Assuming two identical transistors, having equal collector currents with zero input voltage,
with input Vi the differential output current, (I1 - I2), is proportional to tanh(Vi/2Vt)
where Vt =26mV at 25 deg.C.
Expanding the tanh function as a power series:
tanh x = x - x3/3 + 2x5/15 - 17x7/315 + ...... an infinite series. (for |x| less than pi/2)
If the current source It in the diagram is 1mA, then the first two terms of the power series are:
(I1 - I2) = Vi/52 - Vi3/421824 where Vi is in millivolts.Suppose Vi = A sin(wt) where A is in millivolts.
Using the formula sin3(x) = 3sin(x)/4 - sin(3x)/4
(I1 -I2) = A sin(wt)/52 - A3[(3sin(wt)/4 - sin(3wt)/4] / 421824.As an example, with 2mV peak input signal the 3rd harmonic component will be 0.025% of the fundamental level.
The fundamental terms are all we need to work out the stage gain, and then
(I1 - I2) = Asin(wt)[1/52 - 3A2/1687296]
The gain is then proportional to (I1 - I2)/Vi which is
1/52 - 3A2/1687296Substituting various values for signal amplitude A we find:
At 1mV input, the gain falls by 0.009%
At 5mV input, the gain falls by 0.22%
At 20mV input, the gain falls by 3.6%
These figures become increasingly inaccurate at higher voltages because we have only used the first two terms of the power series, and the higher order terms can be expected to become increasingly important as signal level increases.
