Question about diode connected transistor (BJT) with current source
If I have done my math correctly, to a very good approximation,
$$\Delta V \approx \frac{kT}{q} ln(n) = V_T ln(n)$$
where

k is Boltzman's constant
T is the absolute temperature in Kelvins
n is the ratio between the reverse saturation/leakage currents of the two &quot;diodes&quot;.
q is the charge of an electron
\$V_T\$ is the temperature equivalent voltage

Thus, the output voltage will quite accurately reflect the absolute temperature.
The derivation I used is as follows:
For a silicon diode, (and approximately for a diode connected silicon transistor)
$$I_d = I_s(e^{\frac{qV_d}{kT}}-1) \approx I_se^{\frac{qV_d}{kT}} $$
Since the diodes have the same current,
$$I_{s1}e^{\frac{qV_{d1}}{kT}} \approx I_{s2}e^{\frac{qV_{d2}}{kT}}$$
Taking logarithms on both sides
$$ln(I_{s1}) +\frac{qV_{d1}}{kT} \approx ln(I_{s2})+\frac{qV_{d2}}{kT}$$
Rearranging gives
$$ln(\frac{I_{s1}}{I_{s2}}) \approx \frac{qV_{d2}-qV_{d1}}{kT}$$
$$kT \cdot ln(n) \approx q\Delta V$$