6.301 Solid-State Circuits
Recitation 18: “Pythagorators,” and other circuits
Prof. Joel L. Dawson
Way back at the beginning of the term, in Recitation 2, we talked about what it means to do
engineering design. Quoting from those notes:
“In engineering design, we make use of nature’s laws to build useful machines.”
The Gilbert Principle is a classic example of how we use the exponential I C vs. VBE relationship to
build analog computation elements. Before examining this in more detail, let’s use the class exercise
to look at another intriguing case.
CLASS EXERCISE: Consider the following
Perfect Current
Mirror
↓I
↓I
Let
Q1
( I S1 )
Q2
IS2
>1
I S1
( IS2 )
V0
R
Using KVL, write V0 as a function of I, IS1, and IS2. Can you think of any useful function served by
V0 ?
(Workspace)
6.301 Solid-State Circuits
Recitation 18: “Pythagorators,” and other circuits
Prof. Joel L. Dawson
How about that?! Note that while the temperature dependence of VBE is complicated, the
temperature dependence of a ΔVBE is simple:
ΔVBE =
If
kT ⎛ I C1 ⎞ kT ⎛ I C 2 ⎞ kT ⎛ I C1 ⎞
ln
−
ln
=
ln
q ⎜⎝ I S ⎟⎠ q ⎜⎝ I S ⎟⎠
q ⎜⎝ I C 2 ⎟⎠
I C1
is independent of temperature, we say that ΔVBE is “PTAT,” or
IC 2
Proportional to Absolute Temperature.
So…we took advantage of our detailed knowledge of bipolar transistors to build a thermometer. The
Gilbert Principle takes advantage of our knowledge in a different way, and for a different end.
Final Note: Do not use circuits like the class exercise without some sort of start-up circuitry. Note
that I=0 is a valid state.
As we saw in lecture yesterday, the Gilbert Principle is a kind of mathematical shorthand for KVL.
Let’s review.
Current square-root circuit:
I0
I1 ↓
Q3
Q2
Q1
IB ↓
Q4
 Loop of VBE s
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6.301 Solid-State Circuits
Recitation 18: “Pythagorators,” and other circuits
Prof. Joel L. Dawson
Write out KVL:
VBE1 + VBE 2 − VBE 3 − VBE 4 = 0
⎛I ⎞
⎛I ⎞
⎛I ⎞
⎛I ⎞
VT ln ⎜ 1 ⎟ + VT ln ⎜ B ⎟ − VT ln ⎜ 0 ⎟ − VT ln ⎜ 0 ⎟ = 0
⎝ IS ⎠
⎝ IS ⎠
⎝ IS ⎠
⎝ IS ⎠
⎛ I02 ⎞
⎛ I1 I B ⎞
VT ln ⎜ 2 ⎟ = VT ln ⎜ 2 ⎟
⎝ I ⎠
⎝I ⎠
S
S
I 0 = I B I1 = k I1
Pretty neat. From this example, simple though it is, we can draw a couple of very general
conclusions.
(1)
(2)
Translinear circuits are fast.
Translinear circuits always involve an even number of VBE s .
To understand (1), look at all of the Cµ s. None of them get Miller multiplied. Cµ1 Even gets
bootstrapped, while Cµ 4 gets shorted out altogether.
To understand (2), consider a tempting but extremely wrong way to implement the square root
function.
↓ I0
I1 ↓
Q2
Q1
Q3
Translinear principle:
I1 = I 0 2
I 0 = I1
(The units don’t even work.)
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NO!!
6.301 Solid-State Circuits
Recitation 18: “Pythagorators,” and other circuits
Prof. Joel L. Dawson
KVL:
⎛I ⎞
⎛I ⎞
VT ln ⎜ 1 ⎟ = 2VT ln ⎜ 0 ⎟
⎝ IS ⎠
⎝ IS ⎠
⎛I ⎞
⎛I 2⎞
ln ⎜ 1 ⎟ = ln ⎜ 0 2 ⎟
⎝ IS ⎠
⎝ IS ⎠
I0 = IS
I1
Vitally important that all of the IS s
cancel out. This is only possible if the
number of CW VBE s equals the
number of CCW VBE s.
To finish off the basics, we recall that we sometimes have the freedom to change emitter areas. If we
write
I S = AE J S
We have
∑V
BEm
=
CW
I Cm
∏I
CW
∑V
BEn
CCW
=
I Cn
∏I
CCW
Sm
Sn
I Cm
I
= ∏ Cn
CCW AEn J S
En J S
∏A
CW
I Cm
∏A
CW
Em
=
I Cn
∏A
CCW
This is the most general translinear principle.
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En
6.301 Solid-State Circuits
Recitation 18: “Pythagorators,” and other circuits
Prof. Joel L. Dawson
Now, at last, a Pythagorator. One such cell was shown to you in lecture yesterday. Here is a seven
transitor version.
I0
↓ Iy
Ix ↓
Q2
Q1
Q4
Q6
Q5
Q3
Q7
How to analyze? Start with
And
The left Gilbert loop gives:
I0 = I5 + I6
I7 = I0
I1 I 2 = I 5 I 7 = I 5 I 0
I x2 = I5 I0 ⇒ I5 =
Ix2
I0
Right Gilbert loop gives:
I6 =
I y2
I0
So for the output:
I0 = I5 + I6 =
2
Ix2 Iy
+
I0
I0
I02 = I x2 + I y2
I0 = I x2 + I y2
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6.301 Solid-State Circuits
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