The problem I was given is:
A base-emitter voltage of 480 mV is measured on a test transistor with a 100 um x 100 um emitter area at a collector current of 10 uA.
Calculate the punch-through voltage.
Express your answer in Volts, accurate to 0.1 V
Then the hint I was given is:
Punch-through is reached when the base is completely depleted due to the reverse base-collector voltage. Under these circumstances, 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 in the collector. From the data provided, the necessary depletion-width in the collector, and hence the punch-through voltage, can be calculated, assuming the base doping is much larger than the collector doping.
Use Dn = 13 cm2/s, mobilityp = 150 cm2/V-s, Nepi = 1015 cm-3, silicon permittivity = 1.04 x 10-12 F/cm, and the collector-base built-in potential is 0.55 V.
How I went about doing the problem,
So at first after looking in all of my text books I found that with the parameters I have been given I am unsure unsure as to how to find the Length of the channel. I have found
Xdo + Xd = L
I have enough information to solve for Xdo if i assume the channel length is 100um I can solve for Xd. I feel this is wrong using the emitter area and assuming W*L = 100μm*100μm. If I can find the channel length then I can use the equation below to solve for Vds.
Vbi +Vds = ((Xd)^2*e*Na)/(2*εs) and then using Vbi = 0.55 V
The reason I think this is wrong is I get a Vds of 185.04V but every other problem I have encountered has not had a Vds over 10V
Any help would be appreciated. The other thing that I am worried about is I never used the base-emitter voltage or the collector current given in the problem.
A base-emitter voltage of 480 mV is measured on a test transistor with a 100 um x 100 um emitter area at a collector current of 10 uA.
Calculate the punch-through voltage.
Express your answer in Volts, accurate to 0.1 V
Then the hint I was given is:
Punch-through is reached when the base is completely depleted due to the reverse base-collector voltage. Under these circumstances, 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 in the collector. From the data provided, the necessary depletion-width in the collector, and hence the punch-through voltage, can be calculated, assuming the base doping is much larger than the collector doping.
Use Dn = 13 cm2/s, mobilityp = 150 cm2/V-s, Nepi = 1015 cm-3, silicon permittivity = 1.04 x 10-12 F/cm, and the collector-base built-in potential is 0.55 V.
How I went about doing the problem,
So at first after looking in all of my text books I found that with the parameters I have been given I am unsure unsure as to how to find the Length of the channel. I have found
Xdo + Xd = L
I have enough information to solve for Xdo if i assume the channel length is 100um I can solve for Xd. I feel this is wrong using the emitter area and assuming W*L = 100μm*100μm. If I can find the channel length then I can use the equation below to solve for Vds.
Vbi +Vds = ((Xd)^2*e*Na)/(2*εs) and then using Vbi = 0.55 V
The reason I think this is wrong is I get a Vds of 185.04V but every other problem I have encountered has not had a Vds over 10V
Any help would be appreciated. The other thing that I am worried about is I never used the base-emitter voltage or the collector current given in the problem.
