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7 segment decoder

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koolguy

Active Member
Hello guys,
What to do for 7 segment decoder, actually i have done project with 7447 with small/ 1 chip red leds at 5v.
But now i have 5 leds in series with 120ohms.
How to decodesthis voltage is at 12V .
 
Hello guys,
What to do for 7 segment decoder, actually i have done project with 7447 with small/ 1 chip red leds at 5v.
But now i have 5 leds in series with 120ohms.
How to decodesthis voltage is at 12V .

The CD4055B can run at 12 volts. There will be some pin/function differences so you need to check the datasheet.
 
Here's https://www.google.com/url?sa=t&rct...=08NzkUBSna5wadLLQSeR1Q&bvm=bv.91427555,d.cWc

the datasheet for the part you used and the alternatives. You did not supply Vf or the current needed,

But (5 leds)*(2.1 Vf)+0.6 > 12 V, so it really won;t work for 12 V. You might be able to use two strings, if the current limitations of the driver is not exceeded. Other chips can use up to 30 V for the open collector side of it. You could possibly use two strings, but the current for each string would have to be less than 20 mA. 40 mA/2

There are other options, like chose an active high version and use N-channel FET;s to switch the segments. Use a logic level FET and about a 100 ohm resistor in series with the gate. The 0.6 drop would be less to compensate for. Rds on would likely be in milliohms resulting in a much lower voltage than a transistor.

Still you can drive one string unless you increase the supply voltage. You would have to use two strings and two resistors. Say a 3 LED and a 2 LED.
 
Your basic problem is with 5 LEDs, you run out of voltage. Depending on the color, the voltage drop is different. 2.4 V is a guess.

5 * 2.4 = 10.8 and then there is a voltage drop of the driver. Darlingtons will drop about 1.2 V and your at 12V with no headroom for a resistor. That 2.4 is a number I picked out of the hat. There is a Vfmin and a Vfmax for your LED.

You can certainly use it, just not for the entire string.
 
40 mA is cutting it close. I bet, they wanted you to use the 28xx series https://elabz.com/wp-content/uploads/2011/03/ULN2801-2803-2805-datasheet.pdf

which is an 8 pin version of the ULN200x.

To use the decoder you have, you;ll need pull ups, a TTL inverter and the ULN2803 (8 outputs) or the ULN2003 (7 outputs)

or pull ups, a TTL inverter and n channel logic FETS and a 100 ohm resistors for each FET.

Depends on where you are (side of the pond) and what you can get easily. The ULN2003 are very widely used chips. The FETs might be harder to work with and might be surface mount.

The pull-ups could be done with a resistor network.
 
Sure can. I had to do it with a 6800 uP back in the late 70's.

It's generally based on multiplexing and a look-up table. Your drivers (ports) still has to be bale to source/sink the current. You either have common anode or common cathode. You have to watch Max average current for the PIC.

One trick is you have to implement inter-digit blanking. You actually have to turn off the previous digit before selecting the next digit otherwise you get a blur.
 
Because of the voltages, you would use a high side and low side drivers: e.g. high side for segments and low side for digits or vice versa.
 
I'm not going to be up much longer, but you will need both high side and low side switches.
But I'm too tired to be coherent.
 
Hello,
Here is Display with 12V directly 120Ohm series Resistance taking 20mA each segment!
with 9.5V/1.9V leds
 

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The 4511 takes 0000 and give 1111110 //0
and the ULN2003/8 will convert 1(high input) to 0(low output) and 0(low input will be give what) will convert x??
 
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