2
9.7kviews
What is bandgap reference. In short describe various methods of implementation of bandgap reference.
1 Answer
3
464views

Bandgap ref.

  • Objective of reference generation is to establish a dc voltage or current that is independent of the dupply and process and has a well defined behavior temperature.
  • In most of the applications ,the required temperature dependence assumes one of the 3 forms.

    (a) PTAT - proportional to absolute temperature.

    (b) constant $g_m$ behavior.

    (c) Temperature Independent.


A] Temperature-Independent References

  • Reference voltage /current that exhibits little dependence on temperature prove essential in many analog circuits.
  • Most process parameter vary with temperature,if a ref is temperature-independent,then it is process independent as well.
  • if 2 quantities having opposite temp co-efficient(TCs)are added with proper weighting , the result displays a zero TC.

For e.g: 2 voltages $\to\ V_1 \ \&\ V_2\ \to$ vary in opposite directions with Temp. choose $\alpha_1\ \&\ \alpha_2$ such that $\alpha_1 \frac{\partial V_1}{\partial T_1}+\alpha_2 \frac {\partial V_2}{\partial T_2}=0$,obtaining a reference voltage ,$V_{Ref}=\alpha_1 V_1+\alpha_2 V_2$ ,with zero TC.

  • WKT,$V_{BE}$ of a transistor has -ve TC $\&\ \ \Delta V_{BE}=V_T\ln(n)$ has a +ve TC.
  • Now,we can develop a reference voltage having a temp. coeff. as :

    $V_{Ref}=\alpha_1 \alpha_2 V_{BE}+(V_T\ln(n))..........(1)$

where $V_T\ln(n)=\Delta V_{BE}=$ change in $V_{BE}$voltage of 2 BJTs operating at different current densities

At room temp,$\frac{\partial V_{BE}}{\partial T}\approx -1.5\ mV/K\ \ \&\ \ \frac{\partial V_T}{\partial T}=+\ 0.087\ mV/K$

Set $\alpha_1 =1\ \&$ calculate $\alpha_2 \ln (n)$ which in above case is equal to $\alpha_2\ln(n) \approx\ 17.2$

$\therefore V_{Ref}=V_{BE}+17.2 V_T............(2)$

Circuit shown below shows implementation of temperature independent reference:-

enter image description here

$V_{01}=V_{02}$

thus $V_{02}=V_{B{E_2}}+V_T \ln(n)$

$V_{02}$ serve as a Temp-Independent ref. if $ \ln(n)\approx 17.2$

The above circuit requires 2 modifications

  1. Mechanism that guarantee $V_{01}=V_{02}$
  2. Since $\ln(n)=17.2$ translates to a prohibitively to largen the term $R_1=V_T\ln(n)$ must be scaled up by a reasonable factor.

$\to$Op-amp based implementation of Temperature Independent bandgap ref.

enter image description here

  • $A_1$ senses $V_x\ \ \&\ \ V_y ,$ driving the top terminals of $R_1\ \&\ R_2$.

    $(R_1=R_2)$ such that X $\&$ Y will be approx equal to each other.

  • The ref voltage is obtained at the output of the amplifier

$V_{out}=V_{B{E_2}}+\frac{V_T\ln(n)}{R_3}(R_3+R_2)$

$\quad \quad =V_{B{E_2}}+V_T\ln(n)(1+\frac{R_2}{R_3})$

For s zero TC, we must have $(1+\frac{R_2}{R_3})\ln(n)\approx 17.2$

Foe eg:$n=31\ \&\ \frac{R_2}{R_3}=4$

$\to$Various Issues:

(a)Collector Current Variation

(b)Compatibility with CMOS technology

(c)Op-amp offset and O/P Impedance

(d)Feedback Polarity


B] Supply Independent References :-

  • In CM ckts we have assumed that ref. voltage is available.
  • We need to generate $I_{ref}$ independent of supply voltage.
  • As an approximation to current source in CM, we tie a resistor from $V_{DD}$ to gate $M_1$.
  • O/P current $\to$ quite sensitive to $V_{DD}$.

    $\Delta I_{out }=\frac{\Delta V_{DD}}{R_1+\frac{1}{g_m}}\frac{(\frac{W}{L})_2}{(\frac{W}{L})_1}$enter image description here

In order to make $I_{out}$ less sensitive ti $V_{DD}$

$I_{Ref}\ \to$ derived from $\to\ I_{out}$

enter image description here

$M_3\ \&\ M_4\ \to$ Copy $I_{out}$ thereby defining $I_{Ref}$ with 2 back to back CM ckts.

$I_{out}\ \&\ I_{Ref} \to $ independent of $V_{DD}$,To define $I_{out}\ \&\ I_{Ref} $ uniquely ,we add another constraint in fig(3).

enter image description here

$R_s\ \to\ \downarrow$ current of $m_2$, whereas pmos device requires $I_{out}=I_{Ref}$ because they have identical dimensions.

$\therefore$ we can write

$V_{GS_1}=V_{GS_2}+I_{D_2}R_S$

$\sqrt{\frac{2I_{out}}{\mu_n(\frac{W}{L})_nC_{ox}}}+V_{TH_1}=\sqrt{\frac{2I_{out}}{\mu_n(\frac{W}{L})_nC_{ox}}}+V_{TH_2}+I_{out}$

Neglecting body effect

$\sqrt{\frac{2I_{out}}{\mu_n(\frac{W}{L})_nC_{ox}}}(1+\frac{1}{\sqrt{k}})=I_{out}R_S$

$\therefore I_{out}=\frac{2}{\mu_n(\frac{W}{L})_nC_{ox}}(1+\frac{1}{\sqrt{k}})^2\frac{1}{R_S^2}$

$\therefore$ current -independent to $V_{DD}$

Please log in to add an answer.