I am experimenting with driving two common cathode 7 segment leds (sc52-11ewa) using a 74ls595 and two 74ls48.
Each output of the 595 is 4.7v when high but when it comes out of the segment drivers of the 74ls48 it is only 2.1v so when I put even a small resistor , i.e. 100ohms, in series with the 7 segment leds the led are very dim.
The 74ls48's are new, I purchased 10 and they are all the same so I suspect they are fine.
Has anyone any experience of this and any idea as to what is happening?
I am using a pro mini and a usb to serial converter which is providing the power for the chips and leds. If I just connect one led and one 74ls48 the values are the same.
Thanks.
after opening this thread I have continued to look for an answer and wonder if using a 2003 would solve the problem? is this possible?
Why is the 74ls48 dropping the voltage so badly?
The 74ls48 has a very limited sink capability IOL in the data sheet.
The 2003 would work but they invert the logic levels.
that would just mean altering the program or could I put an invertor like a 74ls04 (only hex)?
Either/Or would work.
A 74ls04 has only 6 inverters.
Why not just You could use the 74ls595 directly, Or with 2003 or ULN2803
only 8 outputs from a 595! 14 needed for two seven segment leds.
I suppose I could multiplex the leds and have them switch on each from the eighth output? I might try that. I am only playing on my bench and getting programming practice.
Is the problem just due to the small sink capability of the 74ls48?
74ls04 (only hex)
A 74ls04 has only 6 inverters.
Yes two 595s would work, keep in mind the IOL is QH 16, Q 24ma.
I just wanted to say 6 inverters < 7 segments.
Some think hex is eight for the chip.
matelot:
I am experimenting with driving two common cathode 7 segment leds (sc52-11ewa) using a 74ls595 and two 74ls48.
Each output of the 595 is 4.7v when high but when it comes out of the segment drivers of the 74ls48 it is only 2.1v so when I put even a small resistor , i.e. 100ohms, in series with the 7 segment leds the led are very dim.
The 74ls48's are new, I purchased 10 and they are all the same so I suspect they are fine.
Has anyone any experience of this and any idea as to what is happening?
I am using a pro mini and a usb to serial converter which is providing the power for the chips and leds. If I just connect one led and one 74ls48 the values are the same.
Thanks.
LarryD is right in that the device alone doesn't have enough drive current, but I wanted to address what you are seeing in terms of the voltage drop. The voltage drop is normal and it is equal to the forward voltage of the LED--this is the amount of voltage the LED consumes in generating the light. In this case, the voltage presented to the LED, referenced to ground, is 4.7VDC. This means that the LED and the resistor must consume 4.7VDC since the algebraic sum of the voltages around a closed loop is zero--see Kirchoff's Voltage Law. The LED is consuming 2.1VDC, in your case, and so the resistor must consume 2.6VDC. I like to use the simple calculator at the following link in order to find out the proper value for the LED resistor. http://led.linear1.org/1led.wiz In your case, the LED forward current is 20mA--the maximum is 30mA, but 20mA is used for test conditions so I like to use that. The forward voltage is 2.1VDC and the supply voltage is 4.7VDC. According to this calculator, you should use a 150-ohm, 1/8-watt resistor--assuming that the driver could supply such current, which we know that it can't. Alternatively, you could take the 6mA that the circuit can supply and punch that into the calculator to see what sized resistor to use at this current level--it's 470-ohm, 1/8-watt. When LEDs were first being developed, it was rare to be able to operate the LED under the specified forward current rating. As the materials that made up the LED became more refined, and pure, it became possible to operate a LED at a lower than optimal forward currents. I've worked with Kingbright LEDs in some of my circuits and I have found that they are really great at operating over a range of current levels. That said, they do need more current to become brighter. Once you figure out which method you wish to go with so that more current will be supplied to the LED, you will find that the voltage drop is still there even as your LED is bright. Magic.
Laser-Lance.
matelot:
The 74ls48's are new, I purchased 10 and they are all the same so I suspect they are fine.
They may be new, but they are old - obsolete - the "LS" series should in most cases, not even be used for replacement of the same IC in old equipment. Sold essentially as "disposals".
In my (very) limited experience, I deliberately replace them with HCMOS versions to improve performance and in particular, reduce heat build-up and improve reliability.
Paul__B:
In my (very) limited experience, I deliberately replace them with HCMOS versions to improve performance and in particular, reduce heat build-up and improve reliability.
You might replacement for that reasion but only one out of those three are correct.