The datasheet of the Texas Instruments SN54LS47 says that this part is for use with common anode 7-segment displays on pg 3 para 1 of the datasheet.

Common anode dispay is where all the segments are collectively connected to VCC and we sink current from the one we wish to turn on. The SN54LS47 has open collector outputs to do this.

The 7-segment display being used is the TDSG5150. It is a common anode display. It has forward current of 10 mA as per its datasheet. It has typical forward voltage of 2.4V.

Now if all the segments are on in the display that would be 100mA. The confusion is that the pg 11 of the SN54LS47 gives a "peak output current" of 200mA subject to tW <= 1 ms and duty cycle being 10%.

Q: What does tW mean?

Q: This is a common collector output part, what does it not say "peak input current" instead?

Q: How do I know if this part can drive the 7-segment display continuosly with all displays on and not overheat and be destroyed?

Q: The front page of datasheet says "open collector output" for the SN54LS47. However, the pg 3 says that it features active low output. What does this mean?


2 Answers 2


It's common to have several of these displays connected to the same driver and switching between the displays with the common anode; i.e. each of the A-G segments are bussed to the 7447 and a transistor is connected to the anodes of each display. Relying on persistence of vision, the segments are quickly (~kHz) turned on one at a time, then the next, and so on. This way fewer components are needed and the current required may be reduced, albeit with lower brightness.

A. tW probably means the period of these switching cycles. At 200mA peak current, the switching period must be <=1ms (i.e. f>=1kHz) at <=10% duty means the current can be pulsed at 200mA for a maximum of 100us.

A. Potatoes, potatoes. Physically it's outputing electrons, but yes, they're a bit inconsistent with terminology in this datasheet.

A. On page 9, under 'recommended operating conditions' it lists a maximum 40mA per segment ('On-state output current, IO(on)'). You'll be fine with 10mA per segment.

A. Active-low means that the desired output will present a low value -- here it's a ground signal to sink current. If it were active high, such as with the CD4511 for common-cathode displays, the desired output would be a high value (+Vcc).

  • \$\begingroup\$ I found in a TI databook that \$t_W\$ measures a pulse width, between the leading and trailing edges. So the requirement is that you can drive the display at peak current for no more than 1ms, with a 10% duty cycle. (e.g. 1ms on, 9ms off) \$\endgroup\$ Jan 5, 2017 at 2:56
  • \$\begingroup\$ Ken, Could you describe what you mean by leading and trailing edges? I do not understand this term from the datasheet \$\endgroup\$
    – quantum231
    Jan 5, 2017 at 8:39

TI data sheet specs the seven segment driver outputs are active low. The spec is pulling 12 mA to ground, with a measured 0.24v (typical) and 0.4v (maximum). Thus 4.8 mW would be dissipated inside the chip for each segment turned on, worst case.

All seven segments turned on would dissipate 33.6 mW in addition to the static power for the rest of the chip. Static power is 13 mA x 5v = 65 mW worst case. Total chip power is 98.6 mW
Thermal resistivity is not given in that data sheet, but for epoxy DIP packages commonly used to house TTL logic, temperature rise of the chip is roughly 100 degrees C per watt. So 100mW will cause your chip to rise in temperature about 10 degrees C (worst case) for every segment pulling 12 mA. No sweat.
Where you might run into trouble, is where you multiplex many display digits, using one chip. Since each digit is ON for a fraction of the multiplex cycle, the single 74LS47 must sink more current to get each digit displaying brightly.

The 1 millisecond spec for tW likely refers to thermal time constant of the chip heating up.

  • \$\begingroup\$ Provided only one digit is on at a time, why should that result in the 74LS47 sinking more current than when not multiplexing? \$\endgroup\$
    – quantum231
    Jan 5, 2017 at 8:43
  • 1
    \$\begingroup\$ Suppose you MUX two digits using one 74LS47. First digit is ON while second digit is OFF. Then the situation reverses. 74LS47 is conducting current for both situations. But each digit is ON for half the time, so it is less bright. So current must be increased (doubled) to make digits appear brighter. To do so, the current-setting resistors in series with LED diodes are reduced in value. So 74LS47 must work harder. See calcium3000 first paragraph. \$\endgroup\$
    – glen_geek
    Jan 5, 2017 at 14:32

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