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I am currently reading Digital Computer Electronics, and on page 4 of the book (page 7 of the PDF), I am confused by this diagram: diagram

The diagram shows that the transistors with no (or almost no) current being supplied to the base have their outputs on.

The authors even describe the leftmost transistors as being cut off, despite showing 5V output:

The transistors on the left are cut off because the input base voltages are 0 V. The dark shading symbolizes the cutoff condition. The two transistors on the right have base drives of 5 V. The transistors operate at either saturation or cutoff. A base voltage of 0 V forces each transistor to cut off, while a base voltage of 5 V drives it into saturation.

I think this may be a difference between NPN and PNP transistors, but I'm not totally sure because of the way the authors described the left ones as being cut off despite showing 5V.

Any help would be appreciated. Thanks!

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    \$\begingroup\$ The transistor being "off" means it is not conducting any current from collector to emitter. So no current flows through the 1K pullup. So the voltage across the pullup is zero. So the voltage on both sides of the resistor is 5V. This is an "inverting" circuit, so a low input on the transistor yields a high on its output (even though the transistor is off). \$\endgroup\$ – td127 Jan 30 at 18:00
  • \$\begingroup\$ @td127 So if I understand correctly, the transistor is off as you described it, however there is no current actually flowing because the circuit isn't complete, and the potential difference across the transistor from collector to emitter is 5V from 5 to 0? \$\endgroup\$ – applemonkey496 Jan 30 at 18:03
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    \$\begingroup\$ Yes. Because the base-emitter voltage is zero the transistor is off and effectively out of the circuit. What you're left with is a 1K resistor dangling from a 5V source. A voltmeter from collector to emitter (ground) would show 5V. The transistors on the right are 'on', the collector and emitter become essentially shorted - a voltmeter would read about 0.2V (the collector-emitter saturation voltage). Their 1K resistors now see nearly 5V across them so 5/1K = 5mA flows through the resistor, into the collector, and out the emitter to ground. \$\endgroup\$ – td127 Jan 30 at 18:27
  • \$\begingroup\$ have their outputs on ... transistors do not have outputs per se ... the circuit that contains the transistor has an output \$\endgroup\$ – jsotola Jan 30 at 20:57
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The transistors in the diagram are all NPN.

The two greyed out transistors are effectively switched off. But the outputs are still connected to the 5V line through the 1k resistors. So the outputs are pulled up by the resistors to 5V.

The two white transistors are switched on. This pulls the output down to (approximately) 0V. The important thing is that the transistor is between the output and ground (0V). So turning the transistor on connects the output to ground, not to 5V.

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