Emitter Coupled Differential Amplifier

Transcription

1 Emitter Coupled Differential Amplifier Returning to the transistor, a very common and useful circuit is the differential amplifier. It's basic circuit is: Vcc Q1 Q2 Re Vee To see how this circuit works, let's first assume power is / 12V and =. First, assume > (say, =, = 1V). In this case, the diode in Q1 is on and Q2 is off. This results in getting pulled up to Vcc (12V). 12V Vcc 12V Q1 0.7V Q2 off 1V Re 100 Vee 12V page 1 February 21, 2018

2 Next, assume <. ( = 0, = 1V) In this case transistor Q2 saturated and = 9.6V. 12V Vcc 9.6V 12V Q1 off 0.3V 0.7V Q2 off on 1V 1V Re 100 Vee 12V The net result is that this is a differential amplifier which swings from 9.6V when < to 12V when > When <, the output saturates at 9.6V (approx) When >, the output saturates at 12V (approx) When, the output is proportional to the difference, times a large gain 12 8 Saturated where you operated for digital electronics 4 0 Active Where you operate for analog electronics 4 8 Saturated where you operated for digital electronics page 2 February 21, 2018

3 The actual schematic for a 741 opamp (shown below) has a differential amplifier at its heart (upper left corner). The rest of the circuit amplifies the output and sets up some constant current sources. Note that the heart of this circuit is Emittercoupled logic with Q1 and Q2 creating a differential input A pushpull amplifier at the output (Q14 and Q20), and A Darlington pair (Q15 and Q17) to increase the gain The net results is the inputoutput characteristic being V o k(v V ) where k is a large number. For short, the following symbol is used for an differential amplifier: V V page 3 February 21, 2018

4 Operational Amplifier Characteristics LM741 LM V V V V V.. 22V V V V V LM741 LM833 MCP602 Ideal General Purpose Audio Amp up Amp (ECE 376) Input Resistance 2M Ohms 20M Ohms > 1G Ohm infinite Input Offset Current 20nA 25nA 1pA zero Output Short Circuit 25mA 5mA 30mA infinite Current: Operating ltage / 1.. / 22V / 2.5V.. / 15V V to 6. infinite tlage Range (/15V) / 13V / 13.7V railtorail infinite Diffential Mode Gain 200, dB 115dB infinite Slew Rate 0.5V/us 7V/us 2.3V/us infinite Gain Bandwidth Product 1.5MHz 15MHz 2.8MHz infinite Price (qty 100) $0.49 $0.40 $0.45 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. Operating ltage: A LM741 needs at least /12V 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 = 20 log 10 (gain) Slew Rate: The ouput can't change from 1 to 1 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 2 (V V ) page 4 February 21, 2018

5 where k1 = 200,000 (the differential gain) and k2 = (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 LM741 1V (a) 741 Op Amp Analysis: First, replace the opamp with a model taking into account the input, output resistance and gains: 10k 2M 1V 2M Vx ,000( ) 6.37( ) Since = V =, this simplifies a little: the gain of the opamp works out to 199,994. Now, write the voltage node V m 1V 2M Vx 10k75 = 0 page 5 February 21, 2018

6 @Vx: V x = 199, 994V m Solving ,994 2M 10,075 V m = 50.4μV V m = 1V V x = 199, 994V m = 10.07V V o = 75 10,00075 V o = V (b) Ideal Op Amp: V m 10,000 10,00075 V x Note that many of the terms don't affect the output all that much: 2M Ohms in parallel with is about 1 199,994 is about 199, uV is about zero. If you approximate these terms, you're essentially using an idealop amp. The circuit simplifies to: 10k I=0 1V = k(v V) Now the voltage node equations V m 1V 10k = V o = k(v p V m) Note that: ou 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 page 6 February 21, 2018

7 V o = 10.0 which is very close to what you get for a 741 opamp. Some useful opamp circuits (will cover again in ECE 321) Analog Amplifiers: If you have negative feedback, the amplifier is probably an analog amplifier. In this case V p V m Buffer: A buffer provides a large input impedance and a low output impedance. It's useful when you don't want to load a circuit. Buffer: = NonInverting Amplifier: Provides a gain from (typically). If you need a gain large than 100, cascade several in a row. NonInverting Amplifier: = 1 R 1 R 2 page 7 February 21, 2018

8 Inverting Amplifier: Provides a gain from to 100. This does load the input,, however with resistor Inverting Amplifier: = R 1 R 2 Instrumentation Amplifier: Provides a gain and an offset Va Vb Instrumentation Amplfier: = R 1 R 2 (V a V b ) = page 8 February 21, 2018

9 Digital Amplifiers (will cover next lecture). If you remove the negative feedback or apply positive feedback, the output will slam to the power supply (V and V). If you make V equal to zero volts, the ouput becomes logic levels. Usually, you use an MCP602 for digital amplifiers. This opamp has railtorail outputs, meaning if you power the opamp with and 5V, the output can go from (logic 0) to 5V (logic 1). Comparitor 5V A B MCP602 A > B Comparitor: = (A > B) Schmitt Trigger A 5V MCP602 B Schmitt Trigger: = (A > B) with hysteresis is no longer equal to with positive feedback page 9 February 21, 2018