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 Forum index » DIY Hardware and Software » Lunettas - circuits inspired by Stanley Lunetta
getting started with binary/decade counters, flip flops
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Joined: Feb 02, 2010
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PostPosted: Wed Feb 10, 2010 11:38 am    Post subject: getting started with binary/decade counters, flip flops
Subject description: that dm7490A I have had sitting in my parts drawer for 12 years
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I tried breadboarding a few 40106 oscillators with this thing (7490) last night.

http://webpages.ull.es/users/fexposit/7490.pdf

I thought I could get maybe some octave dividing or some sort of division, but I couldn't really get my head around the logic tables to figure out how to get it to count.

I connected separate oscillators to pins 6&7 (R9 1&2) and took the output from 11 (QA), and I got an on/off gating of one signal by the other, but that was all I was able to get out of it, and I am not sure what or how I was getting or why (just sort of semi-randomly connecting stuff).

Any suggestions on how to connect to this chip? The language in the data sheet is not making sense to me yet. Hopefully I can learn by some examples.


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Sam_Zen



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PostPosted: Wed Feb 10, 2010 7:15 pm    Post subject: Reply with quote  Mark this post and the followings unread

Why connect the oscillators to pins 6&7, while pins 1 and 14 obviously are marked as 'input' ?
The pins 2&3 and 6&7 are reset inputs and should be kept low for proper counting, as you can see in the function table of the PDF.
If you want to try these things, it's better to get used a bit to a data sheet and its truth tables.

The 7490 is a decade counter made of 4 flipflops (A,B,C,D), each dividing by 2.
Normally these 4 would mean a division by 16 (2^4), but this counter is reset internally when 10 is reached.
If you use input A (pin 14), the output QA (12) has a division by 2 (one octave lower)
If you want the full decade count, you have to connect QA with input B (pin 1) so other divisions are available at the outputs of B,C and D (pins 9,8,11).

But I noticed that you use a 40106 as clock oscillator. This is an IC of the CMos types, which is different from the TTl 74xx series.
Directly connecting these two types is electronically somewhat unreliable. Normally a special buffer/driver is used in between.

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Uncle Krunkus
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PostPosted: Wed Feb 10, 2010 7:41 pm    Post subject: Reply with quote  Mark this post and the followings unread

Yes, definitely decide on all CMOS or all TTL first. To cut a long story short (I lost my mind!): -
CMOS - Wide operating voltage, (3-15V). Slightly susceptible to static damage.
TTL - Must operate on 5V only. Not susceptible to static damage. Reduced fanout.

If you go with CMOS, remember to hold all unused inputs either high or low with a 20K resistor. ie, pin -> resistor -> +V or Gnd

Either way, it's a real good idea when starting out to have a LED (with 1K current limiting resistor in series) on a probe to ground, so you can quickly and easily see what any pin is doing. You can also get complete "logic probes" which do the same thing.

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PostPosted: Thu Feb 11, 2010 11:16 am    Post subject: Reply with quote  Mark this post and the followings unread

Thanks for the responses.

Yea, about the 7490, I was only trying it out because I had it in a collection of parts that I got from a tv repair shop that was going out of business many years ago.... it's just been sitting in with my parts since then. I wasn't aware that you couldn't mix them. I am making an order soon to get some more CMOS chips, as they seem to be more flexible and use voltages I would more commonly use.

Sam_Zen, the reason I tried the pins I did was because I tried the inputs and got nothing out of them on any of the other pins, so I just started randomly trying other things. As I said, I am new to logic chips and the logic tables don't make much sense to me (I tried), so your bit of explanation about the truth tables and what they means helps a lot.
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electri-fire



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PostPosted: Fri Feb 12, 2010 7:50 am    Post subject: Reply with quote  Mark this post and the followings unread

Uncle Krunkus wrote:
If you go with CMOS, remember to hold all unused inputs either high or low with a 20K resistor. ie, pin -> resistor -> +V or Gnd


Is there an advantage of the inserted 20K resistor over directly connecting inputs to +V or ground?

If so, can other values be used also? What minimum/maximum value?
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Sam_Zen



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PostPosted: Sat Feb 13, 2010 12:57 am    Post subject: Reply with quote  Mark this post and the followings unread

Quote:
..an advantage of the inserted 20K resistor over directly connecting inputs to +V

More than one advantage.
1. If a resistor is enough to keep the input high, why connect it directly ? Such a connection would use more current with the same result.
2. The resistor is also a protection of the system.
If the input of the IC would break down, causing a shortcircuit to ground, it would mean a connection between your V+ and ground.

The difference in nature about input pins of CMOS and TTL (supposing they are 'open' so unconnected) :

- CMOS inputs have a rather high impedance, so they have the tendency to 'float' qua level, so they have to be forced high or low.
By either be connected to some CMOS output or connected via a resistor to V+ or gnd.
This 'floating' also means a bigger sensitivity for picking up external HF signals.
To decrease this sensitivity, the safe value range of the resistor is quite big, upto 470K, but dependent of the circumstances.

- TTL inputs, if open, are bound to be 'high', due to a resistor connected to V+ inside the IC.
So to use them, a signal or a switch has to 'pull down' the input to a low state.
A CMOS output is usually too weak to directly pull down a TTL input.

About the truth tables :
See it as a table of "what if" shown in a matrix.
Where "X" means : 'it doesn't matter whether H(igh) or L(ow)'.

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electri-fire



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PostPosted: Sat Feb 13, 2010 4:36 pm    Post subject: Advantages of inserting resistors for pullup/down. Reply with quote  Mark this post and the followings unread

Another advantage occurred to me later. I don't have a lot of CMOS building experience. So now I have a feel for what it does I want to add other IC's in the same case, and change use of others.

Some input pins I tied to ground directly, and now I want to use them I have to desolder and insert a pulldown resistor, soldering between a mess of wiring.

I'm thinking to start anew, and have tiedowns with resistors as standard procedure.
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