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Differential Amplifier with Constant Emitter Current

### Circuit Description In the circuit shown to the right, we have replaced RE with a current mirror. As a result, the total emitter current IE is now much less dependent on emitter voltage, VE. There will still be some variation due to the internal resistance of Q4, but its effect will be reduced.

To help calculate voltages and currents in this circuit, we will assume all transistors are matched, and assign some typical component values:

• +VCC = +12 V
• -VEE = -12 V
• VBE = 0.7 V
• RREF = 20K
• RC1 = RC2 = 10K
• β = 200
• VA = 50 V

If we now ground both inputs, we can begin to establish a quiescent operating point for this circuit, as follows (we will ignore base currents for now, in view of the high value of β):

• IE = IREF = [+VCC - (-VEE) - VBE]/RREF = 23.3 V/20K = 1.165 mA
• IC1 = IC2 = IE/2 = 0.5825 mA
• VE = -VBE = -0.7 V
• VC1 = VC2 = VCC - IC1RC1 = 6.175 V

Now let's include the internal resistance of Q4 as well as the effect of β, and see just how much of an effect that has on the quiescent operating conditions of the overall circuit:

• IREF = [+VCC - (-VEE) - VBE]/RREF = 23.3 V/20K = 1.165 mA
• IC4 = IREF × β/(β + 2) = 1.165 mA × 200/202 = 1.153 mA
• rO4 = VA/IC4 = 50 V/1.153 mA = 43.35K
• VCE4 = VE - (-VEE) = -0.7 + 12 = 11.3 V
• IE = IC4 - (VCE4/rO4) = 1.153 mA - (11.3 V/43.35K) = 0.892 mA
• IE1 = IE2 = IE/2 = 0.446 mA
• IC1 = IC2 = IE1 × β/(β + 1) = 0.444 mA
• VRC1 = IC1 × RC1 = 0.444 mA × 10 kΩ = 4.44 V
• VC1 = VC2 = VCC - VRC1 = 12 - 4.44 = 7.56 V

Well! That makes quite a difference. Clearly we need to take these factors into account if we are to have an accurate model of this circuit. Note that this doesn't make the circuit useless by any means. On the contrary, this particular circuit finds many practical uses. We simply need to recognize and allow for the parameters that affect its operation.

### Common Mode Operation

If Q4 was an ideal transistor, it would exhibit no internal losses and we would have a truly constant total emitter current regardless of the applied common mode voltage. Unfortunately, this is not the case. Therefore, we need to determine the common mode gain of this circuit. To this end we note that any change in VIN(CM) will cause an equal change in VE for Q1 and Q2, since VBE will remain essentially constant. This in turn changes VCE4, so we will need to recalculate IE and its effects on Q1 and Q2, for each new value of VIN(CM). The following table shows our results:

VIN(CM) VE VCE4 IE IE1 IC1 VRC1 VOUT
+5 V +4.3 V 16.3 V 0.777 mA 0.389 mA 0.387 mA 3.87 V 8.13 V
+4 V +3.3 V 15.3 V 0.801 mA 0.400 mA 0.398 mA 3.98 V 8.02 V
+3 V +2.3 V 14.3 V 0.824 mA 0.412 mA 0.410 mA 4.10 V 7.90 V
+2 V +1.3 V 13.3 V 0.847 mA 0.423 mA 0.421 mA 4.21 V 7.79 V
+1 V +0.3 V 12.3 V 0.870 mA 0.435 mA 0.433 mA 4.33 V 7.67 V
0 V -0.7 V 11.3 V 0.893 mA 0.446 mA 0.444 mA 4.44 V 7.56 V
-1 V -1.7 V 10.3 V 0.916 mA 0.458 mA 0.456 mA 4.56 V 7.44 V
-2 V -2.7 V 9.3 V 0.939 mA 0.469 mA 0.467 mA 4.67 V 7.33 V
-3 V -3.7 V 8.3 V 0.962 mA 0.481 mA 0.479 mA 4.79 V 7.21 V
-4 V -4.7 V 7.3 V 0.985 mA 0.493 mA 0.490 mA 4.90 V 7.10 V
-5 V -5.7 V 6.3 V 1.008 mA 0.504 mA 0.502 mA 5.02 V 6.98 V

If we now determine the ratio of ΔVOUT/ΔVIN from the above table, we find it to be about 0.11. This means that by using a current mirror to supply emitter current. we have reduced the common mode gain of the circuit almost five-fold. This is a significant improvement over using just a common emitter resistor.

### Difference Mode Operation and CMRR

Since Q1, Q2, and the collector load resistors haven't changed from the previous circuit, the differential behavior of the circuit also remains effectively unchanged. However, because the common mode gain has been reduced, the common mode rejection ratio (CMRR) has improved. We will use a maximum differential gain of 219.8 and the equation:

CMRR = 10 log10(Ad/Acm)² = 20 log10(|Ad/Acm|)

With these values, we get a maximum CMRR of about 66 db. As long as the differential input voltage remains under about 0.1 V, CMRR remains above 60 db. This amounts to a 10 db improvement over the basic differential amplifier.

As it stands, our differential amplifier has some practical applications. However, it also still has some shortcomings, especially if it is to be put into an integrated circuit for best matching of the transistors. The biggest problem at this point consists of the two collector load resistors, RC1 and RC2. These are high-value resistors (10K in our example) and will occupy considerable space on a chip. Our next step, then, will be to find a way to replace them with something different, that will still do the job.

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