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In this image When calculating the fan out of the TTL circuit the current in the TTL drive output is 16mA and is divided into the emitters of the load each emitter has 1.6mA because the fan out is 10. My question is why he makes the current in the base of the load is also equal to 1.6mA, should it be equal to 16uA because we divided 1.6mA of the load emitter by beta if we assume that beta is 100.

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    \$\begingroup\$ There is no current through the collector of the transistor so all of the base current flows through the emitter. \$\endgroup\$
    – Kevin White
    Commented May 12, 2021 at 0:03
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    \$\begingroup\$ When the emitter is pulled down close to ground, so also is the collector. In fact, that's the whole point, as the collector pulls down hard on the base of another transistor, not shown, but without any collector current associated with it except for some leakage. It's a saturated BJT with negligible collector current and \$\beta\ll 1\$. So the base current is the emitter current in this case. See some discussion here. \$\endgroup\$
    – jonk
    Commented May 12, 2021 at 1:42

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