# Employment Test

Discussion in 'Electronic Design' started by Jim Thompson, Dec 30, 2003.

1. ### Jim ThompsonGuest

Posted at....

Newsgroups: alt.binaries.schematics.electronic
Subject: Employment Test - EmploymentTest.pdf
Message-ID: <>

...Jim Thompson

3 0

3. ### Jim ThompsonGuest

You're supposed to show your supporting equations ;-)

...Jim Thompson

4. ### John LarkinGuest

By symmetry. We don't need no stinkin' equations.

John

5. ### Jim ThompsonGuest

I happen to live "by symmetry".

We don't need no education... dum... di... dum ;-)

"The Wall", one of my all-time favorites... best played at 90+ dBm ;-)

I also liked the DISCO era... even developed a few sub-bass schemes.

...Jim Thompson

6. ### Walter HarleyGuest

I heard a Muzak'ed version of it in a shopping mall once. Honest! Still
chills me to think of it. Sort of a "resistance is futile - you will be
assimilated" message.

7. ### John SmithGuest

I thot the two diodes on the left had a total of -4.4 mV/C while the Vbe on
the right was -2.2 mV/C giving a net of -2.2 mV/C at the emitter.

explain why the TC is 0?

Thanks,
John

8. ### Spehro PefhanyGuest

Ummm.. measured from *where* to *where* ?

Best regards,
Spehro Pefhany

9. ### analogGuest

Since Jim specified infinite forward beta (current gain) the right
side transistor does not load the left side circuitry. Therefore,
by symmetry the center point on the left is always at exactly half
supply (3 volts) in spite of the TC of the transistors.

Jim also conveniently specified an Is (reverse bias saturation
current) for the transistor base emitter junctions that when plugged
into the diode equation, just happens to yield a 0.7 volt forward
drop at room temperature (27C) and the 1ma operating current. Note
that the 3k resistor on the right has exactly 3 volts across it and
the 2.3k resistors on the left have exactly 2.3 volts across them,
thus making all currents 1ma. Remember that base currents are all
zero with the stipulation of infinite beta. By the way, the two
transistors on the left are hooked as diodes since the bases and
collectors are shorted.
It is because the base of the right side transistor is connected to
the base of the upper left hand transistor. Since both base emitter
junctions experience identical temperature induced voltage changes,
the emitters will both be exactly the same voltage below the shorted
together bases and stay at 3 volts regardless of temperature induced
changes in the junction voltages.

Well, almost. There is a second order effect that results in the TC
only being reduced by a couple of orders of magnitude, so it's not
actually zero. The left and right hand currents don't stay exactly
the same when the temperature changes. The right side always
maintains the 3 volts on its resistor so its current holds at 1ma,
whereas the left side resistor voltage changes by approximately -2mv
per degree C due to the transistor TC. This results in a +0.87ua
per degree C drift in the left hand side current which upsets the
match in left and right hand side junction drops.

In a real circuit other effects would probably mask this small
residual drift, so, for all intents and purposes, the TC can be
considered to be zero.

PS to Jim: Does this makes up for my initial terse answer? 10. ### John WoodgateGuest

I read in sci.electronics.design that Jim Thompson
The only way I know to see this is to:

1. set up my newsreader to subscribe to a.b.s.e in addition to what I
subscribe to now;

3. Search for the PDF.

Is there an easier way? I have a dial-up connection, if that makes any
difference.

11. ### Active8Guest

I can e-mail it to you. Let me know.

12. ### Keith R. WilliamsGuest

such things (I'm not familiar with Turnpike).
Yes, someone could email it to you (it's a 10K .pdf), but you
specifically ask people not to. ;-)

13. ### RickGuest

Current down left-hand branch:
Ie(Q2) = (3 - Vbe(Q2)) / 2.3K (1)

CUrrent down right-hand branch:
Ie(Q1) = (3 + Vbe(Q2) - Vbe(Q1)) / 3k (2)

If we *assume* that the Vbe's do indeed cancel out to produce
exactly 3V at vout regardless of the temperature, then
the above equation simplifies to:

Ie(Q1) = 3 / 3k (3)

You can see that the assumption has NOT produced the same equation
for the currents in Q1 and Q2. If the equations aren't the same
then neither are the Vbe's, and hence Vout is NOT a nicely cancelled
3V.

