I would.You just wouldn't believe the stack of papers I have ! ! !
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.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...
Just can't leave this thing along ! ! !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.
Even though I haven't numbered/lettered any of these papers, when I do number things I use 01, 02, 03, etc........I would believe the stack of papers you have, too!
Ask me how I might know...
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...
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.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...
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.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 ??????
what ???????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.
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
That's not the way it works.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.
He added a -20mA curve to the graph, and simply extrapolated.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;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 ??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
You are absolutely correct. That is what I was doing.Perhaps you're looking at the 1ua line and thinking it is the 1ma line ??
See !You are absolutely correct. That is what I was doing.
hgmjr
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