Solving for an unknown function

Question

I'm having trouble with the mathematical aspect of a journal article I'm researching. I need to figure out how the equation in the curly brackets in Vref turned into an exponential function in Id.

Basically the article states that the terms in the curly brackets should be constant with variable T. I have checked that it is true:

$ln(f(T)) + \frac{T}{f(T)} \frac{\partial}{\partial T}(f(T))=\frac{A}{C}$ (1)

where

$f(T)=exp(\frac{AT+B}{CT})$ (2)
and A, B and C are constants.

But the question I like to ask is how to get (2) from (1). If anyone can give a reference for this or a guide, I'm extremely greatful.

If you're interested in the article:
L. Magnelli, F. Crupi, P. Corsonello, C. Pace, G. Iannaccone, "A 2.6 nW, 0.45 V Temperature-Compensated Subthreshold CMOS Voltage Reference," IEEE J. Solid-State Circuits, vol. 46, no. 2, pp. 465-474, Feb. 2011.

Answer 1

You want to solve the differential equation

$$\ln(f) + \dfrac{T}{f} \dfrac{df}{dT} = \frac{A}{C}$$

This is a separable differential equation: rewrite it as

$$ \dfrac{df}{f (A/C - \ln f)} = \dfrac{dT}{T} $$

and integrate both sides, then solve for $f$.

Answer 2

Define a function $$ f(T) = \exp{\left(\frac{AT + B}{CT} \right)} $$ Its derivative is, by the chain rule $$ f'(T) = \exp{\left(\frac{AT + B}{CT} \right)} \frac{\partial }{\partial T} \left( \frac{AT + B}{CT} \right) = \exp{\left(\frac{AT + B}{CT} \right)} \frac{A}{CT} $$ Now the first equation is $$ \begin{split} \ln{f(T)} + \frac{T}{f(T)}f'(T) =& \frac{AT + B}{CT} + \frac{T}{\exp{\left(\frac{AT + B}{CT} \right)}}\exp{\left(\frac{AT + B}{CT} \right)} \frac{A}{CT}\\ =& \frac{AT + B}{CT} + \frac{TA}{CT}\\ =& \frac{2AT + B}{CT} = \frac{AT + B}{CT} + \frac{A}{C}\\ \end{split} $$ This result is not exactly the answer we're looking for ...

Answer 3

You need to solve $$\ln(f) +T\dfrac{ f'}{f} = \frac{A}{C}$$ Let $f=e^g$ to make $$g+T g'= \frac{A}{C} \implies (Tg)'=\frac{A}{C}\implies Tg=???\implies g=???\implies f=???$$