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In this scenario, let's say that XMM2 has a reading of 0 A and XMM3, being a voltmeter, has a reading of ten volts.

I presume that XMM2 reads the base current, so since there is no IB, there will also be no IC by the relation: $$I_{C}=\beta I_{B}$$ It is an npn transistor, so it is going to amplify. Depending on the supply voltage VBE, the transistor is going to work in three states: active, cutoff, and saturation.

If VBE is less than 0.7, then the base-emitter is not configured properly, which can then be seen probably via the input characteristics. Hence, the depletion region is not going to reduce as the electrons hardly move to the base. In other words, no current is flowing. Thus, since IC does not collect any electron from the base current, IC is 0.

And since VCC, VCE, and Vout are at max (10 volts) and IC = 0, the state of the transistor is at cuttoff. It almost looks like an open circuit; no emitter current, and the base voltage must be less than both the emitter and collector voltages.

Is my interpretation correct?

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  • \$\begingroup\$ OK until "... and the base voltage must be less than both the emitter and collector voltages." The emitter is at zero. Your circuit can't drive the base below zero. \$\endgroup\$
    – Transistor
    Feb 6, 2021 at 9:15

1 Answer 1

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Your interpretation is correct, but with 1V applied to the base, the current is unlikely to be zero unless Rb is infinite.

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  • \$\begingroup\$ If RV is 1M0 then In will be 0.3uA or so. With an Hfe of 100 the collector current will be 30uA. If Rc is 10k then XMM3 will read 9.7V. Spoiler alert: if the ratio of the input to output resistors is the same as the Hfe value then the voltages dropped across both will be the same. \$\endgroup\$
    – Frog
    Feb 6, 2021 at 9:52

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