NDSU PushPull Amplifiers ECE 321 JSG Background: PushPull Amplifiers The amplifiers we have been looking at are termed CassA amplifiers. They bias the transistor in the active region, and vary the operating point on the loadline according to the AC input. Ice Load Line AC Input Q Point Class A Amplifier AC Output Vce A problem with ClassA amplifiers is they are very inefficient: with no input signal (and hence no output power), they still consume power to bias the transistor. A Class B amplifier (termed a pushpull amplifier) biases the transistor so that it is off when there is no AC input. This results in a much mor efficient amplfier but it also results in a sine wave that is clipped. Ice Saturated Load Line Active AC Output AC Input Q Point Class B Amplifier Vce Off To prevent clipping, a second transistor is used to suply the missing half of the sinewave output. page 1 October 6, 2014
NDSU PushPull Amplifiers ECE 321 JSG Push Amplifier Problem: Drive an 8Ohm speaker where the voltage at the speaker (Vout) follows the voltage at the input () Input: 0..10V analog signal, capable of 20mA (i.e. a function generator) Output: speaker Relationship: Y = X Solution: (Take 1: Push Amplifier with Crossover Distirtion) First, use a transistor capable of driving s at 10V (1.25A). Assume a TIP112 NPN transistor: Vbe = 1.4V (Darlington pair) Vce(sat) = 0.9V β = 1000 max(ic) = 4A Next, use the following circuit: 12V c 10V b 1.07mA 1.4V e 8.6V 1.07A 4W By placing the load at the emitter, the voltage following, minus the 1.4V drop across the baseemitter diode. The transistor provides current gain so that only needs to supply 1mA to drive 1A at the load. To remove the 1.4V drop across the diode, add an opamp: page 2 October 6, 2014
NDSU PushPull Amplifiers ECE 321 JSG 12V 10V c 11.4V b 10V 1.25mA 1.4V e 1.25A 10V 4W With negative feedback for an opamp, V equals V. This results in the voltage across the load ( speaker) following the input in spite of the 1.4V drop across the baseemitter diode. Solution: First, use The basic pushpull amplifier is as follows: 12V NPN Vout PNP RL 12V page 3 October 6, 2014
NDSU PushPull Amplifiers ECE 321 JSG First, let's ignore the PNP transistor (assume it's off). The smallsignal model for the black part of the circuit is then: Ib Vb Vc 0.7 200Ib Ve 12V Ib 200Ib RL Assuming the transistor is on, and V in 0.7 R L(1 β)i b = 0 1 I b = (1β)R L (V b 0.7) V e = V out = V in 0.7 Circuit model with β = 200 For the transistor to be on, Ib>0, meaning Vb > 0.7V. Note that when the traansistor is on, the input impedance is (1 β)r L. This means that the speaker you're driving looks like a 1.6k Ohm load as far as the input is concerned. (The previous CE amplifiers can drive a 1.6k Ohm load, but not an load.) If the transistor is not on, it's off and Ib=0. This results in clipping when Vb < 0.7V: Volts 0.7V Vout Time page 4 October 6, 2014
NDSU PushPull Amplifiers ECE 321 JSG The PNP transistor operates when < 0. It's circuit is almost the same as the NPN side only flipped: RL Ib 200Ib Ve 12V 0.7 200Ib 12V Ib Vb Vc For the PNP transistor, the arrow signifies the diode from the emitter to the base. When < 0.7V, current flows from ground to through this diode, turning on the transistor. The resulting current is amplified by β. The equations for VL are then: V in R L(1 β)i b 0.7 = 0 1 I b = (1β)R L (V in 0.7) The impedance as seen by is again (V in 0.7) > 0 V in < 0.7 When on, the output is then V L = V in 0.7 or the output follows the input, with 0.7V removed.. Further, this equation is valid only for Ib > 0, or (1 β)r L Putting the NPN (push) and the PNP (pull) transistor together gives the following output for a sinewave input. Note that the output is 0.7V below the input. There is also a small part of the sine wave where the output is stuck at 0V while V in < 0.7V. This is called crossover distortion. page 5 October 6, 2014
NDSU PushPull Amplifiers ECE 321 JSG Time Vout NPN on Time PNP on Darlington Pairs (TIP112 TIP117) A problem with the above amplifier is you need a transistor with A high gain (to reduce the current draw from the source), and A high current capability Both are difficult to get in a single transistor. To fix this problem a Darlington Pair can be used the schematics shown below Take for instance the NPN on the left If the base current is 1 ua, that current is amplified by the first transistor This amplified current then feeds the base of second transistor This is again amplified by the second transistor The net results is A high gain (500 1000 for a TIP 112 / T117), but page 6 October 6, 2014
NDSU PushPull Amplifiers ECE 321 JSG Vbe = 1.4V (since you have two diodes to go through), and The saturation voltage (Vce:sat) is no longer 0.2V. It is instead about 2.0V. Darlington pairs make very poor switches Darling pairs make for good pushpull amplifiers, however. You just need an extra 2V on the power supply to drive the load. (i.e. a 10V source can only supply 8V to the load.) page 7 October 6, 2014
NDSU PushPull Amplifiers ECE 321 JSG Improved PushPull Amplifier If you're willing to splurge, you can eliminate the crossover distortion with an opamp. 10V @ 1A NPN B E C VL = 2 PNP 4 Watt 10V @ 1A TIP112 / 117 Darlington Pairs note: Use external / 10V Supply (CADET max current = 300mA) With negative feedback, the voltage at V = V. This results in V L = 2V in The pushpull amplifiers works by Turning on the NPN when > 0 (push), and Trning on the PNP transistor when < 0 (pull) For example, the voltages and currents for = 2V are: page 8 October 6, 2014
NDSU PushPull Amplifiers ECE 321 JSG 10V @ 1A 499mA 2V 5.4V 1mA NPN Active VL 4V PNP Off B C 500mA E 4 Watt 10V @ 1A TIP112 / 117 Darlington Pairs note: Use external / 10V Supply (CADET max current = 300mA) The voltages and currents when = 2V are: Push: VL = 4V 10V @ 1A 2V NPN OFF 4V B C E 1mA 5.4V 499mA PNP 500mA 4 Watt 10V @ 1A TIP112 / 117 Darlington Pairs note: Use external / 10V Supply (CADET max current = 300mA) Pull: VL = 4V page 9 October 6, 2014