counter
you see, i have this circuit
simulate this circuit – Schematic created using CircuitLab
however when i add another more 74161
the first one stops working and it just alternates between Qa and Qb LEDS i think it's because of the current needed, however im using a pc 600 watts power supply, any idea of what's going wrong?
You could do it with your SN74LS393 counter chip by using the correct output but you cannot simply connect the crystal like that. You need to turn the output of the crystal into a digital pulse.
You can do this in any number of ways but the most simple is a "Pierce Oscillator":
U1 is an inverter, something like a 74HC04 or you can find tiny SMD parts in SOT/SC-70 cases with a single inverter. C1 and C2 should be chosen according to the capacitive load your crystal expects, check the datasheet. R1 is a feedback resistor, you could start with a high value and work downwards until you find it is stable.
Another common way to do this is using D flip-flops, which is essentially how the internals of your counter chip will be working. The Wikipedia page on frequency dividers illustrates a divide-by-four counter nicely:
I have previously done this using 74HCT74 ICs.
At first glance, this appears to be a power problem. It appears that your "5 V" power supply droops to less than 4 V in operation.
Perhaps the power supply is inadequate. So power your circuit from some other wall-wart or other power supply that can supply at least 10 times as much power as you think you need (at least 10 times as much current at 5V). (Another possible explanation: perhaps the wires and connectors and traces between your power supply and the chip's GND and power pins have too much resistance; so replace them with thicker wires and less oxidized connectors).
No matter how good the wall-wart is, it can't do anything about the inductance in the wires between the wall-wart and your PCB. That inductance creates an impedance significantly larger than measuring the resistance of those wires might suggest. You need a "bulk capacitance" capacitor on your board. There are ways to calculate how much you need, but many people just throw on a 1000 uF capacitor on their first prototype -- usually overkill, but not worth trying to find out how much less we actually "need". "Logic Supply Decoupling Techniques", "General PCB Design Guide", "Decoupling Capacitance Design Rules", "Power Distribution System Design Methodology and Capacitor Selection for Modern CMOS Technology", etc.
No matter how good a bulk capacitor is, it can't do anything about the inductance in the traces between that capacitor and each IC. That inductance creates an impedance significantly larger than measuring the resistance of those traces might suggest. Put a 100nF decoupling capacitor across the GND and power pin of every IC. "Decoupling Capacitors Explained -or- If You Like It Then You Shouldn’t Let It Ring Like That", "What is a decoupling capacitor and how do I know if I need one?", "Decoupling by Example", etc.
Unfortunately, these 3 problems bite extremely often, and cause all kinds of strange and not-quite-working-right behaviors.
(How do we add "decoupling and bulk capacitance" to the Electrical Engineering site FAQ ?)
Best Answer
OK, first things first. Multiply R1 and R2 by a factor of 100, and reduce C2 by a factor of 10. Frankly, I'm surprised that your 555 will work reliably with such small resistors.
With the 555 done right, look at the counters. As has been suggested, you need a decoupling capacitor for each counter. You also should add a larger capacitor, like 10 to 100 uF between +5 and ground near the counters. Depending on how you have your circuit wired up (long, small-diameter wires from the power supply?) this may or may not be important, but it won't hurt - as long as you don't connect the caps backwards, of course.
You are driving your LEDs wrong. TTL and LSTTL do not do well sourcing current. That is, they don't provide much current for loads to ground. What they do much better is sinking current, so your LEDs should be connected like
simulate this circuit – Schematic created using CircuitLab
Here I've shown the first LED as being On when the logic from the chip is 0, which you may find confusing. So the obvious solution is to add an inverter and drive the LED that way. If you do add an inverter, you don't need the first LED.
Another thing to be careful about is your ground connections. I'm not sure what you're using for construction. If you're using a solderless breadboard, make sure your connections from pin 8 to the ground strip is as short as possible. Also, make sure your 0.1 uF caps (ceramic, please) are connected directly to the sockets immediately adjacent to the IC pins. Do not put the capacitor somewhere else on the board and then run jumpers to the ICs.