6.301 Solid-State Circuits
Recitation 19: More on Translinear Circuits
Prof. Joel L. Dawson
We’re going to continue today with more on “Pythagorators,” and then talk about an industry design:
The LH0091 True RMS to DC converter. It’s a very interesting application of some of these
translinear ideas. In the middle of the class, though, we’ll try our hand at a small translinear design
problem.
Five-Transistor Pythagorator
I0
Ix ↓
Q3
Q2
2AE
↓ Iy
Q5
Q4
Q1
2AE
What is I 0 in terms of I x and I y ? Well, we see a Gilbert loop right away in Q1 − Q4 . Must use the
more general form:
Ix
I
I I3 + Iy
⋅ x = 3
2AE 2AE AE
AE
(
1
4
(
Ix2 = I3 I3 + Iy
)
)
6.301 Solid-State Circuits
Recitation 19: More on Translinear Circuits
Prof. Joel L. Dawson
Need the relationship between I 3 and other variables of interest. We can write
I0 = I3 + I5
Q4 and Q5 form a current mirror (which, incidentally, is the simplest possible Gilbert loop):
I5 = I4 = I3 + I y
I 0 = 2I 3 + I y ⇒ I 3 =
I0 − I y
2
Substituting back into our first expression:
1
4
⎛ I0 − I y ⎞ ⎛ I0 − I y
⎞
I x 2 = ⎜
+ Iy ⎟
⎟
⎜
⎝ 2 ⎠⎝ 2
⎠
=
1
4
(I
2
0
)
− 2I y I 0 + I y 2 + 12 I 0 I y − 12 I y 2
2
I x2 I02 I y
=
−
=
4
4
4
1
4
(I
2
0
− I y2
)
I02 = I x2 + I y2 ⇒ I0 = I x2 + I y2
Question of the day: How did someone come up with that? Now let’s try some on our own.
Page 2
6.301 Solid-State Circuits
Recitation 19: More on Translinear Circuits
Prof. Joel L. Dawson
CLASS EXERCISE 1 (to be worked on individually):
Design a translinear circuit that performs the function
i0 = 4 ⋅ ii
(Workspace)
CLASS EXERCISE 2 (to be worked on in pairs):
Design a translinear circuit that gives the following input-output relation:
i0 = ki 3
Page 3
6.301 Solid-State Circuits
Recitation 19: More on Translinear Circuits
Prof. Joel L. Dawson
Now let’s turn our attention to National’s LH0091 True RMS to DC converter. It is called this
because its output is
v0 =
1
RC
∫v
2
I
dt
To see how this happens, turn your attention to Q1 − Q4 indicated on the schematic:
v0
vI
R
R
Q1
i4
i3
−
+
Q4
Q2
Q3
See the Gilbert loop? I1 I 2 = I 3 I 4
But I1 = I 2 =
vI
R
, and I 4 =
v0
R
. This gives
vI
R
2
2
= i3
( ) .
v0
R
Looking back at the schematic, we can tie v0 and i3 together:
C
i3
C
−
+
v0
Page 4
dv0
dt
v0 =
= i3
1
C
∫ i dt
3
6.301 Solid-State Circuits
Recitation 19: More on Translinear Circuits
Prof. Joel L. Dawson
So now we can fully analyze the circuit:
2
vI
v
= i3 0
R
R
Integrate both sides
1
R2
∫v
2
I
dt =
1
i3v0 dt
R∫
Now assume v0 changes slowly compared to vI , so that we can write
1
R2
∫v
1
R2
∫v
v0 =
2
I
dt =
2
I
v0
i3dt
R∫
dt =
v0
Cv0
R
1
2
vI dt
∫
RC
Page 5
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6.301 Solid-State Circuits
Fall 2010
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