Digital clock concept

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Goldenshuttle

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This 6 digit LED clock at 4.5$ from EBAY. I am trying to replace the small LED display with big ones 2.3 inch common Anode type. I came up with a simple tweak is to remove the resistor (marked with red X ) and replace it with drive with S8050 NPN transistor to provide more volatge to each segment.
is this going to work ?
:?:

 

No. The transistor will invert the signal so the segments are on when they should be off. You can fix that by changing the code in the code in the IC to 'pre-invert' the segment signals so the transistor reverses them back again or you can use a two transistor circuit to double-invert the existing signals.

It might be possible to use PNP transistors with the collector and emitter swapped but there are potential dangers in doing that.

Brian.
 
Thanks Brian..I am in not capable of PIC code as I barely understand it. Being slow minded can barely work some Arduino short sketch.
But I understand from you that using PNP transistor in a different way may work without code touching. I ask you if you can provide sketch for one transistor and I will solder it and give it a try?
 

on the other hand, is it possible to use on ULN2003 instead of adding 7 transistor? if yes, please can you help any link to a schematic I can try ?
 

Same problem. ULN2003 also inverts signal polarity. ULN2003 has high voltage drop, so it's no good choice for a 5V LED driver anyway.

To add another point, it's necessary to talk about designed LED currents, current limiting and driver strength. Your NPN transistor solution lacks a clear current limiting means, except for the loosely defined limiting by base resistor and transistor current gain. It runs at risk to damage LED displays.

Secondly the digit driver strength must be possibly increased.

To make stronger segment drivers with same polarity, NPN transistors driven by inverter gates (e.g. 74HC04) may be a solution. Provide base and collector resistors for current limiting.
 

Between the 7 resistors and 7 segment displays see if 75452 can be used.

https://www.ti.com/product/sn75452b

Two inputs of one NAND gate in 75452 is shorted and connected to out of resistor and output of 75452 is connected to a,b,c,d,e,f,g, and dp of 7 Segment display.
 

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Use common cathode, ULN2003 and 4 high current positive drivers on P-channel mosfets.

Thanks for the tip, but I have like 30 pcs common Anode and I need to dispose them on a feasible project. While buying common cathode will add up cost.

- - - Updated - - -


Thanks Milan, if I understood you right, schematic would look like this (below)...yet if we look at the display spec's, it is not a current guzzler, its current need is low, just voltage is 5 - 7 Volt.
 

If 8051 pins can handle the current and if you are providing 7V to display then use resistor in each channel to drop the excess voltage across resistor.

Maybe the device (75452) has additional circuit other than nand logic because logic for NAND with two inputs shorted is

A B Y
1 1 0
1 0 1
0 1 1
0 0 1

So, when input is 1 output of nand (with shorted inputs) is 0 and when input is 0 output is 1. See the datasheet of 75452. I think it has additional transistors (open collector) at the output of NAND gate. They are used as buffers and buffers don't invert the logic.



https://www.seeedstudio.com/depot/7-Segment-Display-8-Inches-Red-p-1194.html

V SSD is 10.8V

Use a 12V 500mA power supply.

Some 0.4V drop across EN transistor's CE junction.

limit current to 15 mA per channel.

12V - 10.8 - 0.4 = 0.8V

I = 15 mA
V = 0.8V

R = V/I = 0.8V/15 mA = 53.3E

Use a 50E resistors between AT89C2051 pins and a,b,c,d,e,f,g,dp of SSDs.





Edit: Are you sure current per channel is 20 mA ?

https://www.sparkfun.com/products/8530
https://shop.evilmadscientist.com/productsmenu/partsmenu/107-sevensegment
 
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