Task №3 Transistors.

3.1 Compare the maximum possible thickness of the base of a bipolar transistor and the gate length of a field-effect transistor when operating at a frequency . Estimate the angle of the span in both cases. Link it to the solution of the task №3 from the 1st assignment.

3.2 Compare the advantages and disadvantages of using HEMT devices and transistors with ballistic transport in the microwave range. What should be the thickness of the high-alloyed area of the HEMT with , if the contact potential difference is . How far can an electron move from the equilibrium position in this layer at Т=300 K?

3.3 Justify the trends in the use of materials such as GaN, InP, SiC, diamond C in modern transistors using the concepts: band gap width, low-field mobility, maximum drift velocity, crystal lattice constant.

3.4 Draw (qualitatively) the input and output VAC of three Schottky Barrier Gate Field Effect Transistors (SBGFET) with the same size, doping level, but made of Si, GaN, GaAs. Justify the dependencies. How will the characteristics change if the gate width is increased?

3.5 Draw and justify a family of input and output VAC and noise factor on one graph. Explain why SBGFET, despite the high electronic temperature of the media at the output, are classified as low-noise devices?

When analyzing, use the solution of the task №6 from the first assignment.

3.6 How are low-frequency noises related to the transistor manufacturing technology?

3.7 Draw a low-signal equivalent scheme of SBGFET. How is such a scheme better or worse than S-parameters?

Given:

f₀ = 8 GHz

N_d = 8,15 10¹⁷ cm^-3 = 8,15 10²³ m^-3

= 0,63 V

T = 300 K

Solving:

1. Compare the maximum possible thickness of the base of a bipolar transistor and the gate length of a field-effect transistor. Estimation of the span angle in both cases

1.1 Calculation of the thickness of the base of a bipolar transistor

Since it is necessary to calculate the maximum possible thickness of the base of a bipolar transistor, we will use the formula for the limiting frequency [22]:

f₀ = (20)

At the same time, (total delay time) consists of the following components:

= + + + (21)

where –charging time of the barrier capacity of the emitter junction; –diffusion time of carriers through the base; – delay time in the collector junction associated with the time of flight; – charging time of the collector junction capacity [22].

It is worth considering that the main delay is the time of diffusion of carriers through the base. Given this, we rewrite the formula (20):

f₀ = (22)

At the same time, the time of diffusion of carriers through the base is determined by the following formula [22]:

= (23)

where D – diffusion coefficient.

Hence, from formula (23) we obtain expressions for calculating the thickness of the base of a bipolar transistor:

W_b = (24)

Then, using formula (22), we write down the expression for the diffusion time of carriers through the base:

= (25)

Calculate by the formula (25):

= = = 1,989 10^-11 s

The electron diffusion coefficient in GaAs is assumed to be equal to [23]:

D = 0,01 m²/s

Then, using the formula (24), we calculate the maximum possible thickness of the base of a bipolar transistor:

W_b = = = 0,446 10^-6 m = 0,446 µm