ClassABampDesign Richard Cooper October 17 2016 When the input signal is positive, the NPN transistor Q1 turns ON, the PNP transistor Q2 is OFF, and the output voltage is positive. The NPN transistor (emitter follower) is sourcing (pushing) the current into the load resistor during the positive cycle of the input voltage. When the input signal is negative, the PNP transistor Q2 turns ON, the NPN transistor Q1 is OFF, and the output voltage is negative. The PNP transistor (emitter follower) is sinking (pulling) the current from the load resistor during the negative cycle of the input voltage. This were it gets its name Push Pull stage. ClassB and ClassAB Find ro, and β from the 2N3906 characteristic curves (used for both). Use for both NPN and PNP calculate at peak values Ic = Iload, and Vce = Vcc Vout. Use β DCmin = 100 for calculating bias (worse case) Use β AC from curves to calculate gain A V, frequency response (capacitors), Rin, and Rout. Solve for the Peak values of Vout, and Iload. You need the peak values to design the bias circuit. First step for lab is calculate Vout peak and Iload peak from the given Pload (power in load) and Rload. Pload = (Vout rms )^2 / Rload solve for Vout RMS Now convert Vout rms to Vout peak next solve for Iload peak Iload peak = Vout peak / Rload Iload max peak = (Vout peak +20% Vout peak) / Rload Do not design for an edge. Class B push pull stage. Vdd = - Vee Vin = Vb (Ri + Rin2) / Rin2 The output voltage can be expressed as Vb = Vin (Rin2 / (Rin2 + Ri)) For Vin > Vbe (Ri + Rin2) / Rin2 Vout = Vb - Vbe where Vbe = 0.7V DC bias For Vin < Vbe (Ri + Rin2) / Rin2 Vout = 0V At Max Vout Av = (Vin Vbe ) / Vin Note: changes with Vin values. We design the positive half (NPN) and copy to the negative half (PNP).. Use the curves for 2N3906 (PNP) for 2N3904 (NPN) they both have the same characteristics but just opposite polarity. Use β min = 100 Page 1 of 10
Using the value of the output voltage peak during the positive cycle for the NPN transistor. We will design Rb1 = Rb2. The base bias resistorsrb1, and Rb2 are not required for the ideal case, but will we add them to hold both base voltages to zero with Vin = 0. This will get you practice for the class AB. +Vcc Rb1 Q1 NPN Rin Vin Ri Cin Vout Rgen 50Ω Rin2 Chi Q2 PNP Rout Rload Rb2 -Vee Class B push pull stage Page 2 of 10
+Vcc Rb1 Irb1 Ic Vb Q1 NPN Rin Vin Ri Vin2 Cin Ib Ve Iload Vout Rgen 50Ω Rload Rin2 Rb2 Rout -Vee Class B positive half cycle NPN at peak output voltage. To start we must find the Vout in Vpeak and Iload as a peak curren peak Rgen Vgen Vin Vb Ve Iin ib Ri B E Rπ Rb βib Rin ic C ie ro Rout Vout Iload Rload Small signal Class B positive half cycle 2N3904 (NPN) at peak output voltage Page 3 of 10
