Propeller - Sinking Current from the Seven-Segments
John A. Zoidberg
Posts: 514
This topic is about the Propeller, and also the Seven Segments, so I will post it here. Otherwise, feel free to move it. 
I just had constructed a little clock with the Propeller, and with a pack of the common-anode 7-segments.
However, the displays are a bit too dim - it seems that I'm not sinking enough current. The thing is, I could have connected a ULN2803 inside, but what I feared is I will sink too much current into the Propeller, which is not a good idea after all.
Here is my current schematic. If my calculation isn't wrong, the current is sinking 10mA, which is within the range of the allowed input current of the Prop.
So how do I properly sink the current without sinking those back into the Prop pins?
I just had constructed a little clock with the Propeller, and with a pack of the common-anode 7-segments.
However, the displays are a bit too dim - it seems that I'm not sinking enough current. The thing is, I could have connected a ULN2803 inside, but what I feared is I will sink too much current into the Propeller, which is not a good idea after all.
Here is my current schematic. If my calculation isn't wrong, the current is sinking 10mA, which is within the range of the allowed input current of the Prop.
So how do I properly sink the current without sinking those back into the Prop pins?
931 x 519 - 20K
Comments
Connect emitter to ground.
From Vcc go to Display and collector.
The current is controlled by the resistor between the prop pin and the base.
Current across this resitor (base current) is (3.3v - 0.7V)/R
0.7 is the base/emitter voltage drop.
At the collector you'll have the base_current X GAIN
Gain is usually around 100-150, so the base current will be 0.1-0.2 mA.
Massimo
A 20k resistor should drive the LED with 13-20mA, depending on the gain (supposing 100 and 150).
A PNP transistor like the 2n2907 would allow you to pull the anode much closer to the positive rail. In that case I'd change the 330 Ohm resistor to about 160 Ohms.
If you stay with your existing NPN design, I'd change the 330 Ohm resistor to about 120 Ohms
Well, with all segments on, you'd have to supply the anode with 70 mA - a bit over Prop specs. You could always tie multiple pins to the anode, but you'd soon run into Prop dissipation limits that way.
It's got Wesley Snipes and Bruce Willis written all over it.
Haha, it's a funny coincidence!
How about interfacing the segments, from A-G to the sink driver, ULN2803? I just need to isolate the Propeller input pins.
Why do you need to isolate the Propeller pins? A Propeller pin can sink 40 mA. You certainly don't need a ULN2803 to drive the individual cathodes.
Oh, because the new seven-segs I bought requires a bit more current, it looks dim here on the board I'm working on.
By the way, if I connect the LED + resistor and a 5V supply into the prop pin, I guess it is already beyond its specifications?
I see.
However, I have a 4-digit all-in-one 7-segment module by Avago, quite a medium sized one.
The thing is it that the display is very dim despite being driven by ULN2803, and the digit selecting transistors.
Or I would just go ahead, and do the one you suggest? I'll have to try on this one.
John, don't you see that the ULN2803 is part of your problem? It is a darlington array. A darlington transistor has two base-emitter drops. When you add two base-emitter drops to the voltage drop of the LED and the voltage drop of your anode transistor, you're out of voltage to work with.
If you placed a milliamp meter in series with a digit, you'd see that you are achieving nowhere near 10 mA. And if you're not reaching 10 mA, what is the point of using a 500mA driver? The driver is part of your problem, not part of the solution.
Ouch, no wonder the display looked starved. I forgot all about the minusing 2-BE drops there.
So how to get around to making the display brighter? I searched around the net but they don't have much schematics for it.
I'll have to drop the driver then.
What is the best solution for driving even bigger displays?
Isn't the common anode tied to Vdd? That should be coming from the 3.3v reg and not the prop. and then you sink the cathodes with the prop pins. What am I missing?
edit. ok...he is hooking up multiple of these up so he deactivates one of the segments by having the common anode on a prop pin. I get it now. lol
The anodes are multiplexed. If you didn't multiplex the anodes, you'd need 28 Prop pins to drive all the cathodes. This way you only need 7 for the cathodes and as few as two for the anodes if you decode them. Four if you don't.
First of all, I don't know what Avago display you are using now, but none that show up in the catalog are very big by my standards. And if you're dealing with a 4-digit module, it's smaller still. So there should be no need for anything other than the Prop pins themselves to drive the cathodes. 40 mA is a lot of current for an LED, even in a pulsed situation.
For the anode, simply use a PNP transistor like a 2N2907 or 2N3406 that you switch to saturation. That only has a voltage drop of about 0.2V. That will give you a pretty good voltage overhead to work with. At that point it's simply a matter of selecting the right current-limiting resistor. But that depends on your LEDs and the degree of multiplexing.
If you're doing 1-in-4 multiplexing, you may as well use the whole 40 mA. LED's don't mind being pulsed.
Here's the datasheet - http://www.avagotech.com/docs/AV02-0568EN
It's a very casual 4-digit seven-segments with two decimal points for the clocks and whatever it is.
I'll just get these transistors you mentioned and let the Prop pins sink the current instead then.
BTW, with this specific display, a 3.3 V supply, and saturated PNP transistors driving the common anode lines, you'd need current-limiting resistors of about 33 Ohms to get 40 mA through an LED segment.
(p.s: If you buy transistors by the bag like I do, they're only about $0.05 USD, each.)