74174 working/ and how to configure it acc. to your circuit

Audioguru said:

Your first post said you have to use 74174, not 74HC174 that is completely different.

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in this post ,i later replied i will you use cmos version........

KlausST said:

Hi,


In a real circuit this will not work.
You may ask yourself if such a simulation tool is useful. On the other hand every designer should know that inputs shouldn´t be left floating.

pull up / pull down resistor:
Meaningless how many ICs you have. Just look into the datasheet for the one you want to use the pull up.
Read: V_IH, VIL, IIH, IIL. this gives you thel max limit of the resistor.
Usually 10k is a good value to start. For very noisy environment or lenghty traces use a lower value. To reduce supply current you may use higher value resistors.

Klaus

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i tried to make the saving section on breadboard.....

i used 4013 to save outputs of push buttons using manual setting......
i applied +ve to set pin (by using a push button like the push button outputs for game) and grounded the clk and d pin. used the reset pin.... but the results are ambiguous....
when i apply the +ve to set pin , it makes the Q pin high... by pressing resets button it becomes zero( i had verified values by using Digital multimeter....... but when i place leds to Q and Q' outputs....only Q led remains on all the time pushing reset button is also useless...... why its not working as desired.... if its not able to differentiate between 2 leds connected on outputs, how is it going to make correct changes in game further which is more complicated.... am i wrong ?

Hi,

nobody can answer this without a complete schematic.
* maybe you have the wrong supply voltage
* maybe you didn´t use reistors at the LEDs
* maybe you made wrong LED connection
* maybe you missed the power supply capacitors.
* maybe hundreds of other issues...

Klaus

KlausST said:

Hi,

nobody can answer this without a complete schematic.
* maybe you have the wrong supply voltage
* maybe you didn´t use reistors at the LEDs
* maybe you made wrong LED connection
* maybe you missed the power supply capacitors.
* maybe hundreds of other issues...

Klaus

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brother here is my circuit schematic which is workable on proteus............

View attachment Tic Tac Toe.pdf
View attachment Tic Tac Toe 2.pdf
View attachment Tic Tac Toe 3.pdf
View attachment Tic Tac Toe 4.pdf

brother i have uploaded schematics......

before i make pcb ,i want to confirm every thing...but now d-flip flop is not working even by grounding its "d" and "clk" pins as you told....

Hi,

Maybe you posted the wrong schematics...

A general rule: "Don´t leave (unused) inputs floating."
* but I see about 60 inputs left unconnected.

maybe you didn´t use reistors at the LEDs
* I see no current limiting resistor at the LEDs

maybe you missed the power supply capacitors.
* I so no power supply capacitor

A 2.4V TTL high cannot make a 3.5V Cmos high.
* but you are continously mixing CMOS and TTL

I´m feeling fooled. I´m out.

Klaus

KlausST said:

Hi,

Maybe you posted the wrong schematics...

A general rule: "Don´t leave (unused) inputs floating."
* but I see about 60 inputs left unconnected.

same thing i was asking earlier what to do with them... you said connect them with ground using pull down resistor 10k. which i did but i didn't get the answer...
look i only made the latch portion on breadboard .... with a push button that is used to save by 4013 (as in circuit) and i grounded the unconnected pins with resistors but it doesn't work.
here also ttl and cmos logic which you said later will not work as i don't used both... i only used cmos 4013 to just save an output from manually formed circuit much alike to that of tic tac saving section.(it is made to check whether this circuit is working or not).


maybe you didn´t use reistors at the LEDs
* I see no current limiting resistor at the LEDs
i will use 220 or 330 ohm resistor. otherwise may not work longer...

maybe you missed the power supply capacitors.
* I so no power supply capacitor

what are power supply capacitors.... would you like to please explain some thing about it......

A 2.4V TTL high cannot make a 3.5V Cmos high.
* but you are continously mixing CMOS and TTL
all the Logic ICs used are of TTL..... except cd4013...... is there any TTL available for d-flip flop

I´m feeling fooled. I´m out.

please brother.... i am trying to learn something and i had learnt ... you told me about pull up and pull down resistors... now you had told me about cmos and TTL could not work together. and i will continue to learn i you continue to help.

Klaus

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brother i am waiting you reply.

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KlausST said:

Hi,

A 2.4V TTL high cannot make a 3.5V Cmos high.
* but you are continously mixing CMOS and TTL

Klaus

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* may i use 7474 TTL instead of 4013.

Hi,

Especially pages 12 and 13, hope it helps.

Digital Logic Families

LED resistors: (Output voltage - LED forward voltage)/ LED current you want. e.g. (4.5V - 1.7V)/0.005A = 560 Ohm resistor.

