Hello all,

I am planning to use the ICM7566IPD 555 timer chip from maxim, and wanted to measure its power dissipation, and needed some help in how to do it.

I know that P=IV, so I was imagining putting a known voltage through the VDD and gnd terminal and measuring its current through a multimeter (I am using the 34401A digital multimeter). However, current is showing as 0 when I put in 5Vdc through.

Is there a more accurate way? Am I using the wrong method?
Any help will be appreciated. Thanks!

 

Why do you want to do this?

A modern 555 timer, CMOS version, will draw an infinitesimally small current, on the order of microamps.

Power consumption will be measured in microwatts.

Unless you have a significant load connected to the OUT terminal?

 

It should be straightforward; I do it routinely to measure the power consumption of low power circuits. Just put the ammeter in series with the device, as you indicated you're doing. Then put a voltmeter across the power supply lines to the chip. I doubt the ammeter's burden voltage (spec is < 0.1 V) is causing a problem, but you might have to increase the power supply's output voltage to counteract it if it is. The meter you referenced appears to read to the nearest 0.1 μA, so hopefully that's enough resolution to tell you what you want to know. If not, get a more sensitive ammeter or build an amplifier.

 

hello, thanks for your reply everybody-

just to reply, tom66, I wanted to measure the power consumption of this chip, because according to the datasheet, the maximum power consumption rating for this device was 500mW. Although this was the absolution maximum, not knowing the value of power consumption, I was thinking that a typical power consumption for this device was in the order of some mW, and therefore wanted to check.

My load will be a load resistor for now- in the end, it will connect to an RLC network.

Thanks for your feedback.

And someonesdad, thanks for your reply as well. So it seems that my multimeter is not reading it properly. However, I actually tried hooking my multimeter up to a test circuit as seen in this forum-
http://www.allaboutcircuits.com/vol_6/chpt_2/7.html, to literally a simple setup like the one in this website, with a 1Kohm resistor. However, the multimeter still gave a reading of 0 current. I hooked up the multimeter as seen in the attached image.
http://imageshack.us/photo/my-images/200/multimetercurrent.jpg/
thanks for the continued help!

 

hello, thanks for your reply everybody-

just to reply, tom66, I wanted to measure the power consumption of this chip, because according to the datasheet, the maximum power consumption rating for this device was 500mW. Although this was the absolution maximum, not knowing the value of power consumption, I was thinking that a typical power consumption for this device was in the order of some mW, and therefore wanted to check.

My load will be a load resistor for now- in the end, it will connect to an RLC network.

Thanks for your feedback.

And someonesdad, thanks for your reply as well. So it seems that my multimeter is not reading it properly. However, I actually tried hooking my multimeter up to a test circuit as seen in this forum-
http://www.allaboutcircuits.com/vol_6/chpt_2/7.html, to literally a simple setup like the one in this website, with a 1Kohm resistor. However, the multimeter still gave a reading of 0 current. I hooked up the multimeter as seen in the attached image.
http://imageshack.us/photo/my-images/200/multimetercurrent.jpg/
thanks for the continued help!

Power dissipation isn't the power that may be drawn by the chip and put into a load, it's the power wasted by the chip when it drives a load.

A 555 only dissipates power when it drops a voltage with a current flow on its output; ignoring internal logic current which is negligible for most purposes. So say you put 5V into it and get a 4.5Vp-p wave out, and you draw 200mA. This would be 0.5V * 0.2A = 0.1W so allowable dissipation. Next say you get a 2Vp-p wave out by drawing over the maximum; around 300mA (notice the output has sagged because you are drawing a lot of current.) That would dissipate 3V * 0.3A = 1W and your chip would begin frying!

At high frequencies the discharge pin current may no longer be negligible; check that you are not exceeding ratings here, too.

A simple way to check if you are exceeding ratings is using your fingers, if it is too hot to touch it is dissipating too much power.

 

thanks tom66.

Yes, what I meant to say was power dissipation. My mistake.

I will be dealing with frequencies lesser than 600Hz, at max up to 1Khz, but that will be all.

I will use the "finger" trick as well

meanwhile, more help on what i might be doing wrong with measuring current in my multimeter will be great as well.

thanks again.

 

thanks tom66.

Yes, what I meant to say was power dissipation. My mistake.

I will be dealing with frequencies lesser than 600Hz, at max up to 1Khz, but that will be all.

I will use the "finger" trick as well

meanwhile, more help on what i might be doing wrong with measuring current in my multimeter will be great as well.

thanks again.

The problem is the 555 may be drawing less than 1µA.

