ok, I think im getting confused, where now using D1 for the switch? and where connecting the other end of the switch to ground?
Alright, time to back away from the breadboard and splain what's going on here. What is a D flipflop? when it gets a clock pulse, the value at the input appears (and remains--that's the part you're interested in) at the output. It is a one-bit memory. All you are accomplishing with the momentary switch in this circuit is sending clock pulses. That's it. The flipflop is providing you with a convenient complementary output (/Q1, read "not Q1" or "complement Q1"), which you are feeding back to the input. It is this very feedback that will accomplish the toggleing you want to happen. It guarantees that the next state will always be the complement of the current state. Again, the switch is just to provide the clock. And yes, the other end connects to ground. With these 74xx ICs, an unconnected pin typically 'floats high.' when you ground it you are forcing it low. Remember too my comment about this device being edge triggered. it will interpret the release of the switch as its clock pulse. D1 is an input, Q1 is the corresponding output, and /Q1 is the corresponding complementary output. These three items comprise a flipflop, of which there are four on your chip sharing a common power connection and enable input. So stop thinking about switches--what you've constructed is a trivial state machine--it just transitions from ON to OFF when it gets a clock pulse. If anything, it's the transistor you're going to tack on to the output that accomplishes the switching, based on the input it gets from this handy state machine. Now we gotta get the state machine working before we deal with switching your load.
ok, the LED starts out dimly lit but visibly lit I hit the button once the led goes off untill I release the button then it goes back to being dimly lit I hit it again it gets brighter but then I let go and it goes back to being dimly lit I hit it again, same thing happens I hit it once more, same thing I hit it again, it goes completely off until I release the button then it goes back to being dimly lit. and the cycle repeats itself
Hey, I think we have joy. Next take your transistor, and wire the base to your output through a resistor. Stick the LED in between +5V and the collector, ground the emitter, let's see if we can switch your LED with thte transistor now. Use the usual current limiting resistor as well, please.
im basically getting the same results as I was getting without the transistor except the LED is fully lit and gets dimly lit when I push the button
ok, next step, tkae the transistor off the flipflop, connect your load as before, and see if it switches properly when you connect the base resistor directly to +5V and ground.
I dont quite understand what you mean I took the transistor away and You want me to connect the 1k resistor to pin7 and the other end of the resistor goes to +5V and then what?
can you use just the transistor to switch the CCFL? No flipflop in the circuit? I'm trying to test your switching stage separately, so I know where to look for the problem.
ok so the base goes to the 1k resistor which goes to the +5v and the collector goes to the LED and the emitter gets grounded?
ok, I connected the 1k Resistor the base, the collector to the CCFLs inverter and the emitter to ground and the CCFL is lit up
right, the transistor seems to be properly working, let's focus on the flipflop then. Any way you can take a meter and measure the voltage at your flipflop output in the ON and OFF states? I would expect values withing a few tenths of 5 and 0 volts respectively.
its about 4.5 normally and when I hit the button it went to 4.61 and stayed there untill I let my finger off the switch
you should se a zeroish voltage in the off state. Please check just the flip flop with no transistor or other load on the output.