exponential conversion with both feet

[SION IVOR JONES]

Hello,
this my first post and going to jump in with both feet!
Seems to be a bit of mystery surrounding exponential convertion
and temp compensation ect.. I've been trying to nail it down
and think I grasped it, so I thought I'd share my thoughts on
the subject, and hopefully recieve some thoughts back?
Theres quite a bit of maths but Ive try to explain each
step, and anyway you can't beat a nice little mathmatical
proof (okay maybey not).

I'm using my brothers email address so I'm not on the list but
I download the archive about once a week to keep up. I did subscribe for a day
and got about 12 emails and my brother flipped. anyway here goes.

Exponential conversion!
=======================

Firstly a quick look at the "Ebers Moll" equation
=================================================

Ic   is the current flowing into the collector
Is   is the saturation current of the transistor
Vbe  is the voltage from base to emitter
q    is the charge of an electron
k    is boltzmans constant
T    is absolute Temperature

and we have

   Ic = Is ( exp( Vbe q / k T ) - 1 )

asumption 1:
            If we assume Vbe to greater than 300mV then the
            exponential term "exp( Vbe q / k T )" is very
            large (greater than 100,000 at room temp!), so
            we can ignore the -1 term, and we can rewrite
            the equation thus:

  Ic = Is  exp( Vbe q / k T )                        eq1

Now look at a long tailed pair
==============================

              |  | Iref
              |  |             <--- Iout
              |  V          *-----------
              |             |
              |             |
           ****             ****
  Vin  ----*Q1*             *Q2*------*
           ****             ****      |
              |             |       -----
              |     Ve      |        ---
              *------*------*         -
                     |
                     |

Two NPN transistors with their emiters tied together.
Base of Q2 at tied to ground.
Vin tied to the base of Q1.
Current Iref flowing into collector of Q1.
Current Iout flowing into collector of Q2.

VbeQ1 = Vin - Ve
VbeQ2 = 0V -Ve

and

VbeQ1 - VbeQ2 = (Vin - Ve) - (0V - Ve)
VbeQ1 - VbeQ2 = Vin - Ve - 0V + Ve
VbeQ1 - VbeQ2 = Vin                          eq2

The current Iref ( from a constant current source)
programs VbeQ1, and VbeQ1 and Vin (eq2) program VbeQ2
which reuslts in current Iout.

Now consider the ratio Iref / Iout and eq1

Iref   IsQ1 exp( VbeQ1 q / k T )             eq3
---- = -------------------------
Iout   IsQ2 exp( VbeQ2 q / k T )

asumption 2:
            If we have well matched transistors and
            close thermal matching ( ie on the same
            chip ) we can assume that;

                   IsQ1 = IsQ2

so eq3 becomes

Iref   exp( VbeQ1 q / k T )
---- = --------------------
Iout   exp( VbeQ2 q / k T )

but  exp( a )
     -------- = exp ( a - b)
     exp( b )

so eq3 now becomes

Iref
---- = exp ( (VbeQ1 q / k T) - ( VbeQ2 q / k T ) )
Iout

so we now have

Iref
---- = exp ( (VbeQ1 - VbeQ2) q / k T )
Iout

but eq2 is

VbeQ1 - VbeQ2 = Vin

so we finally have

Iref
---- = exp ( Vin q / k T )          eq4
Iout

So the ratio of Iref to Iout is an exponential
funtion Vin. And at Vin = 0V drift from temperature
is fully compensated. But the larger Vin the greater
the effects of temperature. When Vin is posetive you
get posetive temp drift and when its negative you get
negative temp drift.

So the long tailed pair compensates for the temp
dependance of the saturation current Is but not for
temp dependance of Vbe. But the temp dependance of
Vbe does cancel out with equal collector currents.

So we need to compensate Vbe. We can do this by
introduce temperature depence to Vin. If we
can make Vin into Vin * T then eq 4 becomes;

Iref
---- = exp ( Vin T q / k T )
Iout

and then

Iref
---- = exp ( Vin q / k )
Iout

Most of this is from chapter 2 
           "The Art Of Electronics" (Harowitz and Hill)
check out fig 2.53 page 91
and sec 4.14 is also relevant.

Is there anything I hav'nt considerd?

Also where can get get the kit/specs the the ASM-1, would like
to check it out.

Oh yeah my name is Chris, and I probably wouldn't say half as much if you met
me in the pub.

-- 


 sion at argonet.co.uk   http://www.argonet.co.uk/users/sion
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