I am working my way through the Art of Electronics and I have today studied the differential amplifier. I aam having issues with Exercise 2.18: Design a differential amplifier to run from ±5V supply rails, with Gdiff = 25 and Rout = 10k. As usual, put the collector’s quiescent point at half of VCC.

I have already seen this this and this questions, which were quite useful but I still have some issues.

This is my design process:

  1. \$R_{OUT}\$ = 10kΩ was given. Given that the collector quiescent point needs to be halfway between supplies (i.e. 0V), that means that the quiescent currenc across \$R_C\$ is 500μA
  2. Given that $$G_{DIFF} = \frac{R_C}{2(R_E+R_C)} = 25$$ I calculated \$R_E\$ to be 250Ω
  3. at 500μA quiescent current \$R_1\$ will see ≈ 1mA of current. To keep the quiescent \$R_C\$ voltage at 0V, I can only use a resistance \$R_1\$ of about 4.2kΩ. (\$R_E = 250Ω, R_e = 50Ω and V_{BE} = 0.65V)\$

I simulated the circuit and it works fine. However, these are my questions:

  1. The circuit only takes a very small input signal: at 0.02V amplitude the circuit already starts clipping. Why is that? What's the benefit of having the quiescent \$V_c\$ set midway between +/-5V, if I can't seem to be able to get an output amplitude greater than about 300mV, which leaves me with about 4.7V before reaching the supply rails?
  2. As the common mode rejection would greatly increase with a larger resistance \$R_1\$, I could increase \$R_1\$ at the expenses of the amount of quiescent current and \$V_C\$ being centered. Would there be any drawbacks if I did that?

EDIT: The second image shows clipping with an input signal of 0.05V amplitude

Differential mode differential amplifier Clipping

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    \$\begingroup\$ To get the collector’s quiescent point at half of VCC and Rout = Rc = 10k. The Ic2 current must be equal to \$I_{c2} = I_{c1} = \frac{2.5V}{10k} = 0.25mA\$ Thus Iee = 0.5mA. and \$r_e = \frac{26mV}{0.25mA} = 104Ω\$, So \$ R_E = (\frac{10kΩ}{25} - 2*104Ω) * 0.5 = 96Ω\$. and the R1 resistor R1 = (5V - 0.7V)/0.5mA - 96Ω/2 = 8.6kΩ \$\endgroup\$
    – G36
    Commented Dec 12, 2020 at 18:10
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    \$\begingroup\$ And your circuit is clipping because you have set Iee at 1mA thus V_RC = 0.5mA *10kΩ = 5V? Thu your BJT's is at the edge of saturation. \$\endgroup\$
    – G36
    Commented Dec 12, 2020 at 18:18
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    \$\begingroup\$ You made a mistake by assuming \$V_C = 0.5(Vcc + |Vee|)\$ instead \$ V_C = 0.5V_{CC}\$. And this is why you are getting a clipped voltage at the collector. \$\endgroup\$
    – G36
    Commented Dec 12, 2020 at 18:28
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    \$\begingroup\$ Notice that your DC emitter voltage is at \$-0.7\$ Thus, the collector voltage cannot be lower than this \$-0.7\$ voltage ( if \$V_{CE(sat)} \approx 0V\$) \$\endgroup\$
    – G36
    Commented Dec 12, 2020 at 18:34
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    \$\begingroup\$ Hence why I asked for the waveform plot. If it's clipping asymmetrically then bias it so that it moves the waveform in the right direction to avoid this. Probably to about +2.5 volts. \$\endgroup\$
    – Andy aka
    Commented Dec 13, 2020 at 11:11

1 Answer 1


The following schematic is your starting point according to the exercise statement:


simulate this circuit – Schematic created using CircuitLab

The value of \$R_\text{C}=10\:\text{k}\Omega\$ comes from the authors' output impedance requirement. The authors also specify \$V_\text{CC}=+5\:\text{V}\$ and \$V_\text{EE}=-5\:\text{V}\$ and that the quiescent output should be \$V_{\text{OUT}_\text{Q}}=\frac12 V_\text{CC}=+2.5\:\text{V}\$.

(Note: There is an assumption that isn't shown in the schematic above or in the examples shown in the textbook. There needs to be some quiescent base current supplied to the circuit. Often, this is provided by a large-valued resistor between ground and each BJT base. You've used two voltage supplies tied directly to the bases. But that's not how these things are implemented in real circuits.)

From this, find that \$I_{\text{C}_\text{Q}}=\frac{5\:\text{V}-2.5\:\text{V}}{10\:\text{k}\Omega}=250\:\mu\text{A}\$ and therefore \$r_e=r_{e_\text{Q}}=\frac{V_T}{I_{\text{E}_\text{Q}}\approx I_{\text{C}_\text{Q}}}\approx 100\:\Omega\$.

From the equation in the book, \$G_\text{diff}=\frac{R_\text{C}}{2\:\left(r_e+R_\text{E}\right)}\$, easily find that \$R_\text{E}=\frac{R_\text{C}}{2\:G_\text{diff}}-r_e=100\:\Omega\$. (To help reduce temperature-dependence, one may prefer \$R_\text{E}\ge 5 \:r_e\$.)

\$R_1\$ needs to supply both collector currents, so approximately \$500\:\mu\text{A}\$. The problem here is that the schematic, as shown, doesn't explicitly state the quiescent voltage values for the BJTs' bases. (It is possible to set them to something other than ground.) But I think it is reasonable to assume that the actual schematic will be something like this:


simulate this circuit

The BJTs are operating at somewhat lower collector currents than is normally associated with a \$700\:\text{mV}\$ base-emitter drop. I'd guess that it would be at least \$60\:\text{mV}\$ less than that. But there is a small drop across \$R_\text{E}\$ of about \$25\:\text{mV}\$, so I'd compute about \$700\:\text{mV}-60\:\text{mV}+25\:\text{mV}=665\:\text{mV}\$ between the base and \$V_\text{E}\$. If you assume fairly stiff values for \$R_\text{B}\$ (say about \$22\:\text{k}\Omega\$), then this would likely bring us right back to about \$V_\text{E}=-700\:\text{mV}\$.

