2
\$\begingroup\$

Please see this circuit.

enter image description here

I'm trying to design a circuit that allows you to use a single button to increment a counter from 0 to 9 (then it will loop). Diode logic will convert the 10 logic outputs to BCD and a chip will then display the number on a 7 segment display.

The outputs from the decade counter will then be used to trigger an off board device with 10 decimal inputs, but only when the 'activate' button is pressed.

So in the circuit everything works. I can use the increment button to find the number I want, and then press the 'activate' button and that output will go high.

However I discover that the off board device has 'active low' inputs! As it is I'm going to need 3 quad AND gates which takes up a lot of space on the board. If I have to add two hex inverter IC's the board will become huge.

Can anyone suggest a way that I can make this design more efficient, such that I when I select the number I want and push the activate button, the appropriate output goes LOW?

EDIT: Following Tim's answer and by way of keeping a complete record here is a revised circuit.

enter image description here

\$\endgroup\$

4 Answers 4

Reset to default
3
\$\begingroup\$

Well, the old fashioned way is a 74LS90 decade counter into a '47 display decoder/driver, and a '154 decoder for the one-of-N outputs (the outputs will be active-low, arrange your loads/switches accordingly).

The '154 has enable inputs so you don't need an array of ANDs.

Most of these are available in CMOS (74HC, etc.), though you may have trouble finding the '47, but that's fine, 74LS inputs are CMOS compatible.

In CD4000 family, consider 4510, 4511 and 4028 or 4067 (the latter is a bidirectional switch, not a decoder; if the common side is assigned logic-high, then one-of-N outputs goes logic-high and the rest are high-Z. Use pulldown resistors to get a logic low, then.)

Further reading:
https://en.wikipedia.org/wiki/List_of_7400-series_integrated_circuits
TTL Cookbook, Don Lancaster
https://en.wikipedia.org/wiki/List_of_4000-series_integrated_circuits
CMOS Cookbook, Don Lancaster
And, go shopping for parts to make sure they're available, and in the family you're looking for. Then look up individual datasheets.

\$\endgroup\$
8
  • \$\begingroup\$ Thank you for this.. I'll look into it. \$\endgroup\$
    – Richard
    Commented Jul 25, 2022 at 19:07
  • \$\begingroup\$ I think this works. Still having a few teething problems in LTSpice, but it's looking good. CD4510 BDC counter into both a CD4511 (display) and a 74HCT154 (do you know the CD prefixed chip?) to convert to decimal active low outs. On the 154 I can use the chip select pins to trigger the output.. When I have a working solution with this, I'll mark your answer as correct... thanks again. also have a small issue getting the CD4510 to go from 9 back to zero.. but the rst pin should somehow do this for me? \$\endgroup\$
    – Richard
    Commented Jul 25, 2022 at 23:47
  • \$\begingroup\$ Ahhh :) The LS90 does cycling (mod 10).. very well recommended! I need the HCT version of it really (CD version).. But again, thanks that's great! \$\endgroup\$
    – Richard
    Commented Jul 26, 2022 at 0:00
  • 1
    \$\begingroup\$ If you're using 5V supply at low frequencies, 74HC(T) and CD4000 are interchangeable. For the latter, check out 4028; and yup, 4510 and 4511 are the classics! \$\endgroup\$
    – Tim Williams
    Commented Jul 26, 2022 at 1:53
  • 1
    \$\begingroup\$ @Richard You've got 9V to "vdd" but also ".param vcc=5", and HC(T) definitely doesn't run much more than 5V; diodes don't really do anything, but reduce R9, R11 to like 100R, that will debounce okay; yes, you need schmitt trigger; chips should be fine no buffering needed. \$\endgroup\$
    – Tim Williams
    Commented Jul 26, 2022 at 17:55
7
\$\begingroup\$

Replace the AND gates with NAND gates.

\$\endgroup\$
3
  • \$\begingroup\$ Yes. That would seem to be the answer. I was hoping someone would say something like.. use a BCD counter, then some BCD to decimal chip that does active low outputs (that I've not heard of)... Or even, here's a circuit that does exactly what you want but designed by someone clever :) Thank you though :) \$\endgroup\$
    – Richard
    Commented Jul 25, 2022 at 16:26
  • 6
    \$\begingroup\$ @Richard i think the space-saving answer nowadays would be use a microcontroller! all the digital logic you could ever want, packed into one chip \$\endgroup\$
    – Stack Exchange Supports Israel
    Commented Jul 25, 2022 at 16:50
  • \$\begingroup\$ Or possibly a FPGA. I haven't really had the need or desire to try them, but they are dirt cheap. Years ago I used Signetics PLDs (PLS153 and PLS173, IIRC). datasheetq.com/PLS153-doc-Philips. Here's a re-programmable 20 pin device for about a dollar: mouser.com/ProductDetail/Microchip-Technology-Atmel/… \$\endgroup\$
    – PStechPaul
    Commented Aug 6, 2022 at 0:06
3
\$\begingroup\$

Want a compact solution that's cheap to implement for a one-off? An Arduino Nano would do this in a smaller footprint than 2-3 DIP ICs. It could handle your switch debounce too, and give flexibility over the 'trigger' output pulse duration.

If you're determined to use MSI logic, a decade counter followed by a 1:16 demux (with enable) would make the gated 1-of-10, with the 7-segment readout driven from BCD using the CD4511 as shown. Yes, 3 ICs in all, but the benefit is that you can do away with the diode logic, which won't work without a pull-up anyway, at least for CMOS.

Smallest footprint and lowest possible cost if you're making a bunch of these? Try a Silego / Dialog / Renesas Greenpak. More here: https://www.renesas.com/us/en/products/programmable-mixed-signal-asic-ip-products/greenpak-programmable-mixed-signal-products

\$\endgroup\$
1
  • 1
    \$\begingroup\$ I am stuck on MSI Logic for reasons that are too long to explain, which is annoying because the off board unit is an arduino based daughter board. Thank you for the Demux idea.. this is exactly the sort of thing I'm looking for. \$\endgroup\$
    – Richard
    Commented Jul 25, 2022 at 19:06
1
\$\begingroup\$

Instead of using three chips 74xx08 with four gates each, you may use only two chips 74xx04 with 6 inverters each.

\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.