Bandgap Reference

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temperature coefficient

The parameter that shows the dependence of the reference voltage on temperature variation is called the temperature coefficient and is defined as: $$ TC_F=\frac{1}{V_{\text{REF}}}\left[ \frac{V_{\text{max}}-V_{\text{min}}}{T_{\text{max}}-T_{\text{min}}} \right]\times10^6\;ppm/^oC $$

Choice of n

image-20221117002714125

classic bandgap reference

bg.drawio

$$ V_{bg} = \frac{\Delta V_{be}}{R_1} (R_1+R_2) + V_{be2} = \frac{\Delta V_{be}}{R_1} R_2 + V_{be1} $$

$$ V_{bg} = \left(\frac{\Delta V_{be}}{R_1} + \frac{V_{be1}}{R_2}\right)R_3 = \left(\frac{\Delta V_{be}}{R_1} R_2 + V_{be1}\right)\frac{R_3}{R_2} $$

OTA offset effect

bg_ota_vos.drawio

$$\begin{align} V_{be1} &= \frac{kT}{q}\ln(\frac{I_{e1}}{I_{ss}}) \\ V_{be2} &= \frac{kT}{q}\ln(\frac{I_{e2}}{nI_{ss}}) \end{align}$$

Here, we assume Ie = Ic

Hence,

$$\begin{align} \Delta V_{be} &= \frac{kT}{q}\ln(n\frac{I_{e1}}{I_{e2}}) \\ &= \frac{kT}{q}\ln(n) + \frac{kT}{q}\ln(\frac{I_{e1}}{I_{e2}}) \\ &= \Delta V_{be,0} + \frac{kT}{q}\ln(\frac{I_{e1}}{I_{e2}}) \end{align}$$

Therefore,

$$\begin{align} V_{bg} &= \frac{\Delta V_{be}+V_{os}}{R_2}(R_1+R_2) + V_{be2} \\ &= \alpha \Delta V_{be,0} + \alpha \frac{kT}{q}\ln(\frac{I_{e1}}{I_{e2}}) + \alpha V_{os} + \frac{kT}{q}\ln(\frac{I_{e2}}{nI_{ss}}) \\ &= \alpha \Delta V_{be,0} + \alpha \frac{kT}{q}\ln(\frac{I_{e1}}{I_{e2}}) + \alpha V_{os} + \frac{kT}{q}\ln(\frac{I_{e2,0}}{nI_{ss}})+\frac{kT}{q}\ln(\frac{I_{e2}}{I_{e2,0}}) \end{align}$$

We omit the last part $$\begin{align} V_{bg} &\approx \alpha \Delta V_{be,0} + \alpha \frac{kT}{q}\ln(\frac{I_{e1}}{I_{e2}}) + \alpha V_{os} + \frac{kT}{q}\ln(\frac{I_{e2,0}}{nI_{ss}}) \\ &= \alpha \Delta V_{be,0} + V_{be2,0} + \alpha \left(V_{os} + \frac{kT}{q}\ln(\frac{I_{e1}}{I_{e2}})\right) \\ &= V_{bg,0} + \alpha \left(V_{os} + \frac{kT}{q}\ln(\frac{I_{e1}}{I_{e2}})\right) \end{align}$$

i.e. the bg variation due to OTA offset $$ \Delta V_{bg} \approx \alpha \left(V_{os} + \frac{kT}{q}\ln(\frac{I_{e1}}{I_{e2}})\right) $$

  • Vos > 0

    ​ Ie1 > Ie2: ΔVbg > αVos

  • Vos < 0

    ​ Ie1 < Ie2: ΔVbg < αVos

OTA with chopper

bg_chop.drawio
bg_chop_shift.drawio

Ie1, Ie2

$$\begin{align} V_{ip} &= V_{im} + V_{os} \\ \frac{V_{bg}-V_{ip}}{R_2} &= I_{e2} \\ \frac{V_{bg}-V_{im}}{R_2} &= I_{e1} \\ V_{ip} &= I_{e2}R_1 + V_T\frac{I_{e2}}{nI_S} \\ V_{im} &= V_T\frac{I_{e1}}{I_S} \end{align}$$ where $V_T = \frac{kT}{q}$

we obtain $$ I_{e1} = \frac{V_T\ln n}{R_1} + V_{os}\left(\frac{1}{R_1} + \frac{1}{R_2} \right) - \frac{1}{R_1}\cdot V_T\ln\left(1- \frac{V_{os}}{R_2I_{e1}} \right) $$

we omit the last part $$\begin{align} I_{e1} &= I_{e0} + V_{os}\left(\frac{1}{R_1} + \frac{1}{R_2} \right) \\ I_{e2} &= I_{e1} - \frac{V_{os}}{R_2} = I_{e0} + \frac{V_{os}}{R_1} \end{align}$$ where $I_{e0} = \frac{\Delta V_{be}}{R_1}$, ΔVbe = VTln n

