6.301 Solid-State CircuitsRecitation 19: More on Translinear CircuitsProf. Joel L. DawsonWe’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 thesetranslinear ideas. In the middle of the class, though, we’ll try our hand at a small translinear designproblem.Five-Transistor PythagoratorWhat is I0in terms of Ixand Iy? Well, we see a Gilbert loop right away in Q1− Q4. Must use themore general form:Ix2AE⋅Ix2AE=I3AEI3+ Iy( )AE14Ix2= I3I3+ Iy( )Q5I0Iy↓Q4Q32AEQ12AEQ2↓Ix6.301 Solid-State CircuitsRecitation 19: More on Translinear CircuitsProf. Joel L. DawsonPage 2Need the relationship between I3 and other variables of interest. We can writeI0= I3+ I5Q4 and Q5 form a current mirror (which, incidentally, is the simplest possible Gilbert loop):I5= I4= I3+ IyI0= 2I3+ Iy⇒ I3=I0− Iy2Substituting back into our first expression:14Ix2=I0− Iy2⎛⎝⎜⎞⎠⎟I0− Iy2+ Iy⎛⎝⎜⎞⎠⎟=14I02− 2IyI0+ Iy2( )+12I0Iy−12Iy2Ix24=I024−Iy24=14I02− Iy2( )I02= Ix2+ Iy2⇒ I0= Ix2+ Iy2Question of the day: How did someone come up with that? Now let’s try some on our own.6.301 Solid-State CircuitsRecitation 19: More on Translinear CircuitsProf. Joel L. DawsonPage 3CLASS EXERCISE 1 (to be worked on individually):Design a translinear circuit that performs the functioni0= 4 ⋅ ii(Workspace)CLASS EXERCISE 2 (to be worked on in pairs):Design a translinear circuit that gives the following input-output relation:i0= ki36.301 Solid-State CircuitsRecitation 19: More on Translinear CircuitsProf. Joel L. DawsonPage 4Now let’s turn our attention to National’s LH0091 True RMS to DC converter. It is called thisbecause its output isv0=1RCvI2dt∫To see how this happens, turn your attention to Q1− Q4indicated on the schematic:See the Gilbert loop? I1I2= I3I4But I1= I2=vIR, and I4=v0R. This gives vI2R2= i3v0R( ).Looking back at the schematic, we can tie v0and i3together: vIR+−i4i3v0RQ4Q3Q2Q1v0−+i3CCdv0dt= i3v0=1Ci3dt∫6.301 Solid-State CircuitsRecitation 19: More on Translinear CircuitsProf. Joel L. DawsonPage 5So now we can fully analyze the circuit:vI2R= i3v0RIntegrate both sides1R2vI2dt =1Ri3v0dt∫∫Now assume v0 changes slowly compared to vI, so that we can write1R2vI2dt =v0Ri3dt∫∫1R2vI2dt