# Current gain transistor

i wonder if any one could tell me ,why does the current gain change as the frequency of ac supply changes?

Base current consists of conduction & displacement components. At very low frequencies, the conduction component dominates, as the displacement current is very small. As freq increases, the displacement component of base current increases while the conduction component does not. Eventually the base current consists of more displacement than conduction.

So, as freq increases, more base current is needed for a specific value of collector current. The current gain, hfe, decreases as freq increases. The dc/lf value of current gain is denoted "hFE", an "h" with upper case "FE" subscript. The ac/hf current gain is denoted "hfe". Note that the "fe" subscripts are lower case.

At a freq known as the "transition freq", denoted "ft", the current gain is 1. At a freq of ft, to establish 10 mA of collector current requires 10 mA of base current. Above this ft value, the device is not providing current amplification, so ft is the useful bandwidth of the device. At half of ft, hfe = 2, at ft/10 freq, hfe = 10, & so on. at freq low enough, hfs reaches its maximum value of hFE. In other words, hfe < hFE.

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Claude

Base current consists of conduction & displacement components. At very low frequencies, the conduction component dominates, as the displacement current is very small. As freq increases, the displacement component of base current increases while the conduction component does not. Eventually the base current consists of more displacement than conduction.

So, as freq increases, more base current is needed for a specific value of collector current. The current gain, hfe, decreases as freq increases. The dc/lf value of current gain is denoted "hFE", an "h" with upper case "FE" subscript. The ac/hf current gain is denoted "hfe". Note that the "fe" subscripts are lower case.

At a freq known as the "transition freq", denoted "ft", the current gain is 1. At a freq of ft, to establish 10 mA of collector current requires 10 mA of base current. Above this ft value, the device is not providing current amplification, so ft is the useful bandwidth of the device. At half of ft, hfe = 2, at ft/10 freq, hfe = 10, & so on. at freq low enough, hfs reaches its maximum value of hFE. In other words, hfe < hFE.

Did this help?

Claude
what is current and displacement component?what does it mean?

uart