Now compute \$R_1=\frac{-700\:\text{mV}-\left(-5\:\text{V}\right)}{500\:\mu\text{A}}=8.6\:\text{k}\Omega\$.

For testing the above idea in LTspice, the following schematic should be used:


simulate this circuit

The addition of the capacitor on the \$V_{\text{IN}_\text{(-)}}\$ side allows the base to appear at ground with almost no impedance from an AC standpoint, despite the fact that there will be a slight quiescent DC voltage drop due to the DC-biasing base resistor there and an actual DC impedance, as a result. The capacitor bypasses the base-biasing resistor.

Of course, a similar capacitor is used on the \$V_{\text{IN}_\text{(+)}}\$ side for similar reasons. This keeps the input signal from directly setting the DC base voltage and instead also allows the DC-biasing base resistor to establish a quiescent DC value while at the same time offering almost no impedance to the AC signal arriving at the left.

Here's the schematic (and other details) that I just now included in a new LTspice schematic:

enter image description here

Since I'm using a semi-realistic model for the BJTs, we should expect a slightly lower gain than expected. This is due to a number of factors, which at least include the fact that BJTs have the Early Effect as well as generally modest values for \$h_\text{FE}\$. That said, it shouldn't be too far afield, either. So let's see what LTspice says:

Direct Newton iteration for .op point succeeded.

inrms: RMS(v(vin))=0.000707135 FROM 0 TO 0.2
outrms: RMS(v(vout)-outavg)=0.0169049 FROM 0 TO 0.2
outavg: AVG(v(vout))=2.49526 FROM 0 TO 0.2
gain: (outrms/inrms)=23.9061

LTspice computes a value of \$\approx 23.9\$. Which is pretty close to the design goal. (If you remove the Early Effect, this value is instead \$\approx 24.2\$, for example.)

(Note: There is another bookkeeping difference regarding the LTspice run and our earlier computations. LTspice uses \$27^\circ\text{C}\$ as the default temperature for the run. This translates to \$V_T\approx 25.865\:\text{V}\$. I'd earlier used \$V_T\approx 25\:\text{V}\$ when computing \$r_e\$. If, instead, we use LTspice's actual value we find \$r_e\approx 103.5\:\Omega\$. This means that the emitter resistors should have been slightly lower, perhaps more like \$R_\text{E}\approx 96\:\Omega\$, instead. If we'd used those values in the circuit and eliminated the Early Effect then the resulting gain is about \$24.7\$ according to LTspice. Yet closer to the predicted value.)

LTspice also computes these quiescent operating point values:

       --- Operating Point ---

V(vout):     2.49526      V
V(ve):      -0.670818     V
Ic(Q2):      0.000250474  A
Ic(Q1):      0.000250474  A

Close enough to the planned values.

  • \$\begingroup\$ this is great, thanks for the brilliant clarification. So in a nuthsell to design such a circuit we have to balance how much gain we want, how much voltage swing at the output we need, and to do so we have to decide where between the supplies we want to set the bias voltage; this will impact the common mode rejection ratio at the differential mode gain required. Plus the output and input impedances and so on... what a fun mess! :D Do electical engineers use mathematical optimisation tools to solve these multiobjective problems? \$\endgroup\$
    – MattiaM
    Commented Dec 13, 2020 at 11:20
  • \$\begingroup\$ I am now slightly confused about the 22kΩ values for the biasing resistors. I calculated the input impedance into the bases of the differential pair as \$β*(2*(R_E + R_e)\$ ≈ 40kΩ. If I go for the rule of thumbs of the bias network to be 1/10th of the input impedance, I would use much lower resistor values, e.g. 8kΩ (and since the two are in parallel with each other, I would obtain 4kΩ. Or am I missing something obvious here? \$\endgroup\$
    – MattiaM
    Commented Dec 13, 2020 at 15:58
  • \$\begingroup\$ @MattiaM Yes, some electronics engineers do use optimization methods -- I've seen them use Excel for that purpose, for example. I'm just a hobbyist, though. For me, it's a matter of knowing a few things that help guide me, in the first place, followed by the use of software tools to compute boundaries and perhaps catch mistakes I make. \$\endgroup\$
    – jonk
    Commented Dec 13, 2020 at 19:31
  • \$\begingroup\$ @MattiaM A more likely choice is to look at the collector load, \$250\:\mu\text{A}\$, and divide that by 10 to work out something to supply up to about \$25\:\mu\text{A}\$. But that's with a resistor divider network. Here, that's not necessary as there is a third supply rail -- ground -- and we don't need a divider arrangement. I selected a "stiff" value that will only drop a few 10's of millivolts at the likely base current so that I could avoid having to discuss that drop, or its impact on \$R_1\$'s value which is to be avoided if possible. \$\endgroup\$
    – jonk
    Commented Dec 13, 2020 at 19:40
  • \$\begingroup\$ @MattiaM It could have been still larger, but that would impact the gain (and I'd have to write a much more detailed answer, too.) For Darlington pairs, such as shown here for the LM380, a much larger value can be used. But for singleton BJT arrangements, about 10 times smaller isn't uncommon. \$\endgroup\$
    – jonk
    Commented Dec 13, 2020 at 19:43

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