My Simple Scoreboard - Need help.

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Wouldn't I need to base the voltage anyway on the type of LED's I shall be using?
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You would either pre-define the number of a given type that youwant in a string and then choose the voltage accordingly, or else you would pre-define the voltage and then decide how many LEDs of a given type you could put in a string.
I think this is the PNP emitter-follower arrangement that dougy83 had in mind:

It has the advantage that the LEDs reduce the voltage which the '47 output transistor is subjected to.
 

I believe the best option for this is to predefine the number of LED's in a string and change the voltage accordingly. Isn't the resistor itself reducing the voltage to '47 so the transistor isn't subject to more, or do you do that just to prolong its lifetime?
 
Ok, so this is what I have now. Do you believe I can now start sourcing the chips and punching in the values to make it function as planned?
 
Isn't the resistor itself reducing the voltage to '47
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The resistor drops the voltage less than 1V. The LEDs drop much more.

With 14 LEDs in series and an average Vf = 2V you will need ~ 32V to drive them. Simulation shows the '47 output transistor could be subjected to ~ 17V, which exceeds its rating and might damage the IC.
BTW the 30k resistors aren't critical values and could be 33k (a standard value). Don't forget 100nF decoupling caps across the supply pins of each IC.
 
If all the LEDs are in series (as in your schematic) then you will probably need a 35v to 50v battery source. Also if 1 LED quits then the whole segment will probably quit. I would go with 2 strings of 7 LEDs if 14 LEDs per segment is your goal.
 
OK thanks for the info guys. I shall take a look at it now and see what I can come up with. BeerBelly, I am looking at sourcing it from the mains. But, thanks I shall put them in parallel.
 

Ok, I'm back with the alterations. This is how I got it now: . Am I not able to just protect the '47 output transistor with an extra resistor after the LED's? If not, cant I just get an uprated transistor that can take the extra Voltage? Thanks a lot guys.
 
R3 is no longer needed due to the fact that it's now an emitter-follower configuration. Remove R4 and place a resistor in series with each of the strings of LEDs to better balance the current between the two strings.
 
dougy83 said:
R3 is no longer needed due to the fact that it's now an emitter-follower configuration. Remove R4 and place a resistor in series with each of the strings of LEDs to better balance the current between the two strings.
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Ok, I fixed what you mentioned. Does this now protect the transistor from being subject to about 17v?
 
Your latest circuit has lower cap values than originally, so the debounce won't work properly. Use at least 1uF (10uF would be better).
Am I not able to just protect the '47 output transistor with an extra resistor after the LED's?
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Only at the expense of a big reduction of LED current.
If not, cant I just get an uprated transistor that can take the extra Voltage?
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No. The transistor in question is the one inside the IC.
 
Well I'd like it as low as possible without compromising the LED's. I plan of adapting the mains here (230v) to the required amount. Well, that is only one segment I have drawn of 28 segments in total. Maybe I can find a better way to make the 7 segment display without as many LEDs or figure how this can be done in its current form. Thanks beer.
 
Maybe an LED configuration like **broken link removed** would be more suitable as its all parallel and using 7 LED's per segment? But, it would need to be altered to have its own powered supply. I shall take a look now with what I can come up with.

i.e. something like this:


I am guessing its still advisable to put the resistors in for each LED to balance the voltage? or a single resistor after the power source to protect it?
 
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Why do you want to put them in parallel? Series strings are perfectly good and reduce the overall current needed, as well as needing only one current-limiting/balancing resistor per string. Think about it: if each LED needs, say, 20mA then 7 in parallel would need 140mA whereas 7 in series would need only 20mA per string. 7 LEDs per string, as you had previously, was fine and would need a supply voltage of around 15V minimum (depending on the type of LED). What type/colour of LED do you plan to use?
I am guessing its still advisable to put the resistors in for each LED to balance the voltage?
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Yes, in a parallel arrangement.
 
Using the 2 strings of 7 would work fine with a 24 volt supply and PNP. Each string needs a current limiting resistor.

Using all LEDs in parallel would work fine with a 5 volt supply and PNP. Each LED needs a current limiting resistor.

Either one of these designs are your choice. Next you may want a dimmer circuit.
 
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Alec, I had no specific reason to put them in parallel other than seeing many doing it. I have been looking at these LED's here . Seeing how bright LED's are nowadays and adding up the costs, I am thinking maybe 4 LED's a segment shall be sufficient? I plan on putting a film of plastic over the segments, so the LED's themselves wont be directly visible. So a series should be sufficient.

BeerBelly, why do you now suggest a NPN circuit over the PNP?

Thanks guys.
 
Hi, I am back. This is what I have come up with now and I think we are nearing the end of the design and punching in the figures.

I believe this configuration is most feasible in the end. I tried to work out the correct resistor required for the LEDs. I put it at 2.4v rather than running it at 2.6v right?

But, regarding the IC's, why does U3 get an extra ground on the 5th Terminal? Also, for the 74LS14N would only have to be 1 chip, and just connecting it on the other terminals, correct?

Cheers, M

p.s. how did you calculate R1 and C1 values?
 
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Your cap values are still way too low. Make them 10uF. You can also eliminate R1, R2 and R18 because the 74LS14 has internal 20k pull-up resistors. You will then have a debounce time-constant of 200ms, resulting in the '14 input being held below the upper Schmitt threshold for ~80ms, considerably greater than likely switch bounce duration.
I put it at 2.4v rather than running it at 2.6v right?
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I've no idea what those voltages relate to, but the Vf of a red LED is typically in the range 1.8V-2V. You still haven't said which LEDs you will be using. White LEDs, for example, have a typical Vf > 3V. For a 12V supply, Vf = 2V and 4 LEDs in series the current through them will be ~33mA, which is excessive if typical red LEDs (rated 30mA max) are going to be used.
why does U3 get an extra ground on the 5th Terminal?
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To blank a leading zero.
Also, for the 74LS14N would only have to be 1 chip, and just connecting it on the other terminals, correct?
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Not understood.
 
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