To find the maximum value for Rb1, Rb2 we will need to calculate the maximum peak current Ib. Use the minimum β (β min = 100) and maximum i c peak to give worst case maximum base current. Ie max = Iload max peak Ic max = (β min / (β min +1)) Ie max ib max = Ic max / β min Minimum β will give maximum Ib The minimum value of Rπ is when Ic is maximum and β is minimum. vt = 26mV Rπ min = (βmin * vt) / Ic max Smallest Rπ Maximum voltage on the base. ClassB Vb max = Vout +20%Vout + Vbe Maximum value of Rb1. Will set Rb2 = Rb1. ClassB Rb1 max = (Vcc Vb max ) / ib max = Rb2 max Minimum value of Rb1, and Rb2 Now for the minimum value of Rb1 we must consider the input impedance requirement. When Vin = 0 both of the BJTs are off i.e. high impedance looking into base, so Rin = Rb1 Rb2 not worst case. So we will calculate at Vout peak the lowest Rin case. Rin2 = Rb1 Rb2 when Vin = 0 will not yield the minimum value of Rb1, Rb2 Where Rin = Ri +Rin2 We are looking for the case where Vin = maximum Vout = maximum and Rin = minimum Consider only positive half cycle NPN on and PNP off. From Rin min required solve for Rin2 min required to meet input requirement. Rin2 min = Rin min Ri Required minimum Rin2. Rbase min = Rπ min + (Rload ro )(β min + 1) Resistance looking into the base of the transistor. Remember Rb = Rb1 Rb2 and Rb1 = Rb2 therefore Rb1 = Rb2 = Rb*2. Now solve for Rb requited to meet Rin. Rin2 = Rb Rbase Therefor Rb = 1/(1/Rin2-1/Rbase). The Rb required. Now solve for Rb1 min and Rb2 min Rb1 min = Rb2 min = 2 * Rb min Page 4 of 10
Check Rin min Rin2 min = (Rb1 min Rb2 min) (Rπ min + (Rload ro )(β min + 1)). Largest Rin2min Rin min = Rin2 min + Ri Now chose Rb1 for a value between the Rb1 max and Rb1 min and set Rb2 = Rb1. Use β from curves to calculate Rin, Rout AC values Check Rin > Rin required. Calculate Rin. Rin2 = (Rb1 Rb2) (Rπ + (Rload ro )(β + 1)) Rin = Ri + Rin2 Calculate Rout. Consider only positive half cycle NPN on and PNP off. Rout = ro ((Rπ + Rb1 Rb2 (Ri + Rgen)) / (β +1)) Looking into the BJT emitter Calculate low frequency cutoff. ClassB FL = 1 / (2π Cin (Rin2 + Ri +Rgen)) at Vout max. Calculate High frequency cutoff. ClassB FH = 1 / (2π Chi (Rin2 (Ri +Rgen))) at Vout max. Class AB push pull stage. The class AB requires Rb1, and Rb2 so the push pull stage will be bias properly. Vdd = - Vee Vin = v b. The input voltage is expressed as Vin = Vout / Av Av at Vout max = Vout / Vin We design the positive half (NPN) and copy to the negative half. Page 5 of 10
+Vcc Rb1 Vin Q1 NPN Ri Cin D1 Re1 Vout Rin 50Ω Rgen Rin2 Chi D2 D3 Re2 Rload Rout Q2 PNP Rb2 -Vee Class AB push pull stage Page 6 of 10
+Vcc Rb1 Ib1 Q1 NPN Vin2 D1 Vd Re1 Cin D2 Vd Loop Ie1 Vload Rin2 D3 Vd Ie2 Re2 Q2 PNP Rb2 Ib2 -Vee Write the loop equation around the loop to solve for Re1 and, Re2. 0 = -Vd Vd Vd + Vbe1 + Ie Re1 + Ie Re2 + Vbe2 Given: I C = 4ma when Vin = 0V this is the quiescent current the amplifier in class A range Solve for Re1 + Re2 set Re1 = Re2. At I C = 4ma, I E = I C (β + 1) / β Rgen Vgen Vin Vb Ve Iin ib Ri B E Rπ Rb βib Rin NPN ic C ie Re1 ro Rout Vout Iload Rload Small signal Class AB positive half cycle 2N3904 (NPN) at peak output voltage Page 7 of 10
