Physics 623The Difference AmplifierJune 7, 20061 Purpose• To construct a difference amplifier, to measure the DC quiescent point and to compareto calculated values.• To measure the diff erence mode gain, the common mode gain, the common moderejection ratio (CMRR) and to compare to calculated values.• To understand the role of the common emitter resistor (R1) in providing feedbacknecessary for differential operation.• To explore examples of applications of difference amplifiers.2 DiscussionNotation:In general, we use the upper case symbols, such as V and I, for the static or Q-point valuesof the voltages and currents. Similarly, we use the lower case symbols, such as v and i, forchanges in the voltages and currents. Typical symbols for difference amplifiers, coupled foramplifying the difference between two AC signals, are shown.Read Horowitz & Hill (2nd Ed) page 98. That description is for a DC coupled differenceamplifier using +15 volt and -15 Volt power supplies. Our amplifier is for AC signals, anduses only one power supply but we use the same notation.If a linear amplifier has two receiving input voltages viand v0i, then its output may be alinear combination of the two inputs:vout= X.vi+ Y.v0iwhere X and Y are two constants. We can manipulate this equation to givevout=(X−Y )2.(vi− v0i) + (X + Y ).(vi+v0i)2.In other words, voutcan be written as the linear combination of the difference (vi− v0i) andcommon mode or average(vi+v0i)2.The difference amplifier provides a number of advantages which make it one of the mostuseful circuit configurations, particularly as an input stage for high gain and DC amplifiers.By choosing matched transistors, often on the same piece of silicon of an integrated circuit,very stable and drift free operation may be obtained because of the symmetry of the amplifier.The differential input is particularly useful in cases where the desired signal is the differencebetween the voltages on two wires, which might be masked by a large and varying voltage,with respect to ground, that is common to both wires. Operational amplifiers often consistof a cascaded series of difference amplifiers which provide excellent stability and high gain.An ideal difference amplifier has a large gain for difference mode signals (vi= −v0i), and1zero gain for common mode signals (vi= v0i). We may write the output voltage (taken fromeither collector) for a real difference amplifier as:vout= Advd+ Acvcwhere Adis the difference mode gain and vd= (vi− v0i) is the difference mode signal; andwhere Acis the common mode gain and vc= (vi+ v0i)/2 is the common mode signal. voutisthe voltage from output to ground, not the difference of voutand v0out.If one uses only a difference mode signal (ie vc= 0), then Ad= v◦/vd. For the commonmode signal only (vd= 0), the common mode gain can be measured Ac= vo/vc.A figure of merit for real difference amplifiers is called the Common Mode Rejection Ratio,CMRR, which is defined as:CMR R = |Ad/Ac|CCvivi'voutvout'Q1Q2ARHRLRCRER1RLRHRCRE2N39042N3904 +15V CCFigure 1: Schematic of the Difference AmplifierFor an ideal difference amplifier, CMRR would be infinite since Ac= 0, and for realamplifiers we want CMRR as large as possible. Consider the circuit shown in Figure 1. Inthe calculation sheet the gains and CMRR are derived and the results are:Ad= −RC2(RE+ rtr)Ac= −RC2R1+ RE+ rtrandCMR R =2R1+ RE+ rtr2(RE+ rtr)where rtr= transresistance ' 0.025 ohm − amp/Ic+ 2 ohm. The symbol rtris sometimeswritten as re.[Sometimes, in technical books, this is carelessly given with inconsistent units asrtr=' 25/Ic(mA) + 2 ohm. Please do not use this notation.]2It is apparent that CMRR will become larger as R1becomes larger. The large R1actsas a current source and an active current source, using an additional transistor is often usedwhen a very large CMRR is desired. The effect of R1may be understood by noting that,in the difference mode the currents in the two transistors are 180◦out of phase, resultingin zero net AC current through R1, thus making point A an AC ground (a virtual ground).This means a large difference mode gain. In the common mode, however, twice the ACcurrent flows through R1than through one transistor. In this case it appears that there is aresistor of value 2R1to ground and the common mode gain dramatically decreases. (For atrue current source, the effective value of R1→ ∞, giving Ac∼ 0.)3 ProcedureThis writeup contains a worksheet which will enable you to calculate the component valuesfor the circuit. This must be done before the lab session.1. Construct the circuit of Fig. 1. Check the lead assignments for the 2N3904 on the datasheets which you will find in the lab.(a) Compare the measured values of Vb, Vcand Veto those calculated in the handout.(b) Is your circuit balanced (ie Vc≈ V0c)?(c) Are both transistors turned on?(d) Use trimming resistors in parallel with the calculated bias resistors as necessaryin order to balance – The collector currents should be the same within ≈ 30%.2. Generate a source of common and difference mode signals. One way to do this is touse the supplied transformer. When the center tap is grounded, the signal at the endof the winding is balanced such that the voltage on one side is equal to, but 180◦outof phase with, that on the other. The difference input thus uses vifrom one side andv0ifrom the other. The common-mode input uses both from the same side.3. (a) Using these s ignals, measure Ad, Acand the CMRR.(b) Compare them to the calculated values.(c) Note the relative phases of the signals at collectors of Q1 and Q2 and at point A.4. The difference amplifier can be used with a one-sided input as a phase inverter.(a) Feed an AC signal into the left input (vi= AC signal) and ground the right input(v0i= 0) through the coupling capacitor.(b) Examine the outputs voutand v0outand the signal at point A.(c) Explain why outputs voutand v0outhave opposite polarity.5. If there is time, you might measure the bandwidth of the amplifier.(a) Use a one-sided input vias in 4a above without the transformer. Put a 3.3 kresistor in series with the input.3(b) Measure the bandwidth.(c) Now put a 250 pF capacitor across the collector to the base of Q.(d) Remeasure the bandwidth.(e) Then put a 1 µF capacitor across Rc, examine the output at the collector of Q0and again remeasure the bandwidth.(f) Discuss your observations.44 Worksheet for the Difference Amplifier Laboratory4.1 Calculation of the