A symmetric GaAs/AlAs RTD has a barrier width of 1.5 nm and a well width of...

A symmetric GaAs/AlAs RTD has a barrier width of 1.5 nm and a well width of 3.39 nm. When this RTD is inserted in the base of an HBT with an emitter flux centered on the first excited level of the RTD. Find the cutoff frequency of the HBT with the inserted RTD, if the original f_T was 100GHz.

[Hint:The transit time across the RT structure is given by (d/v_G) + (2h/Γ)where d is the width of the RT structure, v_G is the electron group velocity (l0⁷ cm/s ) , and Γ is the resonant width (20 meV）.]

Expert Solution

The GaAs/AlAs RTD introduced into the HBT acts as a tunneling barrier for the transit of the electrons within the transistor, aiding the production and release of electrons into the system.

Given the formula to calculate the change in the timing is given -

Given the values,
d can be considered as the maximum width that needs to be crossed or the diagonal value of the barrier width and well width, or , which turns out to be 3.7 nm
h = Planck's constant, 4.13 x 10^-15 eV s, v_G = 1.07 x 10⁹ nm/s, = 20 meV

Plugging in all the values -

The frequency for the barrier is the inverse of the transit time calculated, i.e. 0.29 GHz.

This system is analogous to a parallel circuit connection, and the GaAs/AlAs RTD acts as a shunt, this is demonstrated by the parallel inclusion of the transit time calculations as, the frequencies are calculated as parallel -

Plugging in the values, initial frequency ( as 100 GHz) and the barrier frequency as 0.29 GHz, and taking the reciprocal to find the final cut-off frequency, we get it as 0.28 GHz.

This states the improvement in the HBT.

queen_honey_blossom answered 2 years ago

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