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advanced thanks.

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- Thread starter amaresh92
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- #1

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advanced thanks.

- #2

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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

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

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

- #4

uart

Science Advisor

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Think of the displacement current as bypassing the actual base-emitter junction by flowing instead through a parasitic parallel capacitance. This may be a bit oversimplified but for your purposes I think this is the best way to understand it.what is current and displacement component?what does it mean?

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