Seven segment display decoder

Aidanx1

Hi. I'm using a BCD-to-seven-segment decoder/driver with my BS2 to illuminate digits on a seven segment display (common annode). The driver is active low. I have connected pins A, B, C, D to 0, 1, 2, 3, on my BS2 and I'm using seven 1K resistors. I'm simply using LOW 0, LOW 1 etc to light the segments. I just need to know if the current is being sunk through my BS2 or out of the GND pin of the driver? I was planning on adding more displays and drivers and don't want to exceed the curret limits of my stamp.

I have trouble understanding data sheets. The data sheet for the driver is here. The part number is 'sn74ls47'.

Post Edited (Aidanx1) : 12/2/2008 12:40:09 PM GMT

Mike Green

If you look at page 9 of the datasheet, you'll see an item for Low-level Input Current. This is 1.6mA for all inputs except the blanking input where it's 4mA. This current is what the Stamp has to supply. The actual LED current is shown in the table on page 2 and is sunk through the GND pin of the driver, not the Stamp.

Note that there's a High-level Input Current specified as well, but it's only on the order of tens of microAmps.

Usually these decoders are shared among several LEDs and the Stamp has to feed only one decoder. If you use several of them, you'd need to figure in the load of each of them (1.6mA x # of decoders in parallel). A group of 8 I/O pins on the Stamp can sink about 50mA total with any one I/O pin sinking 25mA.

Post Edited (Mike Green) : 12/2/2008 5:07:20 PM GMT

Aidanx1

Thanks Mike, your very helpful. Let me check that I understand this.

So the low level current is the current that the stamp will have to sink?
and thats 1.6mA in total for each decoder?... or per pin in use?
Does this mean there is no need for any resistors between A, B, C, D and the Stamp pins?
and its called low level "input" current even though it's actually being sunk out of the decoder?


Sorry if I'm overloading your brain with questions, but what is the high level current you mentioned used for?

Mike Green

The particular decoder you're using is a TTL (transistor-transistor logic) device.· When an input is high, the device draws a few tens of microAmps.· When an input is low, the device draws about 1.6mA.· It has to do with the type of internal circuitry used.· Each input draws current.· If you apply a logic low from a Stamp·to each of 4 pins (A,B,C,D), the total amount of current drawn from your Stamp is 4 x 1.6mA = 6.4mA.· You don't need resistors between the TTL inputs and the Stamp I/O pins because the current is limited due to the construction of the device.· If you apply a mix of logic low and logic high, the current will be less because the logic high from the Stamp·has to provide much less current.

Aidanx1

Thank you very much. I understand perfectly now...back to the breadboard.