[SOLVED] multiplexing 22 digit seven segment

[ongky70]

hello all electronic senior
i'm new here, i'm trying to multiplexing 22 digit of seven segment 1 inch.

in my design, i use IC hc4511 bcd to seven segment as the source driver,
ULN2003 to sink the current,
and 3 to 8 decoder that connected to the ULN 2003.
I give resistor 68 ohm to limit the current.
the micro controller was atmega128.

so far I've succeed program all the segment..
my problem is the seven segment is very dull

from my calculation :
I = V(IC)-V(svsgmnt)/ R
I =5V - 3.6 V / 68 ohm =~21mA

i thought 20 mA will give a good brightness. but in the reality,the light is very dull., I've search all around to solve this problem, but i didn't find it..

is this because of im using too many multiplexing of seven segment so the brightness is dull?
any idea to solve this problem?

I really apreciate your feedbacks.

 

[erikl]

is this because of im using too many multiplexing of seven segment so the brightness is dull?

Yes. 20mA would be bright enough, if one 7-segment display would get it all the time. Because of the multiplexing, however, it gets this current only about 1/22 (assuming there's only 1 mux-group) of the time, which means its effective mean current is only 20/22 ≈ 1mA .

any idea to solve this problem?

Grant each segment at least 20*22=440mA - your display should stand it, but your driver is at the very limit of sinking such a high current (if all segments=on). So you perhaps need a stronger driver, or split the multiplexing into 2 groups (11+11) - which probably needs 2 ports from your µP, and/or more external logic.
And you have to account for a considerable higher LED forward voltage, plus the ULN saturation voltage of 1.3 .. 2.0V (max). Then, a power supply of 5V won't be enough, you'll probably need ≥ 6V . You'd also have to take care of sufficient input current from the µP port, but I think that's the least problem.

Another solution would call for special "very bright" displays, which are bright enough with a mean current of 2 .. 5 mA, meaning 50 .. 125mA for your group-of-22 multiplexing (factor 25 instead of 22 accounting for some time slot losses between multiplexing). That would allow you to keep the existing mux/driver hardware and power supply.

BTW: Your topic would better be domiciled in this

 

[ongky70]

hello

so far i've tried to increase the supply voltage until 9 volt, the brightness is better than before but it still not good enough.

at supply 9 volt. I measured the forward voltage of the LED is about 1.25 volt and the out- ground of ULN 2003 is about 3.5 volt. I dont know if the multimeter was right or not since the switch on/off is very fast (on for ≈1 ms and off for ≈21 ms).

so the current increase a bit:
I= (9V-3.6V)/22 digit*68 ohm ≈ 3.6mA

when I increased the supply 12 volt the seven segment display is error,??
is there any problems if I decrease the resistor value until 25 ohm so i can get
≈10 mA current ?

thanks a lot for help
sorry btw how to domiciled this post into the neighbor forum?

 

[erikl]

I= (9V-3.6V)/22 digit*68 ohm ≈ 3.6mA

when I increased the supply 12 volt the seven segment display is error,??
is there any problems if I decrease the resistor value until 25 ohm so i can get
≈10 mA current ?

Calculate R = (Vsupply - Vled - Vdriv) / (Iled * digit)
For your 9V supply: R = (9V-3.6V-1.7V) / (0.02 * 22) = 8.4Ω

Vdriv is the typ. saturation voltage of the ULN driver at 400 .. 500mA (max. 2.0V). The ULN2003 can sink a max. of 500mA at your 1/22 ≈ 5% duty cycle.

 

[ongky70]

halo erikl thank u you was right,
I've change my resistor from 68 ohm into 10 ohm. The brightness was much better.

so the current of a digit seven segment is
I=9v-3.6v-1.7v/10 ohmx22 digit = 16.8 mA

then to make sure if its right I checked using oscilloscope , the result is :

V LED "a" to gnd of ULN2003 = 5.4 volt
with ton = 2ms and toff=20 ms

V resistor ≈ 0.7 volt
so that the current can be calculated I = 0.7 V / 10 ohm = 70 mA

there are vary voltage for the Vce saturated the highest one is 2.9 volt, the lowest one is 1.2 volt as u can see from the picture below. so which one is the rigth Vce saturated ?

I confuse with the calculation of the current, it's 16.8 mA or 70 mA?

 

[erikl]

V LED "a" to gnd of ULN2003 = 5.4 volt
with ton = 2ms and toff=20 ms

This means a duty cycle of only 10%. I think for multiplexing 22 digits, you would need ton = 2ms and toff=42 ms ?

V resistor ≈ 0.7 volt
so that the current can be calculated I = 0.7 V / 10 ohm = 70 mA

Depends on the displayed data, pls. s. below.

there are vary voltage for the Vce saturated the highest one is 2.9 volt, the lowest one is 1.2 volt as u can see from the picture below. so which one is the rigth Vce saturated ?

Hi ongky,

the medium (or mean) current from your 9V power supply is (9 - 5.4)V / 10Ω = 360mA . Each digit gets 1/22 of it, i.e. 16.4mA. This is an average value however, depending on the displayed data: For the display of data "1", only 2 segments have to emit light (and so need current), for a data display of "8", all 7 segments are needed and so the digit needs more current - that's why your scope shows various values.

The statistical average value of segments per data for the display of numbers 0 .. 9 on a 7-segment display is 4.9 , i.e. 16.4mA/4.9 = 3.34 mA/segment. To display a "1", the individual display needs 2*3.34 = 6.68mA, to display data "8", it needs 7*3.34 = 23.4mA for the same digit light intensity per segment. And so the saturation voltages of the individual ULN digit drivers vary. For comparison with dataSheet values, always consider that they have to drive the 22-fold value because of the time multiplexing.

Hope I have made this a bit clearer.

I confuse with the calculation of the current, it's 16.8 mA or 70 mA?

The lower value (16mA) is an average value for the display of arbitrary data. I don't know where (and how) you measured the 70mA value. May be this is the current value for 3 digits with a data=8 display?

 

[ongky70]

halo erikl,

That was right that 70 mA is used for 4 digit of seven segment. I just remember that the seven segment is parallel. So the current for each segment is 17.5 mA. and the ton= 1 ms toff=22 ms.
btw do you have a source about how to prove a multiplexing will get current is
1/ number of digit of the calculated current?

 

[erikl]

btw do you have a source about how to prove a multiplexing will get current is 1/number of digit of the calculated current?

This is valid on average. Prove by Kirchhoff's law.