That is, both Ie1 and Ie2 are proportional to Vos

Ie1 and Ie2 can be expressed as $$\begin{align} I_{e1} &= I_{e0} + V_{os}\left(\frac{1}{R_1} + \frac{1}{2R_2} \right) + \frac{V_{os}}{2R_2} \\ I_{e2} &= I_{e0} + V_{os}\left(\frac{1}{R_1} + \frac{1}{2R_2} \right) - \frac{V_{os}}{2R_2} \end{align}$$ i.e., $\Delta I_{e,cm} = V_{os}\left(\frac{1}{R_1} + \frac{1}{2R_2} \right)$ and $\Delta I_{e,dif} =\frac{V_{os}}{2R_2}$

bandgap output voltage is

$$\begin{align} V_{bg} &= V_T \ln \frac{I_{e1}}{I_s} + I_{e1}R_2 \\ &= V_T \ln \frac{I_{e0} + V_{os}\left(\frac{1}{R_1} + \frac{1}{R_2} \right)}{I_s} + I_{e1}R_2 \\ &= V_T \ln \frac{I_{e0} + V_{os}\left(\frac{1}{R_1} + \frac{1}{R_2} \right)}{I_s} + I_{e0}R_2 + V_{os}\frac{R_1+R_2}{R_1} \\ &= I_{e0}R_2 + V_T \ln \frac{I_{e0}}{I_s} + V_T\ln\left(1+\frac{V_{os}\left(\frac{1}{R_1} + \frac{1}{R_2} \right)}{I_{e0}} \right) + V_{os}\frac{R_1+R_2}{R_1} \\ &= V_{bg0} + V_T\ln\left(1+\frac{V_{os}\left(\frac{1}{R_1} + \frac{1}{R_2} \right)}{I_{e0}} \right) + V_{os}\frac{R_1+R_2}{R_1} \end{align}$$

Therefore, the averaged output of bandgap

$$ V_{bg,avg} = V_{bg0} +\frac{1}{2}V_T\ln\left(1-\frac{V_{os}^2\left(\frac{1}{R_1} + \frac{1}{R_2} \right)^2}{I_{e0}^2} \right) \lt V_{bg0} $$

Vbg, avg < Vbg0 due to nonlinearity of BJT

ripple cancellation

rippleCancel.drawio

phase 0:

$$\begin{align} V_{os}[n] &= V_{os}[n-1] - \frac{\Delta I_1}{g_m} \\ V_{os}[n] &= I_\Delta[n] R_E \\ \beta I_\Delta &= I_1[n] + I_2[n-1] \end{align}$$ where IΔ is the variation of Ie1 + Ie2 due to Vos and $R_E = \frac{R_1R_2}{R_1+2R_2}$

obtain $$\begin{align} \Delta I_1 &= G\cdot V_{os}[n-1] - K\cdot I_1[n-1] - K\cdot I_2[n-1] \\ I_1[n] &= G\cdot V_{os}[n-1] + (1-K)\cdot I_1[n-1] - K\cdot I_2[n-1] \\ V_{os}[n] &= K\cdot V_{os}[n-1] + R\cdot I_1[n-1] + R\cdot I_2[n-1]\\ \end{align}$$

where $G=g_m\frac{\beta}{g_m R_E + \beta}$, $R=R_E\frac{1}{g_m R_E + \beta}$ and $K=\frac{g_mR_E}{g_m R_E + \beta}$

and $$ V_{os}[n] = (2K-1)\cdot V_{os}[n-1] = (1-\frac{2\beta}{g_mR_E+\beta})\cdot V_{os}[n-1] $$

phase 1:

$$\begin{align} V_{os}[n] &= V_{os}[n-1] - \frac{-\Delta I_2}{g_m} \\ V_{os}[n] &= -I_\Delta[n] R_E \\ \beta I_\Delta &= I_1[n] + I_2[n-1] \end{align}$$

obtain $$\begin{align} \Delta I_2 &= -G\cdot V_{os}[n-1] - K\cdot I_1[n-1] - K\cdot I_2[n-1] \\ I_1[n] &= -G\cdot V_{os}[n-1] -K\cdot I_1[n-1] + (1-K)\cdot I_2[n-1] \\ V_{os}[n] &= K\cdot V_{os}[n-1] - R\cdot I_1[n-1] - R\cdot I_2[n-1]\\ \end{align}$$

similaly $$ V_{os}[n] = (1-\frac{2\beta}{g_mR_E+\beta})\cdot V_{os}[n-1] $$

That is, for either phase $$ V_{os}[n] = (1-\frac{2\beta}{g_mR_E+\beta})\cdot V_{os}[n-1] $$

reference

ECEN 607 (ESS) Bandgap Reference: Basics URL:https://people.engr.tamu.edu/s-sanchez/607%20Lect%204%20Bandgap-2009.pdf