Flash EPROM Programming Current An n-channel FLASH EPROM cell is programmed by hot-carrier gate current. The device parameters are VT = 0.7 V, xj = 0.2 μm, W = 100 μm, L = 0.5 μm , and (VFB + 2|Φp|) = -0.2 V. For simplicity, treat the device as a MOSFET with an extra capacitor in series with the

Question

Flash EPROM Programming Current

An n-channel FLASH EPROM cell is programmed by hot-carrier gate current. The device parameters are [latex] V_T = 0.7 \ V, x_j = 0.2 \ μm, W = 100 \ μm, L = 0.5 \ μm [/latex], and ([latex] V_{FB} + 2\left|\phi _p\right| ) = -0.2 \ V [/latex]. 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 [latex] C_{GS} [/latex] 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 [latex] V_{GS} = 10 \ V [/latex] and [latex] V_{DS} = 5 \ V [/latex].

Accepted Answer

From Equation 10.5.1, [latex] V_T [/latex] measured from the control gate is

[latex] V_T=V_{FB}+2\left|\phi _p ight| +\frac{\left|Q_d ight| }{\epsilon _{ox}} (d_1+d_2)+\frac{\left|Q_{fg} ight| }{\epsilon _{ox}}d_1=0.7 \ V [/latex]

with the values given above for [latex] V_{FB}+2\left|\phi _p ight| ,d_1,d_2, ext{ and }Q_{fg}, [/latex] we have

[latex] \frac{\left|Q_d ight| d_1}{\epsilon _{ox}} =0.45 \ V [/latex]

Therefore, the threshold voltage [latex] V_{TF} [/latex] measured from the floating gate is

[latex] V_{TF}=V_{FB}+2\left|\phi _p ight| +\frac{\left|Q_d ight|d_1 }{\epsilon _{ox}}=0.25 \ V [/latex]

For hot-electron programming, the MOSFET is biased in the saturation region. Using the model of Sec. 9.1, [latex] C_{GS} = (2/3)C_{ox} [/latex]. Assuming a simple voltage divider, we find the voltage at the floating gate to be

[latex] V_{GF}=10\frac{C_{ox}}{C_{ox}+(2/3)C_{ox}} =6 \ V [/latex]

We calculate the drain current as in Sec. 9.2:

[latex] \xi _{eff}\approx \frac{V_G-V_T}{6x_{ox}} +\frac{V_T+0.5}{3x_{ox}} \approx 1.21 imes10^6 \ V \ cm^{-1} \ \mu _{eff}\approx \frac{\mu _0}{1+(\xi _{eff}/\xi _0)^ u } =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 imes10^4 \ V \ cm^{-1} \ herefore I_{Dsat}=WC_{ox}(V_G-V_T-V_{Dsat})v_{sat}=23 \ mA [/latex]

We can also calculate [latex] \xi _m [/latex] using

[latex] \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 imes10^5 \ V \ cm^{-1} [/latex]

The charging current just after the bias is applied is found by using Equation 10.3.2:

[latex] I_G\approx C \ I_D \exp\left(-\frac{\phi _B}{\lambda \xi _m} ight) [/latex] (10.3.2)

[latex] I_G\approx CI_{Dsat}\exp\left(-\frac{\phi _B}{q\xi _m \lambda } ight) \approx 2 imes10^{-3} imes 23 \ mA \ imes \exp\left(\frac{-3.32}{2.59 imes 10^5 imes7.3 imes10^{-7}} ight) \approx 1.1 \ pA [/latex]

Question 10.4: Flash EPROM Programming Current An n-channel FLASH EPROM ce...

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