Scoreboard Project Cont;

Wendy

Joined Mar 24, 2008
23,415
You just wouldn't believe the stack of papers I have ! ! !:eek::eek:
I would. :D

Bill;
Tx's for the wish......
Wish I had the time to mess with robots, they're cool things.
Yeah, used to watch some of those fighting robot devices on TV, gotta stay close to the floor and destroy the others, etc.
Have seen some amazing videos of the robots the Japanese are making...
There has been serious argument (justified IMO) whether they should be called robots. We tended to drive the FIRST guys nuts, which didn't bother use any, our bots could beat their bots, but they were welcome to try taking us on. :p
 

SgtWookie

Joined Jul 17, 2007
22,230
I would believe the stack of papers you have, too! ;)

Ask me how I might know... :rolleyes:

Seriously though - I've been in the habit of not printing stuff out unless I have to - and when I update a schematic or board layout, I start by saving it with the next higher number or letter at the end of the filename. If you decide on using letters though, avoid Suzy-Q, which is a convenient way to remember SIOUXZY-Q.

"S" can be confused with the number "5".
"I" can be confused with the number '1'.
"O" and "Q" can be confused with the number '0'.
"U" can be confused with "V" and "Y". If only "V" remains, it can't be confused with "U" nor "Y".
"Z" can be confused with the number '2'.

I hope my attempt at clearing up confusion when using letters hasn't confused you even more...
 

Thread Starter

Oxbo Rene

Joined Feb 20, 2009
201
For sourcing 160mA, Ib needs to be 1-2mA.
Change R1 = 9.1k to 6.2k - anywhere in that range.
R2 gets moved to between +12v and the junction of R1/ULN2004A, and changed to 10k Ohms.
R3-R9 get 120 Ohms.
R10 should be 360 Ohms.

You need to update your schematics. I suggest putting revision numbers/letters in them. Otherwise, it will be very easy to get lost as to what's what.
Just can't leave this thing along ! ! !
OK, look at this........
First pic is based on your values and layout for R1 and R2 which puts them in parallel= I'm lost again...
But, this is what it looks like...
Have included data hgmjr added during explanation of Thevenin's stuff which throws off the voltage value at MPSA64 collector.
Please examine and advise
Second pic is layout of all four LEDs, biasing resistors and all.....
 

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Thread Starter

Oxbo Rene

Joined Feb 20, 2009
201
I would believe the stack of papers you have, too! ;)

Ask me how I might know... :rolleyes:

Seriously though - I've been in the habit of not printing stuff out unless I have to - and when I update a schematic or board layout, I start by saving it with the next higher number or letter at the end of the filename. If you decide on using letters though, avoid Suzy-Q, which is a convenient way to remember SIOUXZY-Q.

"S" can be confused with the number "5".
"I" can be confused with the number '1'.
"O" and "Q" can be confused with the number '0'.
"U" can be confused with "V" and "Y". If only "V" remains, it can't be confused with "U" nor "Y".
"Z" can be confused with the number '2'.

I hope my attempt at clearing up confusion when using letters hasn't confused you even more...
Even though I haven't numbered/lettered any of these papers, when I do number things I use 01, 02, 03, etc........
These papers I have been just naming the file what it is, then next one will be same name(+) 02, etc.
After it is all settled I get rid of everything but the best one and just give it the name, etc..
Somethin like that ??????
 

SgtWookie

Joined Jul 17, 2007
22,230
Just can't leave this thing along ! ! !
OK, look at this........
First pic is based on your values and layout for R1 and R2 which puts them in parallel= I'm lost again...
Oops - ok, take R1 and rotate it 90°. On the right, connect it to the base of the transistor. On the left, connect it to the bottom of R2 and the output of the ULN2004.

Start with that, then re-post it.
[eta]
It should look more like the attached.
 

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SgtWookie

Joined Jul 17, 2007
22,230
Even though I haven't numbered/lettered any of these papers, when I do number things I use 01, 02, 03, etc........
These papers I have been just naming the file what it is, then next one will be same name(+) 02, etc.
After it is all settled I get rid of everything but the best one and just give it the name, etc..
Somethin like that ??????
Something like that - but keep the revision numbers in the title of the thing, so if for some reason you have to refer back to it, you will have an idea of where in the chronology the schematic fits - otherwise, you'll lose track of where it fits in.
 

Thread Starter

Oxbo Rene

Joined Feb 20, 2009
201
oops - ok, take r1 and rotate it 90°. On the right, connect it to the base of the transistor. On the left, connect it to the bottom of r2 and the output of the uln2004.

Start with that, then re-post it.
what ??????? :eek::eek::eek::eek: :)

(Tomorrow........... )
(I thought that might be an oops ! )
 

SgtWookie

Joined Jul 17, 2007
22,230
Oh, and remember this? :



If the base is being fed about 1mA current, Vce on the MPSA64 is going to be somewhere on that chart,
but it's definitely going to be <= 0.9v.

