The CD4009 is an obsolete inverting buffer. It has two positive power supply pins and the IC will self-destruct if they power-up in the wrong sequence. Its output current is fairly high when sinking but is fairly low when sourcing.

A CD4049 should be used instead if you want a inverting buffer. Its output current is very high when sinking and is high when sourcing.

A CD4069 is an ordinary Cmos inverter with symmetrical sinking and sourcing current outputs.

 

Kris,
Connect the 4009's Vdd pin to 10v
Connect the 4009's Vcc pin to 10v.
Connect the 4009's Vss pin to ground/0v.
You will need to use a transistor driver like in eblc1388's schematic (R1, R2, R3, Q1); otherwise the input to the 4009 won't go high enough to toggle the output. You could substitute other general purpose NPN transistors for the 2N2222 like a 2N3904, 2N4401, etc.

Audiogurus' comments definitely apply.

 

You will need to use a transistor driver like in eblc1388's schematic (R1, R2, R3, Q1); otherwise the input to the 4009 won't go high enough to toggle the output.

you should read the datasheet and it will greatly help your understanding of the chip.

 

you should read the datasheet and it will greatly help your understanding of the chip.

I have read it.
When you get around to reading the datasheet, you will observe that Vcc needs to be less than or equal to Vdd.

Quote from National Semiconductor's datasheet:

Conversion ranges are from 3V to 15V providing VCC <= VDD.

[eta]
In other words, the IC is capable of translating a higher CMOS level down to a lower TTL level, but not the reverse.

 

I have read it.

you didn't read the relevant part.

When you get around to reading the datasheet, you will observe that Vcc needs to be less than or equal to Vdd.

Quote from National Semiconductor's datasheet:

 

you didn't read the relevant part.

So, what's the relevant part in your opinion?

 

So, what's the relevant part in your opinion?

the part that would help assess if the following is correct:

You will need to use a transistor driver like in eblc1388's schematic (R1, R2, R3, Q1); otherwise the input to the 4009 won't go high enough to toggle the output.

 

With apologies to the OP, this thread is turning into another worthless mess. What is the point to all this?

I have the National Semi CMOS data book from 1978. The relevant parameters are that the input swing to Vcc, and that Vdd be less than or equal to Vcc. The output is also good for 8 ma. Easier to use a 4049 as a buffer.

 

I have the National Semi CMOS data book from 1978. The relevant parameters are that the input swing to Vcc, and that Vdd be less than or equal to Vcc. The output is also good for 8 ma. Easier to use a 4049 as a buffer.

none of that is relevant.

the question is basically how do you go about figuring out what, if any, you need to do to interface the said mcu with a 10v cmos part.

 

Ok sorry for any confusion here, the 4009 I'm using is this one:
http://jaycar.com.au/products_uploaded/ZC4009.pdf

it's by ST Microelectronics. It says in page 2 that typical operating condition is Vdd = 3 to 20V. I'm using 5V. I tested using 9V to VDD but my output level is still the same (ie. the square wave has not increased in value than what is being toggled at the micro). I also put Vss to ground. But Vcc is floating.

 

I guess you didn't see where I suggested to put Vcc and Vdd to 10v?

See reply #22, above.

Unless you wish to become confused and annoyed, simply ignore millwood's "comments".

 

If the Vcc pin is left floating, there is no supply for the output stage. The inverter output can't be driven high.

 

Okay Ill put Vcc to 5V rail then (same as VDD is on).

Of course I must goto uni now so thanks and I'll check it out there!!