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I know that the 74(LS)47 is designed to drive common-anode displays. However, is it possible to make it drive a common-cathode display? What is the simplest (in terms of components) way of doing this?

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Yes, it's possible. One method would be to use two extra resistors (plus the usual per-segment resistor) and one PNP transistor per output.

schematic

simulate this circuit – Schematic created using CircuitLab

You can buy these three parts integrated into so-called "digital transistors" so it would take only 7 additional parts.

Edit: Re comments by sherrellbc, here is a version for a high-voltage LED digit (something like a 3" digit with multiple LED die per segment) that operates the transistor in the linear region. It provides a constant current of about 15mA for any LED voltage from 0 to more than 10V (it will get warm at 0V).

When it is 'on', the base is at 12V - 2.2V, so the emitter is at about 1.5V below +12 and therefore the collector current will be close to 15mA (since emitter current ~= collector current).

schematic

simulate this circuit

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  • \$\begingroup\$ For whatever reason I tend to struggle on the calculations associated with anything other than a basic transistor (without feedback) operated in saturation. So, I can easily see the current through the LED is going to be (5-Vec-Vd)/R2, but Vce is a function of the operating point of the transistor. To get the collector current we have to know the base current, which is (sunk into the 7447) (5 - Veb - Vec)/R3 -> (4.3 - Vec)/R3. But the fact that Vec is a function of the operating point and R1 is present, I tend to not get these correct. \$\endgroup\$
    – sherrellbc
    Jun 25 '14 at 2:20
  • \$\begingroup\$ If we know that the transistor is in saturation then this becomes trivial. But how can we be sure of this? \$\endgroup\$
    – sherrellbc
    Jun 25 '14 at 2:24
  • \$\begingroup\$ @sherrellbc You can assume that the transistor is saturated, so Vec = ~0.1V, and the current is determined mostly by the LED forward voltage drop and the supply voltage- I = (4.9-Vf)/240 ~= 10mA for a 2.4V Vf. \$\endgroup\$ Jun 25 '14 at 2:25
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    \$\begingroup\$ @sherrellbc We can be sure of this because it's designed so that Ib >> Ic/hfe. In this case, I made Ib = 4.3mA, which is enough for a collector current of 50-100mA (way more than necessary). It would work just as well with a 4.7K resistor. \$\endgroup\$ Jun 25 '14 at 2:27
  • \$\begingroup\$ So is it that all transistors acting as switching operate in the saturation region? Is the linear region of a transistor used seldom, if at all? I think it's for transistor amplifier circuits, right? Also, I've heard it both ways - does the current through an LED respond to the voltage across it (seems logical), or does the voltage drop across the device develop due to the current through it? \$\endgroup\$
    – sherrellbc
    Jun 25 '14 at 2:28
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A series of seven 74LS04s on the output.

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