[eta] Refer to the -175mA plot. Your maximum current will be less than that. With -1mA base current at Ic=-175mA, Vce is about -0.9v
 
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Thread Starter

Oxbo Rene

Joined Feb 20, 2009
201
If the base is being feed about 1mA current, Vce on the MPSA64 is going to be somewhere on that chart, but it's definitely going to be <= 0.9v

Good morning;
Have been updating my diagrams.
First pic is the Graph where I have deemed that Vce should be approx 0.7V (at 1ma)
What gets me is that the lines stop before running out to the side of the graph, therefore I assume that anywhere after the line has stopped is a non-operable area, etc.
1Kua = 1ma
In the interest of playing safe, will use the larger resistor values and hook this puppy up today and get some RTV (Real Time Values)
DP resistors forgot on the 4-LED diagram........
Have a nice day,
Oxbo
 

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Oxbo Rene

Joined Feb 20, 2009
201
OK, after hooking the circuit up to a dedicated/regulated 12V power supply,
Here are the real time values measured.........
Everything looks pretty good to me.

How about this;
I'm not too familiar with multiplexing.
Lets say segment "a" on all four LEDS are on at the same time, that would pull 80ma through R3.
For one segment the current is 20ma, for four it would be 80ma (x) 2.9V= 232mw (x) 2= 464mw= use a 1/2w resistor.
10-4 that sounds good.....
 

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hgmjr

Joined Jan 28, 2005
9,027
My thinking is that if you assume 1ma of base current, you go to the point on the X-axis where the -1ma base current point is and then run vertically up the graph along the -1ma line until you intersect with the -20ma current line you have added. Once you reach the intersection, you then move horizontally across the graph to the point on the Y-axis that corresponds to the intersection point. That value is the Vce. I read something like Vce around -1 volt or so.

I am not sure why you have added the 0.7 volt line.

hgmjr
 

SgtWookie

Joined Jul 17, 2007
22,230
How about this;
I'm not too familiar with multiplexing.
Lets say segment "a" on all four LEDS are on at the same time, that would pull 80ma through R3.
That's not the way it works.
You have four 7-segment displays. The common source (for common anode) or sink (for common cathode) for each display is turned on in sequence. No two displays are on at any given time, but the sequence happens so quickly, that persistence of vision makes it appear that they are all on at once.

The sequence goes something like this:
1) All display commons are turned off.
2) The segments to be lit on the next display in sequence to be powered are enabled via lines A-G and DP.
3) The display is turned ON for a period of time via the common.
4) The display is turned OFF.
5) Go back to step 2.

This happens very fast, hundreds or even thousands of times per second.

Since only one display is on at any given time, the most current that will flow through the 120 Ohm (normal segment) or 360 Ohm (DP) resistors is 20mA.

Note that there will be more power dissipated across the 360 Ohm resistor for the DP than the 120 Ohm resistors for the normal segments.
 

SgtWookie

Joined Jul 17, 2007
22,230
My thinking is that if you assume 1ma of base current, you go to the point on the X-axis where the -1ma base current point is and then run vertically up the graph along the -1ma line until you intersect with the -20ma current line you have added. Once you reach the intersection, you then move horizontally across the graph to the point on the Y-axis that corresponds to the intersection point. That value is the Vce. I read something like Vce around -1 volt or so.

I am not sure why you have added the 0.7 volt line.

hgmjr
He added a -20mA curve to the graph, and simply extrapolated.

Not sure how valid it is; but some quick testing would verify/disprove the extrapolation.
 

Thread Starter

Oxbo Rene

Joined Feb 20, 2009
201
My thinking is that if you assume 1ma of base current, you go to the point on the X-axis where the -1ma base current point is and then run vertically up the graph along the -1ma line until you intersect with the -20ma current line you have added. Once you reach the intersection, you then move horizontally across the graph to the point on the Y-axis that corresponds to the intersection point. That value is the Vce. I read something like Vce around -1 volt or so.

I am not sure why you have added the 0.7 volt line.

hgmjr
Well;
The way I see it, 1K= 1ma.
If I move up from 1K to the 20ma line I drew, that intersection appears to me to be horizontally across from the .7V (Actually a tad under).
I'm trying to see where you come up with 1V, can't seem to connect it.
Perhaps you could offer a graph showing your vision of it......

On another note,
Lets say R3 is running four segments at a time = 80ma.
So, 80ma running through a 120 ohm resistor equals a voltage drop of 9.6V across that resistor.
The LED segment requires 8.8V.
Having 11.2V available on the anode leaves 11.2 (-) 9.6 = 1.6V across the LED.= I'm lost again........

OK, just posted this then saw Sarge's post of explanation about multiplexing, etc..
Please disregard, etc.............
 

Thread Starter

Oxbo Rene

Joined Feb 20, 2009
201
My thinking is that if you assume 1ma of base current, you go to the point on the X-axis where the -1ma base current point is and then run vertically up the graph along the -1ma line until you intersect with the -20ma current line you have added. Once you reach the intersection, you then move horizontally across the graph to the point on the Y-axis that corresponds to the intersection point. That value is the Vce. I read something like Vce around -1 volt or so.

I am not sure why you have added the 0.7 volt line.

hgmjr
Perhaps you're looking at the 1ua line and thinking it is the 1ma line ??
 
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