Question 10.4: Flash EPROM Programming Current An n-channel FLASH EPROM ce...
Flash EPROM Programming Current
An n-channel FLASH EPROM cell is programmed by hot-carrier gate current. The device parameters are V_T = 0.7 \ V, x_j = 0.2 \ μm, W = 100 \ μm, L = 0.5 \ μm , and ( V_{FB} + 2\left|\phi _p\right| ) = -0.2 \ V . For simplicity, treat the device as a MOSFET with an extra capacitor in series with the normal gate capacitor. That is, assume that the floating-gate voltage is given by the voltage divider composed of C_{GS} and the control-to-floating-gate capacitor. Assume that the floating gate does not carry any charge initially and estimate the programming current immediately after the MOSFET is biased by applied voltages V_{GS} = 10 \ V and V_{DS} = 5 \ V .
From Equation 10.5.1, V_T measured from the control gate is
V_T=V_{FB}+2\left|\phi _p\right| +\frac{\left|Q_d\right| }{\epsilon _{ox}} (d_1+d_2)+\frac{\left|Q_{fg}\right| }{\epsilon _{ox}}d_1=0.7 \ Vwith the values given above for V_{FB}+2\left|\phi _p\right| ,d_1,d_2, \text{ and }Q_{fg}, we have
\frac{\left|Q_d\right| d_1}{\epsilon _{ox}} =0.45 \ VTherefore, the threshold voltage V_{TF} measured from the floating gate is
V_{TF}=V_{FB}+2\left|\phi _p\right| +\frac{\left|Q_d\right|d_1 }{\epsilon _{ox}}=0.25 \ VFor hot-electron programming, the MOSFET is biased in the saturation region. Using the model of Sec. 9.1, C_{GS} = (2/3)C_{ox} . Assuming a simple voltage divider, we find the voltage at the floating gate to be
V_{GF}=10\frac{C_{ox}}{C_{ox}+(2/3)C_{ox}} =6 \ VWe calculate the drain current as in Sec. 9.2:
\xi _{eff}\approx \frac{V_G-V_T}{6x_{ox}} +\frac{V_T+0.5}{3x_{ox}} \approx 1.21\times10^6 \ V \ cm^{-1} \\ \mu _{eff}\approx \frac{\mu _0}{1+(\xi _{eff}/\xi _0)^\nu } =187 \ cm^2V^{-1}s^{-1} \\ V_{Dsat}=\frac{(V_G-V_T)\xi _{sat}L}{(V_G-V_T)+\xi _{sat}L} =2.45 \ V \\ \xi _{sat}=\frac{2v_{sat}}{\mu _{eff}} =8.55 \times10^4 \ V \ cm^{-1} \\ \therefore I_{Dsat}=WC_{ox}(V_G-V_T-V_{Dsat})v_{sat}=23 \ mAWe can also calculate \xi _m using
\ell \approx 0.22 x_{ox}^{1/3}x_j^{1/2}\approx 0.0984 \ \mu m \\ \xi _m \approx \frac{V_D-V_{Dsat}}{\ell } \approx 2.59\times10^5 \ V \ cm^{-1}The charging current just after the bias is applied is found by using Equation 10.3.2:
I_G\approx C \ I_D \exp\left(-\frac{\phi _B}{\lambda \xi _m} \right) (10.3.2)
I_G\approx CI_{Dsat}\exp\left(-\frac{\phi _B}{q\xi _m \lambda } \right) \approx 2\times10^{-3}\times 23 \ mA \ \times \exp\left(\frac{-3.32}{2.59 \times 10^5 \times7.3 \times10^{-7}} \right) \approx 1.1 \ pA