Unused inputs - if they float they don't know whether they are on or off, so the output will be unpredictable but tends to float to "high" state, so you have to put a resistor, 1k or 10k to V+ or to ground (you'll know which you need, I assume thay all need to be tied to ground), even if you're just breadboarding a circuit.

d123 said:

Hi,

Especially pages 12 and 13, hope it helps.

Digital Logic Families

LED resistors: (Output voltage - LED forward voltage)/ LED current you want. e.g. (4.5V - 1.7V)/0.005A = 560 Ohm resistor.

Unused inputs - if they float they don't know whether they are on or off, so the output will be unpredictable but tends to float to "high" state, so you have to put a resistor, 1k or 10k to V+ or to ground (you'll know which you need, I assume thay all need to be tied to ground), even if you're just breadboarding a circuit.

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thanks for the link.... i am studying it....will ask you i don't get something

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i studied the slides and concludes that the output from TTL logic ICs could not be used in CMOS ICs.

Hi aliahsan951,

How's the tic-tac-toe circuit going? Does the original circuit have a video of it showing power-up all the way through to reset after a game?

I was looking at your schematics and find them quite hard to follow. The circuit looks very complicated, perhaps unnecessarily so - maybe I don't understand it - and all circuits should be simplified when possible - who wants to solder hundreds of connections if 100 is enough?

If this is for you, and not an obligatory project for a course or something, I'm sure you can do this with (much) less components. Instead of individual push buttons and what look like about 50 ICs - and thinking about it quickly - maybe two 4 x 3 keypads, 5 CD4081 quad dual input AND gates, and the flip flops, along with some power-up reset thing, might be more than enough. 4 x 3 keypads cost US$ 4 each; they have 12 buttons, so you could use the six spare ones for something else, if needed. It's just an idea.

Also, I hate to mention this, but if I remember well, in the real world flip-flops on power-up select Q or !Q by default (and in homemade discrete component ones it might even select both outputs as high!). I'd need to read the CD4013 datasheet to see if it has some way of avoiding this happening. How will this affect your circuit when playing?

Hope it's going well.

well right... it could be made easy.... but at this stage, i don't had enough knowledge and so i have to go with what i learnt....

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and which value capacitor should be used in push button section to less the noise effect for further feeding........

Hi,

Not sure what you mean by noise effect - push button debouncing? At a guess, you could try this using a 10k resistor and a 10nF capacitor, but maybe you'd need to up the values to 100k and 100nF.

**broken link removed**

There are several ways. You can also do it with the resistor in series after the button, and the capacitor after the resistor to ground. I doubt more complex ones using nand gates are necessary, they never have been for push buttons in circuits I've debounced.

I'm not sure about the clock rising edge part of this IC, never had to deal with rising edge input requirements, maybe you can't just make CLOCK and Dn high to get a high at Qn, but the 74HC173 looks like a good choice for the circuit I suggested using a keypad, 5 quad AND gates, and 5 HC173s (CMOS compatible!) and for your circuit - so at power-up there aren't random LEDs lit up when they should all be off. Only problem is they costs €4 - €5 each.

"well right... it could be made easy.... but at this stage, i don't had enough knowledge and so i have to go with what i learnt...." - I have a very bad frequency counter - CD4033s or CD4026s, a CD4060, two 555s, a couple of buttons and some seven segment displays - it fills a 16cm by 10cm board where a little PIC would have beena tenth the size and accurate. I will never use it in my life, I made it with the same reasoning as yours ("so i have to go with what i learnt") over two years ago. It is a waste of money and of time to do that. You really should check what the flip flop outputs do on power-up to make sure you aren't wasting money and valuable time on a circuit that might not work correctly...

Also, what do you think? I think with circuits that repeat the same building block over and over, maybe people only want to see stage 1 and the last stage, both from input to output, otherwise it's a spaghetti picture which is hard to understand. Look at schematics where the same thing is repeated many times, it should say and only show, more or less:
tictactoe player a square 1
...
tictactoe player a square 9

tictactoe player b square 1
...
tictactoe player b square 9

None of this is criticism, don't think that. I think these forums are/should be about helping each other and discussing ideas and doubts politely. I have some trashy first circuits and just want to suggest other options.

Hope it's all going really well, aliahsan951. If I can answer any other questions you have, I'll be glad to but I know very little, I'm no expert myself. Look forward to seeing the finished game.

d123 said:

ATTACH=142771

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Sorry, your attachment expired prematurely. Reason unknown.

To add an image, it should be successful to use this method:
1) Go Advanced
2) Manage Attachments, etc.

Thanks Brad.

if i use 7474, how is it different from 4013........

what i had studied is that it works same as 4013 but negative is applied on reset and preset ........

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d123 said:

View attachment 142780

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what should be the value of capacitor.......... 0.1uF?

A 7474 is old fashioned TTL and has a high input current of 1.6mA max to make its input a logic low. Then the resistance to ground at the input must be 0.4V/1.6mA= 250 ohms or less to make it low. The more modern CD4013 is Cmos that has no input current. Then a 10M resistor (or less) to ground will make its input a logic low.

