Lecture 39: Intro to Differential Amplifiers. Context

Transcription

1 Lecture 39: Intro to Differential Amplifiers Prof J. S. Smith Context Next week is the last week of lecture, and we will spend those three lectures reiewing the material of the course, and looking at applications of the material. Today we are going to look at the basics of differential amplifiers. 1

2 Reading All of the reading assignments are done Time to start reiewing for the final! Lecture Outline Why differential signaling important, and becoming more important Introduction to differential amplifiers 2

3 Deice Matching One of the things that we depend on in the design of analog integrated circuits is deice matching. For example, if we make a current mirror, we are depending on the reference and the mirror deice behaing is a ery similar fashion. When a gatesource oltage is deeloped on the reference deice, passing a gien current, the same oltage appearing across the gate to source on the mirror deice will allow the same drain current. Variations The transistors will generally ary due to seeral causes: Temperature ery similar for deices on the same substrate Implant ariations important for small deices Variations in width important for narrow deices Variations in length important for short channel deices Layout ariations Often, analog deices will not be minimum sized deices, so that output resistances will be lower, short channel effects will be smaller, and the effect of ariations is Department reduced. of EECS 3

4 Differential s single ended signals A oltage is only defined between two points Charge times oltage difference gies the energy needed to moe a charge between two points. A single ended signal is actually a misnomer, because all oltages are measured between two points, its just that single ended signals use ground as a reference. What is ground In a low frequency circuit, with a big ground plane aailable, it is possible to approximate that ground is a stable reference which all oltages can be measured with respect to. At high frequencies, and at low oltage swings which are needed for reasonable power consumption, the ariation of the ground can ary by more than the signal leel. This is een a problem at low frequencies, if deices are separated by any distance. ground at one deice can easily be seeral olts different than ground at another deice. (DC and AC) 4

5 Differential signaling One solution to this problem is to use differential signaling, that is two wires, close to each other, and in a symmetric configuration, carry the signal as a oltage difference. They needed to be routed together and twisted to aoid coupling magnetic fields from other wires or sources They should also be drien with the same impedance, so they pick up the same noise oltages If all of these things are done, then noise picked up by the wires is picked up by both in the same amount and with the same sign Common mode Differential input In order to use a differential signaling scheme, we need to produce an amplified ersion of the difference between the input oltages This is called a differential amplifier. The output of a differential amplifier can either itself be differential, or it could conert the signal into a single ended ersion with respect to ground. If the inputs change in oltage together, swinging in the same direction, this is called common mode A good differential amplifier is designed to cancel the common mode: called common mode rejection. 5

6 Differential Amplifiers Differential amplifiers also sole some of the other problems that we hae brushed aside so far- like how to bias the input to the right quiescent oltage. The Differential Amplifier Concept The basic idea: amplify the difference between two inputs and reject the common component + out1 in1 + _ in2 + _ + out2 _ out,diff = A,diff ( in,diff )=A,diff ( in1 in2 ) large out, comm = A,comm ( in,com )=A,comm [( in1 + in2 )/2] small 6

7 Two single ended amps One way you could think of making a differential amplifier is to use two separate single ended amplifiers, as we hae been studying for the last seeral weeks. If the deices happened to be identical oltage amplifiers, we would hae: out1 out 2 out1 (If the gain is the same A= A 1 =A 2 ) out = A = A 1 in1 2 in2 out 2 = A = A in 1 in1 A 2 in2 A First try at a Differential Amplifier Notice that this is just two common single transistor amplifiers with resistor pull ups 7

8 Common Mode The common mode of a differential signal is the aerage of the two oltages. If we hae a oltage + and a oltage - the differential oltage is: = + And the common mode oltage is: CM 1 = ( ) Amplify the difference, not the common mode Since the information is carried only by the difference between the two oltages, and the common mode oltage represents noise, we would prefer not to amplify it. Amplifying the common mode oltage wastes the room we hae between the rails, Minimizing the common mode will also improe our ability to correctly bias the following stages 8

9 Two single ended amps If we look at our simple model using two separate single ended amplifiers: out1 out 2 out1 outcm = A = A + 1 in1 2 in2 out 2 = A = A incm 1 in1 + A 2 in2 So unfortunately, using two single ended amplifiers amplifies the common mode noise as much as it does the differential signal. Howeer, with a simple, ery similar circuit, we can sole most of these problems. A good, simple Differential Amplifier By haing the current from both transistors competing for the same current, we can minimize the amplification of the common mode signal 9

10 Let s say something tries to decrease both sides of the output oltage. The current coming through the pull up resistors would then both hae to increase but that can t happen because the only place for the current to go is through the same current sink. 10

11 If one current goes up, the other current has to go down to keep the same current in the current sink. This is perhaps the most common two transistor configuration in linear integrated circuits. Biasing If our resistors or transistors are not completely identical, or if the current source has a finite impedance, then we won t remoe all of the common mode signal, but we will reduce it. The biasing problem is also reduced, in that a common mode steady state oltage can usually be aoided. Internal stages can be isolated from common mode offsets at the inputs. 11

12 In order to bias this circuit, first look at the output oltages. It is desirable to keep the output oltages close to the middle of the rails, so we want: V = V 0 I I 1out d1 bias = I d 2 = I 2out d1 = + V / R + I d 2 Notice that with our idealization of the current source and identical transistors and resistors, with a zero differential input, the output will be zero on both sides oer the accommodation range of the differential amplifier. D = 2( + V ) / R D What is the accommodation range of this differential amplifier? ( The accommodation range is the range of common mode input oltages oer which the amplifier will continue to work satisfactorily.) 12

13 Accommodation limits If both the inputs go low enough so that the drop across the current source is not sufficient to keep it in its operating range, the transistors will both go into cut off and both sides of the output will rise to the positie rail, which also reduces the gain for the differential mode signal. If the inputs both go high enough, the transistors will go into the triode mode, and the gain for the differential mode will be reduced. This limit can be designed to happen only for inputs higher than V dd by some amount. Terminology This configuration is called a differential source coupled pair. The common current source (or resistor) at the transistor s source terminals is called the tail. 13

14 DC characteristics Let s assume that the resistors hae been chosen small enough so that neither transistor goes into the triode mode if the inputs are separately less than V DD, r 0 V V = V V We can then write: in1 gs1 in2 gs2 V V gs1 gs2 = V Tn = V Tn + + C C ox ox 2Id1 µ ( W n 2I µ ( W n d 2 L) 1 L) 2 DC characteristics V in = V V = V V in1 in2 gs1 gs2 Rearranging, with identical transistors: I d1 + I d 2 Vin = Coxµ n W 2 L We also know: I = d 1 + I d 2 Plugging in, rearranging, and using the quadratic formula, we get: I tail I d1 = I 2 tail Coxµ n W ± 4 L ( V ) in C ox 4I µ n tail ( W ) L ( V ) 2 in 14

15 Similarly: Itail Coxµ n W Id 2 = ± 2 4 L ( V ) in C ox 4I µ n tail ( W ) L ( V ) 2 in V V V out out out = ( V DD = ( I = R d D I d1rd ) ( V 1 I d 2) RD Coxµ n W 2 L DD I ( V ) in d1 R C D ox ) 4I n tail ( W ) L 2 ( V ) in Notice that the oltages are not being offset through this differential amplifier, so that these stages can be directly coupled, without problems with the bias oltages, a big adantage µ 15