4013 ic test cicuit

[BrownOut]

That's a good find. If Iol is only 1.3mA, then the IC may not drive the LED's. You might need a driver circuit.

 

[audioguru]

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

No.
If you allow a voltage drop of only 0.5V at the output of a Cmos IC with a 10V supply then its minimum output current is only 1.3mA but is typically 2.6mA which is not much current.
But if you connect a 2V red LED directly to the output then the voltage drop of the output transistor is 7V and the typical current is 15mA (7.5mA minimum).

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?

The 330 ohm resistors limit the current in the LEDs and are not needed.
When the output tries to conduct 15mA in an LED then the 330 ohm resistor will limit the current to about 10mA because then the 330 ohm resistor will have a voltage drop of 3.3V, the LED will have a voltage drop of 2V and the output transistor in the Cmos IC will have the remaining 5.7V voltage drop.

Look at the graphs of output current on Texas Instruments datasheets.

I wonder why bi-colour LEDs are used? One colour will never light up!

 

[neptune]

i have tested the circuit physically and is working good. and yes i dont require 330 ohm resistance.
and also transistor is not required in cock input.

 

[neptune]

now i am currently using cmos 40174 hex d flip flop. this ic is same as 4013 and is working fine, the circuitry involved is same as above.
but when d=0 then q=1 and vice versa, which should not be the case . also sometimes its output is unpredictable, means sometimes high or low.
i have left its unused pins open which also i had done with 4013, i think that should not create problem. please tell me what is the problem ?

 

[MrDEB]

all unused inputs must be connected

just ground all of the unused inputs.
Never allow an input to float (unconnected) as the outputs will be unpredictable as you found out.
the input dosn't know if its supposed to be high, low, or in between.
yes there are 3 states of inputs. I think its called high impedance??

 

[audioguru]

just ground all of the unused inputs.

No.
Some unused inputs must be high and other unused inputs must be low. Look at the datasheet to see which.

Never allow an input to float (unconnected) as the outputs will be unpredictable as you found out.

True.

Yes there are 3 states of inputs. I think its called high impedance??

No.
Logic ICs do not have 3 states of inputs. The inputs are either high or low.
Some special logic ICs have 3 states of outputs where the 3rd state is a high impedance.

 

[MrDEB]

what is a floating input?

it takes on a state of its own due to internal capatance of the IC?? as I understald it??

 

[audioguru]

it takes on a state of its own due to internal capatance of the IC?? as I understald it??

Logic ICs have two output transistors. One pulls up the output and makes it high. The other pulls down the output and makes it low.
A three-state output turns off both output transistors to make its output a high impedance that does nothing.

Three-state outputs connect to a buss that is used by multilple logic ICs that all have three-state outputs. Only one device is active while all the other outputs are a high impedance. Then there is no interference between outputs.

 

[neptune]

No.
"Some unused inputs must be high and other unused inputs must be low. Look at the datasheet to see which."

so which pins according to you should be connected to high or low(4013).
how do we look for this in datasheet ?

 

[audioguru]

My Cmos Cookbook says that when the CD4013 is clocked then the Set and Reset inputs must be at ground.

 

[neptune]

and what about q and q~ outputs ??

 

[audioguru]

and what about q and q~ outputs ??

Q and Q-not are outputs. You must never short an output to ground.

 

[ericgibbs]

Q and Q-not are outputs. You must never short an output to ground.

hi agu,
I have told him many times on the Chat line not to connect the outputs directly to +V or 0V.

See his other thread 'ring counter'...

 

[neptune]

ok where then on earth should i connect q~, for q is already connected to led.

output given still reversed,meaning D=0 -> Q=1
& D=1 -> Q=0.
please help

 

[ericgibbs]

ok where then on earth should i connect q~, for q is already connected to led.

output given still reversed,meaning D=0 -> Q=1
& D=1 -> Q=0.
please help

The /Q outputs are not connected.

The Q's should be connected to the LED's cathodes, using approx 1K resistors to +V from the LED anodes.