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4013 ic test cicuit

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neptune

Member
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.
 

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BrownOut

Banned
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.
 
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neptune

Member
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 ?
 

MrDEB

Well-Known Member
circuit does not work

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
 

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audioguru

Well-Known Member
Most Helpful Member
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.
 

BrownOut

Banned
"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.
 
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MrDEB

Well-Known Member
here is a test circuit

for the 4013
plain n simple
should work??
 

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  • d flipflop.PNG
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BrownOut

Banned
I think your transistor need some kind of bias on the Emitter. I wouldn't trust the circuit as shown.
 

audioguru

Well-Known Member
Most Helpful Member
Laurier also got it wrong.
You must never let a Cmos input float. It needs a resistor to ground for it to be low.

Your High-Low switch does nothing because your D input is connected to +9V.
 

MrDEB

Well-Known Member
only got half wrong??

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 ??
 

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  • d flip flop2.PNG
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BrownOut

Banned
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.
 
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audioguru

Well-Known Member
Most Helpful Member
It is simple to wire the switch to be high-low and to wire the transistor to pull-up and the resistor to pull-down the CLK pin.
 

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  • CD4013 circuit.PNG
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neptune

Member
if i use +5v as a souce. and if i use flip flop multivibrator as a input to clock. will my circuit work ?

i have understood all of the thing above.
but how u people are using clock circuit. whats the role of capacitor connected to base of transistor ?
 

AllVol

New Member
if i use +5v as a souce. and if i use flip flop multivibrator as a input to clock. will my circuit work ?

That would be ideal. A 555 circuit will work. Since the 4013 is a CMOS device, 9V is okay.

i have understood all of the thing above.
but how u people are using clock circuit. whats the role of capacitor connected to base of transistor ?

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.
 

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MrDEB

Well-Known Member
Data shouldn't float ??

I revised so data isn't floating. I assume it shouldn't.
 

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  • d flip flop2.PNG
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audioguru

Well-Known Member
Most Helpful Member
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.
 

BrownOut

Banned
I'd probably leave it the way it is. Either way works. The resistor insures that the input never floats.
 

neptune

Member
http://C:\Documents and Settings\intex dem\Desktop

i have only pnp transistor available to me. i think i can reverse all polarity and use it in circuit.
this shall not create any problem
 
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neptune

Member
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.
 
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