i created my own test circuit for 4013 ic.
what i have done is use of voltage devider circuit. it devides 9v battery to 3.5v and 1.5 v which are required Vih logic'1' & Vil logic'0'.
for output i have connected pnp transistor because Ioh= -1.36ma. so connected to base which will drive led. i left open other pins.
i have not tested this circuit yet but i want u guys to tell me if this circuit will work or not.
Are your values absolute max/min ratings? ( I'm too lazy to look them up ) A simple switch will not work for the clock input. At a minimum, you need a "debounce" circuit. Your indicator circuit will light when your DUT is in reset. Q=LOW. Also, your divide resistors don't produce the voltage you want.
they are absolute maximum ratings.
"Your indicator circuit will light when your DUT is in reset. Q=LOW." why is that. should i connect npn transistor up there.
and whats wrong with voltage devider circuit is that the value of resistor ?
just for fun I put your circuit into Tina
attached are results.
you never achieve a LOW for data input
your switch won't clock properly as it will bounce
just use a 555 clocking very slow to see how the 4013 works
attach LEDs to the outputs and data in to see what states the pins are
Your 4013 is not powered so it won't work.
Its output current depends on its supply voltage.
Its input voltages of +3.5V high and +1.5V low are only when its supply voltage is +5.0V, not when it is +9V.
Your +3.5V and +1.5V settings will be reduced as your 9V battery runs down and should be powered from a regulated voltage. A low dropout 5V regulator should be used.
"Your indicator circuit will light when your DUT is in reset. Q=LOW." why is that. should i connect npn transistor up there.
and whats wrong with voltage devider circuit is that the value of resistor ?
You definitely need to reconfigure your transistor. I was confused originally by the lable "9V" on the ground side of the battery. I think other's have answered some of your questions.
the high low switch is correct I am pretty sure
the 9v is connected to D input for high
when switched to low the resistor R3 is grounded.
I might should have R3 connected to 9v then ground w/ switch??
revised schematic. Probably have too many resistors connected to transistor.
attempt to keep current low I guess??
started inputting into Tina to experiment w/ different configurations.
not right I am pretty sure but ??
Naw, the hi/lo switch won't work. The D input should be connected between the resistor and switch. Look at how you have the clock connected to R2. That's how it should be done ( think of the transistor as analogous to the switch ).
And R3 is going to prevent the clk signal from getting to a reasonable low. By R3, I mean the 10K on in the emitter, not the 1K one in the base circuit.
The resistor and capacitor together make what is known as a RC timing circuit. During the time it takes for the cap to charge, the mechanical vibrations of a switch are ignored, resulting in a clean signal to the base of the transistor.
You do not need the 10k resistor at the D input of the CD4013. The switch should be wired so it connects the D input to either +9V for a logic high or to 0V for a logic low.
i am new to ic working so please help.
in 4013 datasheet it is given that Ioh= -1.3mA if conditions Vdd=10V, Vo=9.5V .
and Iol = 1.3mA if conditions Vdd=10V, Vo=0.5V
why you all have connected q output directly to +9v through 330 resistor. will not +9v battery push Iol(1.3ma) back current when q is low.
rest of the circuit is understandable.