Here is the link for a current mirror discussion. I am trying to understand the meaning of compliance voltage equation and how it can be dervied: https://wiki.analog.com/university/courses/electronics/text/chapter-11

Attached the same relevant section screen-shot: enter image description here

Please advice.

Thanks VT


Please commit this to memory forever. You will need to thoroughly understand it, literally over and over and over again.

$$\begin{align*} \textrm{Shockley equation applied} \\ \textrm{to simplified BJT model}\\ \textrm{ignoring the Early Effect:} \\ I_C&=I_S\cdot\left(e^{\cfrac{V_{BE}}{V_T}}-1\right) \\ \textrm{move }I_S\textrm{ over to the other side,}\\ \cfrac{I_C}{I_S}&=\left(e^{\cfrac{V_{BE}}{V_T}}-1\right) \\ \textrm{Drop the insignificant -1 term,} \\ \cfrac{I_C}{I_S}&\approx e^{\cfrac{V_{BE}}{V_T}} \\\\ \textrm{Take the logarithm of both sides,} \\ \\ \operatorname{ln}\left(\cfrac{I_C}{I_S}\right)&\approx\cfrac{V_{BE}}{V_T} \\ \\ \textrm{Solve for }V_{BE}, \\ \\ V_{BE}&\approx V_T\cdot \operatorname{ln}\left(\cfrac{I_C}{I_S}\right) \end{align*}$$

You should be able to do the above in your sleep. You should understand the meaning of any of the above in your sleep.

And you should also know that the thermal voltage \$V_T=\frac{n k T}{q}\$ and is approximately \$26\:\textrm{mV}\$ at room temps, that for small signal BJTs \$n\approx 1\$ (but for diodes it is usually larger), and that \$I_S\$ is, itself, highly temperature dependent (roughly proportional to the 3rd power of T) and that it dominates the temperature behavior of \$I_C\$ because its effect is not only larger but of opposite sign to the temperature effects caused by \$V_T\$. (The temperature-dependent \$I_S\$ equation is more complicated and not provided here.)

Getting back to the point of the current mirror discussion, this just means that you don't want to saturate the BJTs in a mirror -- that's a bad thing as they stop being much of a mirror then. So \$\vert V_{CE}\vert\ge\vert V_{BE}\vert\$, as computed above. Just make sure that the load you attach, together with the mirrored current, doesn't violate that.

  • \$\begingroup\$ For understanding purpose, How is the scenario if the BJT is in saturation? \$\endgroup\$ – vt673 Dec 29 '16 at 10:03
  • \$\begingroup\$ @vt673, A diode-connected BJT can't saturate (Because \$V_{be}=V_{ce}\$). \$\endgroup\$ – The Photon Dec 29 '16 at 18:00
  • \$\begingroup\$ @ThePhoton Yeah. But actually that complicates the discussion. There are three regions of behavior and I wanted to AVOID talking about that. In region I, \$I_C\$ is low but still follows a linear line (on a log-lin plot) towards the intercept at \$I_S\$, while \$I_B\$ starts curving towards \$I_S\$, too. (Meaning lower \$\beta\$.) But in region II (normal), there's just a \$\beta\$ factor between them. In region III (crowding), \$I_B\$ continues linearly upwards, while \$I_C\$ curves down (lower \$\beta\$ again.) Do you really think I should explore all that above? I wanted to keep it simple. \$\endgroup\$ – jonk Dec 29 '16 at 18:34
  • \$\begingroup\$ @ThePhoton I mentioned that this was for a "simplified BJT" model and figured that was enough. \$\endgroup\$ – jonk Dec 29 '16 at 18:37
  • \$\begingroup\$ @ThePhoton In Region II (where \$\beta\$ applies well), \$I_C=\beta\cdot I_B\$, which is why \$\beta\$ is actually kind of useful. Ignoring the Early Effect, this doesn't have to be a diode-connected BJT case. More, in the current mirror case this is exactly why the thing works well, at all. (The Early Effect is one of the problems to be dealt with in more complex current mirrors.) I'm not sure I entirely see why the diode-connected bit is needed. But I do see why the Early Effect should be added, so I added that. But maybe my brain is out to lunch today. \$\endgroup\$ – jonk Dec 29 '16 at 20:14

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