Old 7 segment LED needs more than 5 volts. How can this be done?

Thread Starter

Qsilverrdc

Joined Aug 3, 2014
66
Hello,
I found about 20 of these 2.75" tall clock type 7 segment displays in a very old box - early 90's era. I performed some basic testing to determine pin outs and built into a circuit. Unfortunately the display is very dull. Please lecture me here about not doing my homework properly, lol. So I actually did more testing. Here is the set up and results. My digital meter is broken, so this was off an analog meter. Vs was from various bateries.
2020-11-30 11.38.13.png
So from what I can tell, each segment needs about 5.6 volts. Each segment contains 3 red leds. So back to my circuit driven with a MM74HC4543, which works well, except, well it's dull.
2020-11-30 11.39.12.png
So, I tried removing the current limiting resistors. At 5.4 VCC it is minimally acceptable in brightness. Ok, so it's acceptable in a dark room. My question is what would happen if I put 8 Volts or so to the display supply, but keep rest of circuit at 5 Volts?
The 4543 is sinking when a segment is on (phase pin high). I just don't know about input gates on chips and what 8 volts when chip is at 5 volts will do? Also could 8 volts somehow corrupt or leak to the 5v and fry everything else?
Thanks for your help and time.
Rich
PS: http://pdf.datasheetcatalog.com/datasheet/nationalsemiconductor/DS005128.PDF
 
Last edited:

crutschow

Joined Mar 14, 2008
27,464
The 4543 maximum supply and output voltage is 7V, so you can't operate the LEDs at more than that.
To operate the LEDs at a higher voltage/current, you could add add a transistor buffer on each output.
You can buy transistor arrays to minimize package count.
 

ericgibbs

Joined Jan 29, 2010
13,274
hi Q,
The sinking current of the 4543 is very low.
Try this, disconnect the 4543
With the resistors connected and 5V on the Common Anode pin, connect the free end of the 330 ohm to 0V, how bright is is the LED.?
E
 

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Thread Starter

Qsilverrdc

Joined Aug 3, 2014
66
hi Q,
The sinking current of the 4543 is very low.
Try this, disconnect the 4543
With the resistors connected and 5V on the Common Anode pin, connect the free end of the 330 ohm to 0V, how bright is is the LED.?
E
Hi, at 5 VDC and no resistance, is barely on. At 5.4 no resist, is visible and almost ok. I might be misreading the data sheet, but the per pin sink is +/- 50ma.
 

ericgibbs

Joined Jan 29, 2010
13,274
hi,
With an LED, you must have a series resistor to the supply else you risk damaging the LED.

I assume the LED's are RED.???

E

BTW: THe 50mA is the IC's absolute current, don't go there
 

Thread Starter

Qsilverrdc

Joined Aug 3, 2014
66
The 4543 maximum supply and output voltage is 7V, so you can't operate the LEDs at more than that.
To operate the LEDs at a higher voltage/current, you could add add a transistor buffer on each output.
You can buy transistor arrays to minimize package count.
So when the chips PI phase input is high, the segment pins sink when on. If like a switch when low (on) the voltage at pin would be at zero. When a switch is off [pin high] the voltage at the pin would be 8, but no current. If the gate, pin, conducts wouldn't that be below 5 volts?
 

ericgibbs

Joined Jan 29, 2010
13,274
hi Q,
Checking your original test voltages, I would say the LED is designed to work from 12V.
3 off RED LED's per segment at 1.8V each in series is 5.4V , at 20mA current via a 330R.
Thats 330R * 0.2A = 6.6V,,, total ~12Vdc.
E
 

Thread Starter

Qsilverrdc

Joined Aug 3, 2014
66
hi Q,
Checking your original test voltages, I would say the LED is designed to work from 12V.
3 off RED LED's per segment at 1.8V each in series is 5.4V , at 20mA current via a 330R.
Thats 330R * 0.2A = 6.6V,,, total ~12Vdc.
E
That could explain a lot. When I was testing the first board, suddenly the display became very bright. Turns out that my bread board voltage regulator shorted on during my tests. Was from a 12 volt transformer. Nothing, including the 4543's burned out.
 

ericgibbs

Joined Jan 29, 2010
13,274
Hi Q,
The 90's LED's where not super bright, usually required about 20mA for a decent brightness.
E

BTW: the number '8' will have all 7 segs lit, 7*20mA= 140mA, thats three times the abs value for the 4543.!!!
 

dl324

Joined Mar 30, 2015
12,929
I looked at that chip. It's for driving common cathode LED displays. Unfortunately this is common anode.
Missed the fact that the 4543 had an invert function for the outputs. You can invert the outputs from 74HC4511; that gives you the potential for more drive current.
Read that even though 4543 is named for LCD, can act as LED driver.
Only if driving the LEDs with a half mA is acceptable.
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clipimage.jpg
 
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