To find the maximum value for Rb1, Rb2 we will need to calculate the maximum peak base current Ib. Use the minimum β (β min = 100) and maximum i c peak Re1 = Re2 from loop equation. Ie max = Iload max peak Ic max = (β / (β +1)) Ie max ib max = Ic max / β min Minimum β will give maximum Ib Solve for Rb1, and Rb2 the same as the class B but must include Re in the solution. Maximum voltage on the base, must include voltage drop on Re. Vb max = Vout +20%Vout + Vbe + Ie max Re1 Must include voltage across Re1 at maximum load current. Maximum value of Rb1, and Rb2. ClassAB Rb1 max = (Vcc Vb max ) / ib max Rb2 max, Rb1 max set equal to each other. With class AB Rb1, Rb2 cannot exceed Rb1 max, and Rb2 max because there must be enough current to keep the diodes forward biased so they act as voltage sources. The minimum value of Rπ is when Ic is maximum and β is minimum. vt = 26mV Rπ min = (βmin * vt) / Ic max Rπ 4ma = (βmin * vt) / Ic 4ma for Ic = 4ma Minimum value of Rb1, and Rb2 Now for the minimum value of Rb1 we must consider the input impedance requirement. Where Rin = Ri +Rin2 We are looking for the case where Vin = max, Vout = max, and Rin = min Consider only positive half cycle NPN on, and PNP off. Rb min = Rb1 min Rb2 min Remember Rb = Rb1 Rb2 and Rb1 = Rb2 therefore Rb1 = Rb2 = Rb*2. Rbase min = Rπ min + (ro (Rload + Re1)) * (β min + 1) Now solve for Rb requited to meet Rin. Rin2 = Rb Rbase Page 8 of 10
The Rb min required. Rb min = 1/(1/Rin2 min -1/Rbase min). Rb1 min = Rb2 min = Rb min * 2 Rin2 min = (Rb1 min Rb2 min) ( Rπ min + (ro (Rload + Re1))(β min + 1)) Largest Rin2 min. From Rin min required, solve for Rin2 min required, now solve for Rb1 min and Rb2 min Now chose a value for Rb1 between the Rb1 max and Rb1 min and set Rb2 = Rb1. Use β from curves to calculate Rin, Rout, and Av Check Rin > Rin required. And Rb1, Rb2 are < Rb1 max Calculate Rin. ClassAB Rin2 = (Rb1 Rb2) (Rπ + (β + 1) (ro Re + Rload )) Rin = Ri + Rin2 Calculate Rout. ClassAB Consider only positive half cycle NPN on, and PNP off. BJTemitter = ro ((Rπ + Rb1 Rb2 (Ri + Rgen)) / (β +1)) Looking into the BJT emitter Rout = Re1 + BJTemitter Calculate low frequency cutoff. ClassAB FL = 1 / (2π Cin (Rin2 + Ri +Rgen)) at Vout max. Calculate High frequency cutoff. ClassAB FH = 1 / (2π Chi (Rin2 (Ri +Rgen))) at Vout max. Voltage gain Av ClassAB Vin = v b (Ri + Rin2) / Rin2 Vout = v e (Rload / (Re + Rload)) input voltage divider from input to base Output voltage divider from emitter to Vout RloadE = ro (Re + Rload) load seen by the emitter ve = ib (β +1) (ro (Re +Rload)) = ie RloadE signal at emitter Vout = v e (Rload / (Re + Rload)) = i b (β + 1) (RloadE ) * (Rload / (Re + Rload)) vb = i b Rπ + i b (β + 1) (ro (Re + Rload)) = i b Rπ + i b (β + 1) (RloadE) Avb = Vout / v b = ib (β + 1) (RloadE ) * (Rload / (Re + Rload)) / (Rπ + (β + 1) (RloadE) Page 9 of 10
Av = Vout/ Vin = ((Rin2 / (Ri + Rin2) Avb Av = ((Rin2 / (Ri + Rin2)) * (Rload / (Re + Rload)) * ((β + 1) RloadE / (Rπ + (β + 1) (RloadE ) Input Divider Output Divider Gain from base to emitter Current Gain Ai ClassAB Ai = Iload / Iin = (Vout / Rload )/ (Vin / Rin) = Av (Rin / Rload) Page 10 of 10