CMOS buffer

[Bogdan]

Hello everyone,

I'm pretty much a newbie when it comes to electronics, but now i'm making my first steps and i encountered a problem.
I wanted to use a CMOS buffer for something ( i will go in more details if necessary) the 4050B
I connected the Vdd(3.3 V) and Vss(0 V), simple enough. Now i connected on the first input a 3.3V input..here is the problem
My output is 3.3 volts until i connect something, after i connect some LED or something else the voltage drops to 1.7 V or something like that.
I want to know why this happens and what can i do to fix it, or use to get my desired result.

I suspect that what i'm asking here is 4th grade physics but i simply don't understand what is going on..

Thank you

[hamster_nz]

The rough answer is that the output of a CMOS buffer has output impedance - you can't draw an infinite amount of current frm it.

As per page 3.10 of http://www.lns.cornell.edu/~ib38/teaching/p360/lectures/wk09/l26/EE2301Exp3F10.pdf, the output impedance of a CMOS gate is about 1kOhm. (so at 3.3V you might get around 3.3.mA)of current) - even if you connect the pin to 0V. An NXP 4050B looks to be even higher (maybe 5kOhm).

However, the 4050B is good at sinking current (12mA @ 5V for NXP part) - so connect the LED between the positive power rail, and add an appropriate current limiting resistor then the output. But you might want then to use an inverter so the LED is only on when you have a high input.

[alsetalokin4017]

The TI CD4050B data sheet says that at Vcc of 4.5 volts the chip can only sink 5.2 mA per gate in the LOW state at room temperature and can source way less than that in the HIGH state. It can sink 16 mA at Vcc of 10 volts. It's definitely designed for high impedance loads.
http://www.ti.com/lit/ds/symlink/cd4050b.pdf