Power dissipation calculation with CMOS 4093

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Joined Nov 15, 2011
Hi. I'm drawing up a circuit for a closset lamp, not unlike the fridgedoor alarm. This may sound trivial sure, but I have to consider it to be drawing low power the closset door is left accidentally open.

This is what I want:
- door closed, doorswitch cuts battery connection, null power.
- door opens, a led-strip is on for 1minute (accuracy not important).
- door accidentally left open, hopefully the circuit will draw low power.

So I've looked at CMOS 4093, quad nand schmitt triggered gates.
The quiesent supply current is very nice in the sub micro amps for 5V supply, but what I don't understand is the near 1mA output currents (?) as noted in the datasheet. Could someone explain this to me ? How do I read the I_OL I_OH in the electrical characteristic part of the datasheet? 1mA is not low-power if this is to be run by 4x AA battery.

My circuit is preliminary, and has some error with direction of D1 or C1/R1 switched places..


Last edited:


Joined Oct 2, 2009
When reading the I_OL and I_OH specs what you also want to look at is the Vo value under these conditions.

With Vo = 0.4V, this is saying that the output can sink 0.8mA and the output will still be LOW at 0.4V

Similarly, with Vo = 4.6V, the output is a respectable HIGH while still supplying 0.8mA.

These are the currents for a load resistor of 4.6V/0.8mA = 6K ohm.
If you make the value of the load resistor much higher than 6K ohm, lets say 100K, the output current will be much lower, 50uA in this case. For a load of 1M ohm, the current would be 5uA with a 5V supply voltage.