Post your schematic.

Originally Posted by financecatalyst

Hi
I have 4013 which toggles the output Q pin upon applying pulse on the clock pin. D and Qbar are joined. Pin 4 - 6 are joined and earthed. It works just fine when no load is on the Q (output pin) and it goes crazy when there is load attached via transistor which is bunch of LED's requiring 300mA. Tansistor used is 9014 directly attached to Q pin to drive 300mA bunch of LED's.
when load is attached, it has very irregular behaviour when pulse is applied to clock pin.

Can someone help me please?

Thanks (alot!)

hi,
Whats the value of the resistor connected between the 4013 Qpin and the base of the transistor.???

If the Q-bar output drives the D-input then they must not be shorted by the direct connection of the base of the transistor. a series resistor must be used.

A CD4013 has an output current of about only 7mA if its supply is 10V. It can drive a 1k resistor. If the supply voltage is less than 10V then the output current is lower.

But the transistor has a load of 300mA so it needs a base current of 30mA for it to saturate well. It won't work well with a base current of only 7mA.

Where's the base resistor?

With a supply voltage as low as 3V the output of the CD4013 has a current of only 1mA or less which is too low to drive the base of the transistor when the transistor has a load of more than about 15mA. Since the currents are so low a base resistor is not needed.

The transistor needs a base current of 30mA for a load of 300mA.

Very cheap? Transistors & resistors are very cheap. Are you building 1000's?

Use an N-channel MosFet instead of a bipolar transistor. Use a gate resistor of 100Ω.

Not many FETs will turn at 3V?

Originally Posted by blueroomelectronics

Not many FETs will turn at 3V?

That one is perfect for the application:

V(BR)DSS=30V, RDS(ON)=0.08Ω, ID=3.5A, VGS(th)=0.7V, package SOT23

ZETEX ZXMN3B14F.

Boncuk

If you look at the datasheet of a CD4013 you will see that the Q output is completely separate from the Q-bar output and you can short it without affecting the operation of the IC unless the power supply voltage is high enough to cause a high short-circuit current.