TI's data sheet says that the 503 is capable of sinking and sourcing current: http://www.ti.com/lit/gpn/cd4503b and the drawing shows N and P transistors in the output.
So the device isn't open drain. at 5V; the device can sink (low) 2.3mA typical and source 1.9mA typical (high). Even with a minimum beta of 25 for the kathode drivers; IB would be 80uA given a 2mA emitter current. On Thursday, April 4, 2013 5:12:31 PM UTC-5, Zitt wrote: > > On Thursday, April 4, 2013 7:03:19 AM UTC-5, Terry S wrote: > >> The 4503 is an tri-state device -- do you need pullups on the outputs? >> I can't tell from a quick glance how you are using it. How about >> series resistors to the bases of the transistors? I see those on A1 >> thru A4. Without some series resistance your driver can only go to .7 >> (Vbe) and I'm not sure how well it will like being clamped that low. >> > > I'll look again; but pretty sure the 503 is a cmos device - not open drain. > > I actually tried a series resistor on at the base of A5; no impact. It > was the reason I could wire the base to the series resistor to show that > the digit works when the circuitry is bypassed. > -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send an email to [email protected]. To view this discussion on the web, visit https://groups.google.com/d/msg/neonixie-l/-/W4YmPHTmCowJ. For more options, visit https://groups.google.com/groups/opt_out.