Hi aliahsan951,

Use 0.01uF or 0.1uF - You'll have to experiment to see what works best for your circuit. Don't forget about the other ways to debounce, page 13 has the other RC configurarion you may be interested in.

Also, apart from Audioguru's explanation, preset and clear need a low signal to do anything (under 0.8V), they should normally be high (over 2V)/tied to V+ with a pull up of 10k or 100k. You could use it if you had some great need to do so, but then you'd need to use a lot more inverters to get a low from a high signal to trigger preset and/or clear.

Sorry, I misunderstood your circuit, I missed the obvious in whichever pdf it is that !Q has no use, so that's your POR and reset, I assume, to have a blank board. You've obviously thought about it a lot and have the logic right.

How many IC packages does this circuit use? I counted 32. Is that right?

Audioguru said:

A 7474 is old fashioned TTL and has a high input current of 1.6mA max to make its input a logic low. Then the resistance to ground at the input must be 0.4V/1.6mA= 250 ohms or less to make it low. The more modern CD4013 is Cmos that has no input current. Then a 10M resistor (or less) to ground will make its input a logic low.

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i am using TTL gates in circuit ...... people here says you could get the required output while mixing TTL with cmos components.... so that why i intend to use 7474

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d123 said:

Hi aliahsan951,


How many IC packages does this circuit use? I counted 32. Is that right?

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41 . i think so.....

remaining unused input should be grounded???........ which value resistor should i use to ground them

aliahsan951 said:

i am using TTL gates in circuit ...... people here says you could get the required output while mixing TTL with cmos components.... so that why i intend to use 7474

Click to expand...

No. The TTL input low must be 0.4V at 1.6mA maximum. The output low of a CD4xxx Cmos that has the same 5V supply as TTL is 0.4V at only 0.51mA minimum so there will not be a valid TTL input low. Your "people" are WRONG! Maybe they correctly say that a 74LSxxx TTL input can be driven by a CD4xxx Cmos? 74LSxxx has a lower input current low than a 74xxx ordinary TTL.
No again. A CD4xxx Cmos with a 5V supply needs an input high that is at least 3.5V but the output of TTL does not go that high, it goes to 2.4V minimum.

remaining unused input should be grounded???........ which value resistor should i use to ground them

Click to expand...

An unused logic input must be either high or low but never floating. You must analyse the logic to see if it needs to be high or be low. If it is TTL and it needs to be high then connect a 10k resistor in series with the input to +5V. If it needs to be low then calculate 0.4V at 1.6mA (250 ohms or less), or connect to 0V without a resistor. An unused CD4xxx Cmos input can use a resistor as high as 10M or more.

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aliahsan951 said:

i am using TTL gates in circuit ...... people here says you could get the required output while mixing TTL with cmos components.... so that why i intend to use 7474

Click to expand...

No. The TTL input low must be 0.4V at 1.6mA maximum. The output low of a CD4xxx Cmos that has the same 5V supply as TTL is 0.4V at only 0.51mA minimum so there will not be a valid TTL input low. Your "people" are WRONG! Maybe they correctly say that a 74LSxxx TTL input can be driven by a CD4xxx Cmos? 74LSxxx has a lower input current low than a 74xxx ordinary TTL.
No again. A CD4xxx Cmos with a 5V supply needs an input high that is at least 3.5V but the output of TTL does not go that high, it goes to 2.4V minimum.

remaining unused input should be grounded???........ which value resistor should i use to ground them

Click to expand...

An unused logic input must be either high or low but never floating. You must analyse the logic to see if it needs to be high or be low. If it is TTL and it needs to be high then connect a 10k resistor in series with the input to +5V. If it needs to be low then calculate 0.4V at 1.6mA (250 ohms or less), or connect to 0V without a resistor. An unused CD4xxx Cmos input can use a resistor as high as 10M or more.

EDIT: This site is messed up. It takes a long time to open and when I post something one time then it takes a long time to do anything and posts it twice. It has been like this for at least a week.

if i use 74ls74 instead of cd4013

Hi,

Thanks for answering. Are you making the PCB or designing it and having it made? What size packages will you use?

Not a great sicence, maybe, but for pull-down to ground on CMOS inputs I've always used 1k or 10k with no problems.

Read the two attached documents about pull-up/-downs, should be helpful.

I think you might want to read yourself properly about logic families, it seems you need to familiarise yourself with some of the differences. Semiconductor manufacturers have a lot of online resources and downloadable documents and catalogues that are very useful and helpful. Here are three examples:

View attachment Logic Guide sdyu001aa.pdf

View attachment How to Select Little Logic scya049.pdf

View attachment Implications of Slow or Floating CMOS Inputs scba004c.pdf

Enjoy!