slightly OT, but related - processor current measurement

ecoology · Joined: 07 Nov 2006 Posts: 340 Location: USA - California

I am using a mega168, driving a TI CC2500 transceiver [a very complex single chip radio]. When both chips are asleep, the current should be less than a couple microamps. When the processor is running, and the radio is in receive mode, the circuit will draw as much as 35 ma.

Is anyone aware of a way to measure the current draw, so I can verify that the chips are asleep? [obviously if awake, I can verify that they are running]. Seems trivial, but there is a 35,000:1 ratio of current, and a simple resistor in series with the power source won't work - same for a microamp meter, or multimeter.

This circuit will be running on a single coin cell, so current draw is critical - I have to know for sure that it is working as expected.

Any thoughts gratefully appreciated,
George

DToolan · Posted: Wed Apr 16, 2008 1:17 am Post subject:

Um... I am at a loss as to why you can't measure current with a multimeter. Measuring current is done by breaking the power source (seperating it from the object being powered) and placing the multimeter in series. There is no way it "can't" measure the current flowing through it (unless it has a seperate fuse for current measurements and it's internally blown).

ecoology · Joined: 07 Nov 2006 Posts: 340 Location: USA - California

Hi:
If I set the multimeter for its lowest current - 200ua, I can measure the "sleep" current, correct. BUT if I then bring the chips to an awake state, the current will jump up enough that the protective circuitry in the multimeter will interrupt the current flow, resetting the processor and/or the radio. Effectively, the multimeter puts a large resistance in series with the circuit to measure very small currents. The drop across that large R then "eats" up all of the power source voltage.

I want to be able to see current flow, uninterrupted, as the circuit goes from one state to another.
Actually, the multimeter I have just prevented the power supply from supplying enough voltage when in the 200ua setting.
By putting anything in series with the power, I will be interrupting what I want to measure - The Heisenberg Uncertainty principle on a macro scale.

AdrianJ · Joined: 16 Jan 2006 Posts: 2483 Location: Queensland

Not many multimeters will measure 1 microamp successfully. Probably the best way is to pick a resistor which will drop say 100 mV at the expected current ( eg 100 K for 1 uA ) and measure the voltage drop across this resistor, when in series with the load. You need to keep the drop across the resistor so that the supply voltage is still more than the lowest supply voltage you can work at.

You will need a high impedance meter, at least 10 times the sense resistor, but most digital meters have around 100 M impedance these days, which is fine.

You will probably need to start up the system with that resistor shorted out, put it to sleep, and then open the short to do the measurement.

Just saw your post. Maybe you put a couple of different resistors in series, and short them in and out, so you can measure the current in the different states.

ecoology · Joined: 07 Nov 2006 Posts: 340 Location: USA - California

Hi Adrian:
I was hoping for an "elegant" way to do this, where I can see the current change [on a scope preferably] without interrupting the current, thereby resetting either chip. Ideally, I need a current trace so I can see the actual time required to go to sleep, wake up, transmit, etc. Then I can accurately calculate the required battery capacity. By having to use a large R, then short it out, etc, I lose the ability to measure the time required to change state. Theory is great, but measuring actual times in various states allows for much better battery calculations.
a 10k resistor in series, at 1 ua is 100mv. that same resistor, with 35 ma flowing, drops 350 volts! A bit more than my on-board regulator can handle.

AdrianJ · Joined: 16 Jan 2006 Posts: 2483 Location: Queensland

So a 2.7 ohm resistor would drop 100 mV at 35 mA. That is enough to see that the voltage across it falls drastically as you put the system to sleep, and you can watch it with an oscilloscope. Then switch up to a much higher resistor to verify that the final current draw is what you expect. Do your switching or resistors in parallel, then you dont interrupt the supply voltage as you change sense values.

ecoology · Joined: 07 Nov 2006 Posts: 340 Location: USA - California

That gives me an idea.
I could use the processor to do the resistor switching. When it is asleep, doesn't I/O pins become high Z? [have to look that one up, not sure]. If not, I configure the I/O pin to be low during sleep, and high during any wake state.

Anyway, upon waking, the processor could cause a resistor to switch into the circuit, dropping the voltage across the sense resistor. With a 2 channel scope, I can have one channel show the resistor switch, and the other channel show the actual current.
With a large cap across the power supply input, the uP will have time to switch before losing voltage.
Neat, let the processor do the switching for me.
Thanks for thoughts - led to my elegant solution!
George

ecoology · Joined: 07 Nov 2006 Posts: 340 Location: USA - California

That gives me an idea.
I could use the processor to do the resistor switching. When it is asleep, doesn't I/O pins become high Z? [have to look that one up, not sure]. If not, I configure the I/O pin to be low during sleep, and high during any wake state.

Anyway, upon waking, the processor could cause a resistor to switch into the circuit, dropping the voltage across the sense resistor. With a 2 channel scope, I can have one channel show the resistor switch, and the other channel show the actual current.
With a large cap across the power supply input, the uP will have time to switch before losing voltage.
Neat, let the processor do the switching for me.
Thanks for thoughts - led to my elegant solution!
George

AdrianJ · Joined: 16 Jan 2006 Posts: 2483 Location: Queensland

AFAIK, the processor port lines are maintained in whatever state they were in when running, when you enter sleep mode. Only a Reset switches back to Hi-Z input state.

But if you are only doing this for a test, just switch the resistors manually. Once you determine what the currents are at each stage on the bench, they are not likely to change in the field. At most you might have to check each unit in production test.

Beware leakage on bypass caps, I have seen the ordinary 100 nF ceramic bypass caps pass as much as several hundred uA DC current after soldering.

Arera · Joined: 23 Sep 2007 Posts: 386 Location: Wuppertal, Germany

Take care for the IN-Pins!
A pin, switched to "input with pull-up", which is connected to GND will draw current thoug the internal pull-up(30k-50K, I guess).
In my application I set the inputs to "high-Z" and use external pull-ups 220K.
I don't connect unused Pins, but switch then to "input with pull-up".

Maybe that helps to save a few µA.