A base-emitter voltage of 480 mV is measured on a super- test transistor with a 100 m

A base-emitter voltage of 480 mV is measured on a super-β test transistor with a 100 μm × 100 μm emitter area at a collector current of 10 μA. Calculate the Q_B and the sheet resistance of the base region. Estimate the punch-through voltage in the following way. When the base depletion region includes the entire base, charge neutrality requires that the number of ionized acceptors in the depletion region in the base be equal to the number of ionized donors in the depletion region on the collector side of the base. Therefore, when enough voltage is applied that the depletion region in the base region includes the whole base, the depletion region in the collector must include a number of ionized atoms equal to Q_B. Since the density of these atoms is known (equal to N_D), the width of the depletion layer in the collector region at punch-through can be determined. If we assume that the doping in the base N_A is much larger than that in the collector N_D, then (1.15) can be used to find the voltage that will result in this depletion layer width.

Repeat this problem for the standard device, assuming a VBE measured at 560 mV. Assume an electron diffusivity D̅_n of 13 cm²/s, and a hole mobility μ̅_p of 150 cm2/V-s. Assume the epi doping is 1015 cm⁻³. Use ε = 1.04 × 10⁻¹²F/cm for the permittivity of silicon. Also, assume ψ_o for the collector-base junction is 0.55 V.