Does Increasing Hole Density in an NPN Transistor Base Decrease Beta?

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Increasing hole density in the base of an NPN transistor leads to a decrease in beta (β). While a higher concentration of holes may suggest an increase in collector current (IC) due to more electrons passing from the emitter, it also results in a proportional increase in base current (IB). This balance means that the ratio β = IC/IB ultimately decreases. The underlying physics indicates that increased hole density can enhance recombination, further reducing β. Therefore, more hole density in an NPN base negatively impacts its current gain.
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If we make an NPN transistor and change the base doping level so that we get a higher consentration of holes will β increase or decrease? Why?

I know β=IC/IB. Will more holes in the base make more electrons pass from the emitter to the collector increasing IC and thereby β? Or will IB increase propertionally so that β remains the same? What do you think?
 
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More hole density in an npn base results in a decrease in beta.

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