In a UJT relaxation oscillator, the value of RBB is 5 kΩ and stand off ratio n = 0.6 at I= 0. The value of RB2 is: 

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  1. 2.4 ΚΩ
  2. 1.1 ΚΩ
  3. 2.0 ΚΩ
  4. 3.0 ΚΩ

Answer (Detailed Solution Below)

Option 3 : 2.0 ΚΩ
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Detailed Solution

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Concept:

Intrinsic stand-off ratio (η): It is the ratio of RB1 to the sum of RB1 and RB2. It can be expressed as 

\(η = \frac{{RB1}}{{RB1 + RB2}} = \frac{{RB1}}{{R_{BB}}}\)

The typical range of the intrinsic standoff ratio is from 0.4 to 0.8.

RBB: It is the resistance between terminals B1 and B2. In simple words, it is the resistance of the N-Type bar when measured lengthwise. If RB1 is the resistance of the bar from E to B1 and RB2 is the resistance of the bar from E to B2, then RBBO can be expressed as 

RBB= RB1 + RB2.

The typical range of RBB is from 3 kΩ to 10 kΩ.

UJT pin diagram:

F2 Ashiq 26.9.20 Pallavi D3

UJT circuit diagram:

F2 Ashiq 26.9.20 Pallavi D4

Calculation:

Given that, RBB = 5 kΩ and  \(\eta\)=0.6

∴ intrinsic stand-off ratio (η) can be calculated as

\( \eta = \frac{{RB_1}}{{\left( {RB_1+RB_2} \right)k}} = \frac{{RB_1}}{{\left( {RBB} \right)k}}\)     ..................(1)

\(0.6=\frac{RB_1}{5}\)

RB1 = 3 

Putting in equation 1, we get

RB2  = 2 

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