Please can someone suggest to me how to work out the AC Forward current transfer ratio for this BJT Transistor (ZTX650 BJT Transistor) used in common emitter. I looked on the data sheet (herewith is the link http://www.mouser.com/ds/2/115/ztx650-186634.pdf) but it does not mention this. Can someone please give me just a hint/suggestion to work this out? Is it simply hf = ic/ib (i always thought this was for DC beta(hFE))?
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This might help: -
So, look up what the transition frequency is and imagine a line falling at 6dB/octave crossing the point marked \$f_T\$. The horizontal/flat line (at lower frequencies) is the "DC" hFE and, where the 6dB/octave line crosses the horizontal line, that is the place where hFE starts to decrease with frequency.
Basically this means that you can assume hFE applies to AC up to the point where the 6dB/octave line intercepts it.
The typical \$f_T\$ for the ZTX650 is 175 MHz and if hFE is (say) 128 at DC, 7 octaves below 175 MHz is where the two lines intercept i.e. ~1.4 MHz: -
answered Apr 24, 2016 at 14:21
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\$\begingroup\$ I have to simulate this at 1kHz, so does hFE still apply to AC \$\endgroup\$– SarangaApr 24, 2016 at 14:44
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\$\begingroup\$ im sorry i dont quite understand the "im line falling at 6dB/octave". The signal i have been specified to use has 200hz to 10kHz frequency range so can you please tell me does this still apply? \$\endgroup\$– SarangaApr 24, 2016 at 14:47
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\$\begingroup\$ The 6dB per octave line falls from about 1.4Mz and crosses the baseline at fT in other words, the hFE will be flat from DC to nearly 1 MHz with the -3dB point being at ~1.4 MHz. \$\endgroup\$– Andy akaApr 24, 2016 at 15:38
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