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I'm trying to find ΔV for this circuit, but I'm having a bit of trouble doing so. I think the current on the right BJT will be n times greater than the left, but I'm not sure about that either. It looks like a current mirroring circuit to generate a ref voltage that's stable with temperature, but I'm still unsure how to find ΔV.

enter image description here

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  • \$\begingroup\$ In theory both Q’s will have 0 differential Voltage if identical but differential temperature effects on Vbe will make a difference. There is no mirroring here but assumed Io’s are perfectly matched. \$\endgroup\$ – Tony Stewart EE75 Mar 14 at 14:49
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    \$\begingroup\$ The current in both will be \$I_0\$ - how could it be anything else? \$\endgroup\$ – Andy aka Mar 14 at 15:12
  • \$\begingroup\$ Please provide additional context for the circuit in question. To what does size(A) and size(nA) refer? \$\endgroup\$ – ScienceGeyser Mar 14 at 15:44
  • \$\begingroup\$ This technique is used in many semiconductor temperature sensors by using the same transistor time multiplexed between two currents. \$\endgroup\$ – Kevin White Mar 14 at 17:14
  • \$\begingroup\$ Why did you remove the schematic? The question and my answer refer to components, voltages and currents in the schematic, so removing the schematic makes those references unintelligible. \$\endgroup\$ – Math Keeps Me Busy Mar 15 at 1:32
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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 "diodes".
  • 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

$$ln(\frac{I_{s1}}{I_{s2}}) \approx \frac{qV_{d2}-qV_{d1}}{kT}$$

$$kT \cdot ln(n) \approx q\Delta V$$

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