Semiconductor Devices Overview
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New answer posted
a month agoContributor-Level 9
Power gain = (i_c² R_c) / (i_b² R_B) = (i_c/i_b)² (R_c/R_B) = (10²)² (10? /10³) = 10?
i_c/i_b = 100 ⇒ β = i_c/i_b = 100
New answer posted
a month agoContributor-Level 10
Photodiode operate in reverse bias. The photocurrent increases initially and saturates fi
New answer posted
a month agoContributor-Level 9
First part of figure shown is or GATE and Second part of figure shown is NOT GATE So, NOR GATE
New answer posted
a month agoContributor-Level 9
Input are :
(0, 0) ; (0, 1); (1, 0); (1, 1).
Thus, the output y is : (1, 0) s
A | B | P | Q | Y |
0 | 0 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 1 |
1 | 0 | 0 | 1 | 1 |
1 | 1 | 1 | 1 | 0 |
New answer posted
2 months agoContributor-Level 10
When both A and B have logical value 'l' both diode are reverse bias and current will flow in resistor hence output will be 5 volt i.e. logical value '1'.
In all other case conduction will take place hence output will be zero value i.e. logical value 'o'.
So truth table is
A B Y
0 0
0 1 0 (AND gate)
1  
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