Question
The values for $\alpha_F$ are measured for three transistors as $\alpha_F=0.97, \alpha_F=0.98$, and $\alpha_F=0.99$. Calculate the corresponding values of $\beta_F$ for each.
Step 1
The relationship is given by the formula: \[ \beta_F = \frac{\alpha_F}{1 - \alpha_F} \] Show more…
Show all steps
Your feedback will help us improve your experience
Chai Santi and 75 other educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
the Alpha value of given transistor is 0.99 then the beta value is
Consider the transistor described in Problem 12.3. (a) For a common-base current gain of $\alpha=0.9850$, determine the common-emitter current gain [note: $\beta=\alpha /(1-\alpha)]$. (b) Determine the emitter and base currents corresponding to the collector currents determined in Problem 12.3. ( $c$ ) Repeats parts ( $a$ ) and ( $b$ ) for a common-base current gain of $\alpha=0.9940$.
(a) Calculate the base transport factor, $\alpha_{T}$, for $x_{E} / L_{B}=0.01,0.10,1.0$, and $10 .$ Assuming that $\gamma$ and $\delta$ are unity, determine $\beta$ for each case. (b) Calculate the emitter injection efficiency, $\gamma$, for $N_{B} / N_{E}=0.01,0.10,1.0$, and $10 .$ Assuming that $\alpha_{T}$ and $\delta$ are unity, determine $\alpha$ for each case. (c) Considering the results of parts $(a)$ and $(b)$, what conclusions can be made concerning when the base transport factor or when the emitter injection efficiency are the limiting factors for the commonemitter current gain?
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD