I have a TI SN74LS175J its 6 d flip flops on a chip. (datasheeeeettttt) I'm trying to wire it up in the simplest possible way and then step up in complexity from there. But it is failing in a failuremode that is confusing to me.
So, I figure I can just supply power to vcc and ground the ground pin and then all the q pins should be just at some random starting state of low or high. Maybe that logic is flawed? But at any rate I have tried powering at 5v, 6v, and 7v with a solenoid rated for 4.5 v attached on one side to one of the q or -q pins and on the other to ground. but the solenoid never does anything. I checked the q and -q pins with a voltmeter and they seem to be one high and one low. And then presumably the solenoid should be powered? I even put it in series with the voltmeter and it says ~5v. I'm boggled.
I have a feeling either I'm messing up something really simple or I have inadvertently fried this guy. Any tips, advice, or leads would be greatly appreciated!
We can't comment on this issue until you post a schematic of your test circuit.
Is this your first rodeo ?
But at any rate I have tried powering at 5v, 6v, and 7v with a solenoid rated for 4.5 v attached on one side to one of the q or -q pins and on the other to ground.
?
Seriously ?
You're actually trying to power a solenoid from a TTL chip .
Are you crazy ?
I mean. It's my second rodeo. I don't know much about much yet. Still feels like the first rodeo.
So what I'm understanding is that the issue is because the solenoid stores energy and then that discharges through the chip? So for the purposes of just testing this if I was using something that wasn't a solenoid for a read out this setup would be ok?
With logic chips all inputs should be defined. That means every D and CLK and #CLEAR inputs need defining.
The the 74LS family LOW inputs can be directly connected to GND (0V), but HIGH inputs must be connected to
+5V via a 10k resistor (many inputs can share such a resistor).
I'm surprised you don't use a modern CMOS family, such as the 74HC series, where high inputs can be tied
directly to +5V (much easier to use). Also 74HC family can be powered from 2V to 6V, whereas the 74LS series must be powered from an accurate 5.0V (+/- 0.25V).
For quick testing and proof of concept inputs are often left floating, but its not a good idea to do this otherwise,
as various issues can arise (high power consumption, noise pick-up, oscillation even).
An LED typically needs 1 to 20 milli-amps. A 5v solenoid probably needs an amp or so (which is much more than a logic chip can provide.)
If it’s a true “ls” ttl chip, inputs that you want to be a “1” can usually be left unconnected. This isn’t true of newer cmos chip like the 74hc series.
Thanks all! That was very very helpful. And yes, what i was getting at but did not know how to articulate was whether all inputs needed to be defined. I will also definitely look into the 74HC series as that would probably suit my needs better.
In reality I only need one flip flop. Not four but this IC was the one that would arrive the fastest so I just went for it. If I need only a single d flipflop is there a better option?
So I ended up getting the 74hc74. And I have it breadboarded as shown here. But it doesn't seem to be working and I'm not sure exactly where to go with my trouble shooting.
I tested that the IC doesn't have a short using diode mode on my multimeter. The Q and not Q pin do seem to be opposite. I have tried taking out the pushbuttons and testing them separately and they seem fine.
I suppose I will now try to use the multimeter to check that the buttons do actually make the pins go high? Is there anything else I should check?
Not sure if pictures are the best way to show this. But this is what I got. Power supply is set at 4.8v and I think current is 0. Maybe current limit should be higher?
ok. I just tried changing R3 to 220Ω and putting a .1uF capacitor just directly from the negative rail to the positive rail on one side and no cigar. Do I need to put the decoupling capacitor in a specific place?
Why don't you start over and wire it according to the pins in the schematic you posted.
Also, as already mentioned the led needs a resistor and now you need to test it because it
could be fried or backwards.
Why are you using a switch to pull pin-12 to +5V when it's already pulled up by the resistor ?
Move the wire on the switch to GND so when you press it , it pulls pin-12 DOWN.
Also, write down all the voltages before and after pressing the clk button.
