur 7-18
Slew ate Limitations If the output wants to change faster than this, it will not be able to do so This is especially difficult for large signals; e.g. when V is large The maximum switching speed is limited by ω t, which is due to the compensation capacitor in this case, but all capacitors in the circuit in general The open loop roll-off with frequency is due to the limited current sourcing capability of the amplifier and these capacitors V ------ t I max ---------- C So the maximum current sourcing capability and the compensation capacitor, for example, may determine the slew rate S dv o = --------- (volts/µs) dt max A smaller change in voltage can go to higher frequencies before encountering the S limitation Lectur 7-19
Opamp Macromodels We can look at this limited current sourcing capability of the opamp in terms of the opamp macromodel C Vi v id 1 2 G m v id v i2 µv i2 v o V i When change in v id is sudden, G m can only supply a limited amount of current, I max for a real input transconductance amplifier Lectur 7-20
Slew ate Limited esponse At the slew rate limit the output can only ramp up with a slope of I max /C(1µ) C v id 1 2 G m v id v i2 µv i2 v o V i For a sudden change in the input voltage, v id V i v id 1 2 I max v i2 C(1µ) µv i2 v o Lectur 7-21
Slew ate for a 741 is 0.63V/µs 741 Example For a sinusoidal signal, the maximum change occurs near the zero crossing, so this is where we will notice the first signs of slewing dv o --------- dt max ωv What s the maximum allowable frequency for a peak sinusoidal input voltage of 5.0 volts? = VC8-15V - - 741 VIN SIN VC9 15V Lectur 7-22
741 Example Input and output voltage for a 5 volt peak, 10kHz frequency time 0.0 0.1 0.2 ms 5 3 1-1 -3-5 VMOUT VMIN Lectur 7-23
741 Example Input and output voltage for a 5 volt peak, 20kHz frequency 10 time 0.0 0.2 0.4 0.6 0.8 1.0 e-4s 0-10 VMOUT VMIN Lectur 7-24
741 Example Input and output voltage for a 5 volt peak, 40kHz frequency 5 time 0 10 20 30 40 50 us 3 1-1 -3-5 VMOUT VMIN Lectur 7-25
741 Example Note that the frequency response of the opamp does not affect the input signal at 20kHz It is a slew rate limitation that depends on the magnitude of the input voltage (has I max of the input transconductance amplifier been reached?) 0.1 0.0-0.1-0.2-0.3-0.4-0.5-0.6-0.7-0.8 frequency e3 e4 e5 e6 DB(VMOUT/VMIN) Lectur 7-26