Help me Run this digital circuit

[neptune]

i want to store 1bit flag, so i built this.

clock is always runnig as shown , input arrives in pulse so output becomes high.
but when input is zero , it output remains high

 

[ericgibbs]

i want to store 1bit flag, so i built this.

clock is always runnig as shown , input arrives in pulse so output becomes high.
but when input is zero , it output remains high

hi,
It will be high output, because you have the PREset pin held high.

Look at this modified asc

 

[neptune]

ok it is running but , current through R3 AND LED is 3.6mA, if i am not wrong it should, I=V/R which is 9mA.

and also output voltage of Q is 5V, but in datasheet it is shown Min. 9.95V at Vdd-10V

 

[ericgibbs]

ok it is running but , current through R3 AND LED is 3.6mA, if i am not wrong it should, I=V/R which is 9mA.

and also output voltage of Q is 5V, but in datasheet it is shown Min. 9.95V at Vdd-10V

hi,
You are operating the 4013 at 5V, right click on the 4013 symbol and then edit the Spice line from VDD=5 to VDD=9. the LED current will be higher. OK

EDIT:
You are showing a Diode symbol for the LED. Use key F2 , select the LED symbol, then right click the LED symbol and then choose the correct LED from the displayed list.

 

[neptune]

ok silly mistake by me.
now current is comming 7.5mA acrross R3 , does LED also have resistance which is to be considered ?

Also when i change value of R3 voltage value at Q output pin changes.
it decreases with decreasing resistance value.
when R=10ohms Q=6.8V
when R=100ohms Q=8.8V
when R=1000ohms Q=9V

 

[ericgibbs]

ok silly mistake by me.
now current is comming 7.5mA acrross R3 , does LED also have resistance which is to be considered ?

Also when i change value of R3 voltage value at Q output pin changes.
it decreases with decreasing resistance value.
when R=10ohms Q=6.8V
when R=100ohms Q=8.8V
when R=1000ohms Q=9V

hi,
You are still using a diode as an LED, that diode has only a forward voltage drop of 0.7V, use a RED LED, which is ~2V forward drop. [your current results will be misleading with a diode]

The falling voltage, when the current load is increased, is the point I was making yesterday in your 4013 thread, you will not get 15mA at 15V.

 

[neptune]

hi,
You are still using a diode as an LED, that diode has only a forward voltage drop of 0.7V, use a RED LED, which is ~2V forward drop. [your current results will be misleading with a diode]

hey man , i replaced it with LED, i downloaded and checked this it shows LED in it :?

The falling voltage, when the current load is increased, is the point I was making yesterday in your 4013 thread, you will not get 15mA at 15V.

so what this has got to do with falling voltage. the load is controlling voltage drop accross Q and gnd ? , what if i connect many transistors with load like that in parallel or sereis with output Q, then load will change and it will change all voltage drop accross output !
this is not typical characterstics of driver circuitry !

 

[ericgibbs]

Hi,
Because you are using an external transistor to drive the LED, you will only draw the small Base current for the transistor from the Q output

If you tried to drive the led and its resistor directly from the 4013 output, the output voltage would fall as the current increases.

Try as a test only, drive the 100R and LED directly from the 4013 and check the voltage at the Q pin.

 

[neptune]

> Without using transistor the output voltage at Q has fallen from 8.9V to 2.9V and current has fallen from 73mA to 22mA.

if i consider Q as voltage source , this all things seem absurd, breaking the rules. what should i consider Q ?

 

[ericgibbs]

> Without using transistor the output voltage at Q has fallen from 8.9V to 2.9V and current has fallen from 73mA to 22mA.

if i consider Q as voltage source , this all things seem absurd, breaking the rules. what should i consider Q ?

hi,
If it was important to keep the Q output at its highest output voltage, I try to keep the load current to less than 5mA.
This is easily done by using a transistor as a driver for an external LED's in the way you have shown it.

 

[neptune]

But i am not interested in keeping voltage at highest , i am interested in keeping it constant. like Q should act as voltage source. as battery

 

[ericgibbs]

But i am not interested in keeping voltage at highest , i am interested in keeping it constant. like Q should act as voltage source. as battery

The simple answer you cannot keep the output voltage constant while increasing load current, its not a constant voltage source.

Look at this clip from the 4013B datasheet, showing the fall in output voltage for a given current.

This 4013 datasheet is a better quality print than your original.

 

[neptune]

what is Vo in conditions ?
and what is the resistance of LED because current through R3 (1k) is still comming 7.5mA and not 9mA !

 

[ericgibbs]

what is Vo in conditions ?
and what is the resistance of LED because current through R3 (1k) is still comming 7.5mA and not 9mA !

Vo is the output voltage at the stated output current.

The LED 'resistance' is the voltage dropped across the LED at the current thru the LED.

So for a RED led, R equivalent = 2V/0.009 =222R

NOTE: the R equivalent will change with current, its a semiconductor device.

Also are you allowing for the Vbe voltage drop across the transistor.?

 

[neptune]

its not in my hand to allow voltage drop across Vbe.

 

[ericgibbs]

its not in my hand to allow voltage drop across Vbe.

I do not understand this reply.???

 

[neptune]

Also are you allowing for the Vbe voltage drop across the transistor.?

Vbe voltage drop will happen naturally in transistor , it is not in my hand i guess

 

[ericgibbs]

Vbe voltage drop will happen naturally in transistor , it is not in my hand i guess

I understand.
You will always have a Vbe drop of approx 0.65V and a Vce sat of approx 0.15V.

Have you now got the circuit working to your satisfaction.?

 

[neptune]

everything is working right but voltage drop at output Q will be beyond my understanding :-|

 

[ronv]

The 4013 is not strong enough to supply the higher current so it's output voltage drops.