# Possible to use 7447 to drive common-cathode 7 segment display?

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?

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. 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). simulate this circuit

• 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. Jun 25 '14 at 2:20
• If we know that the transistor is in saturation then this becomes trivial. But how can we be sure of this? Jun 25 '14 at 2:24
• @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. Jun 25 '14 at 2:25
• @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. Jun 25 '14 at 2:27
• 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? Jun 25 '14 at 2:28

A series of seven 74LS04s on the output.