CMOS power use?

[cliffclavin]

Hello.

I am hoping one of you smart people could help answer a question for me. I am nearly finished building a three stage timer/ controller out of CMOS 4000 series IC's - I have built & checked each section of the circuit, and it works ok!

A 4093 Schmitt Nand Gate provides clock pulses to some 14 other 4000 series chips. The power is 9v dc from a battery, but I will upgrade to a transformer when its completed

My question is about power requirements / dissipation for the circuit. I have attached a pic from the 4017 datasheet, which shows the power dissipation requirements for the 4017.

Could somebody please look at the calculation and tell me the power requirement for the 4017, and how you worked it out - so I can work out the power requirements for the other IC's in the circuit.
The formula is :

Typical Formula for P (uW) at 10v dc

2200 Fi + SUM OF (FoCl) x Vdd2

where Fi = input freq (Mhz)
Fo = Output Freq (Mhz)
Cl = load Cap (Pf)
SUM OF (FoCl) = sum of outputs
Vdd = supply voltage

This is so the power supply has enough power, and so I can work out a value for a fuse.

Any takers ? Who's able to explain how to work this out?

View attachment 36665

 

[audioguru2]

Hi Cliff,
Cmos draws power only when it is driving something with current (like LEDs), when it is linear (the input Mosfets of your 4093 oscillator) and during switching, or when it is switching at a high frequency (charging and discharging internal and stray capacitance).

You could spend weeks calculating the frequency and supply current of every flip-flop in your counters, then realise that only the first few that are operating in the magahertz range draw a few milliamps, the rest draw only a few microamps or less.

Since the output of your 4093 master clock is driving the input and wiring capacitance of 14 ICs, it will hog most of the low current.

 

[MP1]

What you are looking for is a fanout spec, but output pulses to 14 other gates? You must be ganging them up to get this from a quad output gate. And then, you must also be powering the other gates, too. Have you thought about just hooking up a multimeter between the chip and the battery and reading the current consumption?

MP

 

[audioguru2]

Cmos doesn't have a fanout spec. Its inputs don't draw any current since they are the inputs to Mosfets. A single ordinary gate can easily drive 50 or more other Cmos inputs and is limited only by the capacitance of the wiring if the speed is high.

 

[MP1]

You just contradicted your last post. Capacitance is one of the things you are adding to the circuit by sourcing lots of other chips. Just laying them out on more circuit board space effects capacitance.
Something must power these chips. There will be differences in current consumption for all the chips which is dependent upon how they are used.
Like I said, it would be simplest for his application to just use a DMM to measure the current from his power source.

BTW...Everything has a fan out limitation. Fanout buffers are used in circuits where signal quality and skew are critical.
What a wonderful chip it would be if one could source unlimited amounts of other chips with no changes in the operation of the circuit.

MP

 

[audioguru2]

Yeah,
Cliff is making a timer with a 4093 as a low-frequency RC oscillator, not a high frequency crystal oscillator. It can clock only 14 other Cmos ICs with no problem about fanout.
At the slow speed of the counter ICs, I doubt you could measure any power supply current for them.

Instead of using two handfulls of ICs, maybe Cliff could have used a single 4060 oscillator/counter or a 4541 oscillator/timer IC. At fairly low frequencies their supply current is typically measured in femtoamps! (0.010 uA)