 

I don't want to seem obtuse, but I almost remember a calculation involving [the charge required to raise the voltage on the gates of mosfets, and then the discharge of those capacitances] as a major factor in high frequencies. I think it was in a CMOS oscillators tutorial. Bottom line, any current that flows, contributes.

Not much of a problem at 1Khz.

If you want more leakage current, you can use a hair dryer to heat up the chip.

Still can't get the "Big Grin" smiley to go where the cursor is

 

If your meter can't measure any current, here are some possible things to think about. Your meter might be broken. Or, the current is below the level your meter can measure. Or the chip you're using draws significant power only when switching and the meter can't measure this (though try an AC+DC current measurement anyway).

One technique is to put a resistor in series with the chip's power supply and see if you can measure a voltage across it to quantify the current. You can also use a scope to see if there are current transients. If this shunt resistor has to get large to get a decent signal, be aware that the impedance of the scope or voltmeter might affect your measurements.

 

You can use the impedance of the meter to your advantage, too. I have a 10meg Fluke and I can put it in series with a load, set it for millivolts, and call it a self displaying 10 meg resistor. I can get down to nanoamps with that method.

.1 millivolt reading over 10 megohms = 10na

 

Great idea #12! That's never occurred to me -- but it's obvious after you mention it. I'm writing a guide to DMMs right now and I'll stick it in there and credit your post. Or PM me your name and I'll properly attribute you. By the way, for 0.1 mV at 10 Mohms, I get 1e-4/1e7 or 10 pA resolution. I guess my Fluke meter is better than yours...

 

Now I'm getting 100 picoamps for .1 millivolt across 10 megs!?

I should probably turn the lights on because my calculator is solar powered.

pm with real name has been sent.

OK. Did it with a pencil. 10 pa looks good.

 

Hey guys, thanks for your help, as always.

I actually went ahead and measured the power dissipation using two different methods.

First was using #12's suggested method, but using the impedance of my scope instead.
In my school is a Tektronic MSO3000, and the probes for it can exhibit a 1M or 10Mohm impedance.
Therefore, I set the impedance to 10Mohm, and went ahead and measured the voltage of the power rails (Vdd to gnd).
It was very "fuzzy", and therefore I used the "max" measuring method to measure the highest voltage it produced.
After that, I used v^2/R to calculate the power dissipation:

(11.2mV)^2/10Mohms = 1.25 E10^-11 W
Which is pretty darn small, even smaller than tom66 predicted.

In fact, this seemed TOO small, and therefore I tried the much simpler, stupider method by using the power supply
here (80W Triple Output Power Supply - 25V 1AAgilent Technologies Part #: E3631A). What I did was hook up the positive and gnd ends of the chip and connected it to the power supply, and gave it an output limit of 5V. When I turned it on, It gave me 5.001V with .001 A. I actually waited for a while, and the last .001 digit
didn't go away.

If I calculate this out using P=IV, it gives me a power dissipation rate of about .005mW! Much bigger in this sense- in fact, too big for my application- my goal for the chip's power dissipation is to be in the range of tens of microwatts.

What do you think is the issue? Are any of my values even right?

Thanks in advance.

btw yay mavs are champs.

 

Hey guys, thanks for your help, as always.

I actually went ahead and measured the power dissipation using two different methods.

First was using #12's suggested method, but using the impedance of my scope instead.
In my school is a Tektronic MSO3000, and the probes for it can exhibit a 1M or 10Mohm impedance.
Therefore, I set the impedance to 10Mohm, and went ahead and measured the voltage of the power rails (Vdd to gnd).
It was very "fuzzy", and therefore I used the "max" measuring method to measure the highest voltage it produced.
After that, I used v^2/R to calculate the power dissipation:

(11.2mV)^2/10Mohms = 1.25 E10^-11 W
Which is pretty darn small, even smaller than tom66 predicted.

In fact, this seemed TOO small, and therefore I tried the much simpler, stupider method by using the power supply
here (80W Triple Output Power Supply - 25V 1AAgilent Technologies Part #: E3631A). What I did was hook up the positive and gnd ends of the chip and connected it to the power supply, and gave it an output limit of 5V. When I turned it on, It gave me 5.001V with .001 A. I actually waited for a while, and the last .001 digit
didn't go away.

If I calculate this out using P=IV, it gives me a power dissipation rate of about .005mW! Much bigger in this sense- in fact, too big for my application- my goal for the chip's power dissipation is to be in the range of tens of microwatts.

What do you think is the issue? Are any of my values even right?

Thanks in advance.

btw yay mavs are champs.

I think the 0.001A figure is the meter rounding up.

Use a multimeter with microamps?