REFERENCE CIRCUITS. A reference circuit is an independent voltage or current source which has a high degree of precision and stability. Output voltage/current should be independent of power supply. Output voltage/current should be independent of temperature.
is called the sensitivity of y with respect to xi
Sensitivity w.r.t. x1 * percentage change in x1
+ Sensitivity w.r.t. x2 * percentage change in x2
Goal: Design reference circuits so that the reference’s sensitivities w.r.t. various variations are minimized.
For temperature variation, typically use fractional temperature coefficient:
~6-bit, or 2%
Large static power
for small ro
Power sensitivity =1
Temp coeff depends on material
These can be used as “start up” circuits.
If VCC = 10V, R = 10 kW, and IS = 10-15A, then
Taking ∂/∂T and using: VCC − VREF + kT/q ≈ VCC − VREF:
where VGO = 1.205 V is the bandgap voltage of silicon.
If VREF = VBE = 0.6V, TCF of R = 1500 ppm,
then TCF of VREF = -3500 ppm/oC
If VDD = 10V, W/L = 10, R = 100kW,and using parameters from Table3.1-2,
then VREF = 1.97V and
This is not nearly as good as the VBE reference.
mo = KT-1.5 ; VT = VT0 - aT or VT(T) = VT(To) - a(T-To)
The book has one example of using this.
MOS version: use VGS to generate a current and then use negative feed back stabilize i in MOS
Differentiating wrt VDD and assuming constant VDS1 and VGS4 gives the sensitivity of IOUT wrt VDD.
HW: Show that approximately:
Requires start up
Not shown here
Analyze the sensitivity of the output I with respect to VDD and temperature.
Come up with a start up circuit for the circuit on the previous slide, using only active resisters without RB. Note that you need to make sure that at the desired operating point, the connection from the start up circuit should be turned off.
I1 = I2, J1 = KJ2,
but J = Jsexp(VEB/Vt)
J1/J2 = K =
VEB1─ VEB2 = Vt ln(K)
I = (VEB1─ VEB2)/R
= Vt ln(K)/R Vt = kT/q
Vout = VEB3 + I*L*R = VEB3 + (kT/q)*Lln(K)
Vout/T = VEB3/T + (k/q)*Lln(K)
At room temperature, VEB3/T = ─2.2 mV/oC,
k/q = +0.085 mV/oC.
Hence, choosing appropriate L and K can make
When this happens, Vout = 1.26 V
Generate a negatively PTAT (Proportional To Absolute Temperature) and a positively PTAT voltages and sum them appropriately.
Vt(Vt = kT/q) is PTAT that has a temperature coefficient of +0.085 mV/°C at room temperature.
Multiply Vtby a constant K and sum it with the VBEto get
VREF = VBE+ KVt
If K is right, temperature coefficient can be zero.
Transistors Q1 and Q2 are assumed to have emitter-base areas of AE1 and AE2, respectively.
If VOSis zero, then the voltage across R1 is given as