14. ### John SmithGuest

Symmetry. Yes. I overlooked that.

Thanks, Analog.

John

15. ### Jim ThompsonGuest

Correct. Now write a mathematical solution for Ileft =/= Iright to
demonstrate the variation from ideal.

...Jim Thompson

16. ### Jim ThompsonGuest

You are forgiven, my son ;-)

...Jim Thompson

17. ### Jim ThompsonGuest

In Agent, you simply double-click the Message-ID and you go right to
it, even if you are not subscribed to that particular group. HOWEVER
your ISP must carry the group.

I'll E-mail you a copy.

...Jim Thompson

18. ### RickGuest

Why stop there? Here's a *full*, algebraic derivation of the currents
and output voltage, courtesy Maple and Mr Lambert. As a taster,
I've included the expression for the output voltage just below:

Vout = 6-k*T/q*LambertW(2300*Is/k/T*q*exp((3+2300*Is)/k/T*q))+2300*Is-
log(1/Is*k*T/q*LambertW(3000*Is/k/T*q*exp(-(-6+k*T/q*
LambertW(2300*Is/k/T*q*exp((3+2300*Is)/k/T*q))-5300*Is)/k/T*q))
/3000)*k*T/q;

Did I get the job?

Rick

# current in left branch:
# Kirchoff::
eqn1 := 4600 Ie1 + 2 Vbe1 = 6

# Let's find out what Vbe1 is by starting with the diode eqn...
/ Vbe1 \
eqn2 := Ie1 = Is |exp(----) - 1|
\ Vt /

# ...and solving it for Vbe1:
Ie1 + Is
Vbe1 := ln(--------) Vt
Is

# That gives us an expression for Vbe1 in eqn1, so now let's solve eqn1
# for Ie1:
3 + 2300 Is
Is exp(-----------)
Vt
Ie1 := 1/2300 Vt LambertW(2300 -------------------) - Is
Vt

# And now compute the voltage fed across to Q1's base:
3 + 2300 Is
Is exp(-----------)
Vt
Vbase := 6 - Vt LambertW(2300 -------------------) + 2300 Is
Vt

# compute vout now we know Vbase:
# Vout is just a diode-drop from Vbase. I've called Q1's Vbe = Vbe2...
eqn4 := Vout =

3 + 2300 Is
Is exp(-----------)
Vt
6 - Vt LambertW(2300 -------------------) + 2300 Is - Vbe2
Vt

# eqn4 is nice, but we'll need to eliminate Vbe2 from it in order to
# solve for Vout...
Ie2 + Is
Vbe2 := ln(--------) Vt
Is

# ...that's just introduced a term in Ie2 (the current in the right-hand
# branch), but that's easily eliminated...
Ie2 := 1/3000 Vout

# Now that we've got an expression for Vout that's independent of Ie2,
# let's go back to eqn4...
(3 + 2300 Is) q
Is q exp(---------------)
k T
k T LambertW(2300 -------------------------)
k T
6 - -------------------------------------------- + 2300 Is - ln(
q

/
|
|
|
|
1/3000 k T LambertW(3000 Is q exp(- |
\

/ (3 + 2300 Is) q \
| Is q exp(---------------) |
| k T |
| k T LambertW(2300 -------------------------) |
| k T |
|-6 + -------------------------------------------- - 5300 Is|
\ q /

\
|
|
|
|
q|/(k T))/(k T))/(Is q)) k T/q
/
-14
Is := .176 10

# Compute Vout at a couple of temperatures as a sanity check:
2.997588746

19. ### MacGuest

There are a lot of different news readers out there. On mine I can sample
a specified number of headers for a new group (i.e., one I don't subscribe
to.) Even so, I don't normally follow any of the links to a.b.s.e, but
that's just because I am terminally lazy. ;-)

Mac

20. ### John WoodgateGuest

I read in sci.electronics.design that Active8 <[email protected]
I've subscribed temporarily to a.b.s.e. and now have 106 more articles
to scan!

Thanks for the offer. I am looking for a permanent better method of
getting stuff from a.b.s.e., if there is one.  