Measuring current ???

How does one measure current using a DMM? I see a lot of threads where members are reading current draw from their lights.

It's easy to read the battery draw with a DMM.

Open battery compartment.
Set meter to DC amps or DC ma, depending on how many amps you expect to see.
On some meters, measuring amps require moving one of the probes to a different socket.
Turn on the light.
Touch one lead to the end of the battery. touch the other whereever the battery normally connects (usually the body of the light, but not always).

Read the amps (or ma) on the display.

Most DMM are not polarity sensitive when measuring current.

Daniel

One problem with measuring current that you have to get around is probes do not give a big enough contact area to allow all the current to flow. The point of the probe will act as resistance and you won't get a true reading. For smaller current measurements the probe will work, I'd say about 600mAh, but you need to figure out how to get a good clamp to test "real" amps.

I am not getting any reading on my meter???? I take the tail cap off the batt compartment. I have one of the test leads touching the neg end of battery and the other lead touching the compression spring that is mounted in the tail cap. I turn on the light and presto, nothing???

Obviously I must be doing something wrong...

Did you make sure your positive probe is plugged into the "amp" connection on the meter?

Most meters have a fuse in the current measuring circuit. It's real easy to blow. If you ain't blown one yet, you ain't lived.

A DMM will only measure a limited amount of current. Some will measure only a few hundred milliamps. A few will measure 20 amps.

One way to measure a high current is to place a known resistance in the circuit and measure the voltage across the resistance. You need to understand Ohm's law, but that's not rocket science.

An interesting factoid for those measuring high currents. The resistance of a foot of 20 gauge wire is fairly close to .01 ohms.

Of course you foreigners that live where measuring feet is forbidden, and the possession of 20 awg (American Wire Gauge) wire is a felony, are on your own here. /ubbthreads/images/graemlins/smile.gif

[ QUOTE ]
Dukester said:
I am not getting any reading on my meter???? I take the tail cap off the batt compartment. I have one of the test leads touching the neg end of battery and the other lead touching the compression spring that is mounted in the tail cap. I turn on the light and presto, nothing???

Obviously I must be doing something wrong...

[/ QUOTE ]

Wrong second contact, you need the *tube* where the cap screws into. The current from the battery must flow through the amp meter. In this case, you're replacing the contact between the bottom of the battery stack (minus end) and the tube (normally done by the cap and spring).

You're on the right track when the light comes on (reading or not).

Doug Owen

Doug got it right! You'll need to connect the probes to the battery tube and the battery negative post.

Thanks Doug, it is working now though the reading fluctuates quite a bit but nonetheless I am getting a reading. This will help determine battery runtime...

Yer welcome, Dukester. Now that the easy part's done, on to trying to understand the readings......

Doug Owen

Doug, I'm new to using a DMM too. In the spirit of the original post, why do my readings seem to be significantly higher than the driver is biased for? Example, a Nexgen400 LE reads in the mid 5's, a NG750 goes around 1.1 if I recall. All my sandwiches seem to read much higher too. Whats up?

Dan C

Make sure that amp position is on the 10 amp position.

Bill

[ QUOTE ]
Dan C said:
Doug, I'm new to using a DMM too. In the spirit of the original post, why do my readings seem to be significantly higher than the driver is biased for? Example, a Nexgen400 LE reads in the mid 5's, a NG750 goes around 1.1 if I recall. All my sandwiches seem to read much higher too. Whats up?

Dan C

[/ QUOTE ]

I am not familiar with these sandwiches but if they are boost circuit that take less than the Vf of the LED and boost it up to that voltage it will take more amperage because the power consumed must be the same + any additional loss of power due to heat and circuit efficiencies less than 100% adds to this total current consumed from the battery source.

[ QUOTE ]
Doug Owen said:
Yer welcome, Dukester. Now that the easy part's done, on to trying to understand the readings......

Doug Owen

[/ QUOTE ]

After changing batteries from 1850 mAH to my new freshly charged Energizer 2500 mAH batteries I am getting a lower reading??? Close to a 90 mAH difference? What is this telling me? I am doing the testing using the Streamlight JR Luxeon. Hard to believe my father is a retired Sr. Design Engineer from GE and I don't know squat about electronics.

To do a complete analysis of the situation you could measure the voltage produced by the batteries. Reading the voltage under load as well as the voltage with no load might show something.


It's possible that the 2500 mAH cells have a higher internal resistance than the others, which could have an effect if you are drawing lots of current.

Or maybe these cells didn't get fully charged.

If you have one of those new-fangled flashlights with an electronic switching power supply, or as they say around here "it's regulated", then it is probably switching the current from the battery on and off at a rapid rate.

I don't know if a DMM will tell you much in this case. Maybe you need an oscilloscope, or a simpler flashlight.

The current reading that Dukester and Dan C noted are not what they expected because:

In a simplified explanation, the flashlights' converter takes a set amount of power at one voltage and converts it to the same amount of power at a higher voltage. Power is defined as volts multiplied by amps. This is abreviated as
P = E * I ( where P = power in watts, E = Volts and I = Amps)

Still simplified, the converter is not able to convert 100% of the input power to the output power. The extra is generally wasted as heat.

To put this all together, A nexgen 400 will put out .4 amps (400milliamps) to the LED. Assume a CR123A (3 volt) battery. If the LED's 400ma Vf is 3.6 volts, the outpt power is 3.6 * .4 or 1.44 watts. The battery will have to supply at least 1.44 watts at 3 volts, so the battery draw will need to be at least 480 ma. Converters are frequently only 80% efficient, so around 550 ma may be required from the battery to provide 400ma at the LED.

In dukester's case, he took a partially used cell that was probably around 1.2 volts and replaced it with a fresh cell that was probably 1.45 volts. If the battery voltage goes up the current goes down. Lets assume he used a pair of batteries and a 1 watt LED. 1 watt drawn from 2.4 volts is 416ma, but 1 watt drawn from 2.9 volts is only 345ma

Hope that helps

Daniel

[ QUOTE ]
gadget_lover said:

In dukester's case, he took a partially used cell that was probably around 1.2 volts and replaced it with a fresh cell that was probably 1.45 volts. If the battery voltage goes up the current goes down. Lets assume he used a pair of batteries and a 1 watt LED. 1 watt drawn from 2.4 volts is 416ma, but 1 watt drawn from 2.9 volts is only 345ma

Hope that helps

Daniel

[/ QUOTE ]

My slightly discharged 1850 mAH Batts were reading 1.32v whereas I swapped them out with freshly charged hot off the charger the new 2500 mAH Batts with a reading of 1.46v.

So inconclusion is this a true statement. If the light is consumming 551 mAH than I should get 4.5 hours of runtime off my JR Luxeon using the 2500 mAH Batt's?

For simplicity if your light was running at 100% efficiency then you would get 4.5 (4 hour, 32 minutes) of light before it dropped below the 50% light output. The heat generated by the light is inefficiency. Say the light is 75% efficient because of heat loss then the run time would be under 4 hours.

Thanks for the bit of education you and the rest of the gang took the time to explain. It is nice to more than double the runtime by using the NiMH's over the Alkalines.

Dukester, the 4.5 hour estimate is fair. You are measuring the battery draw, so the converter losses are already included.

To get it more accurate you have to take into account a few more variables;

1) As the battery voltage drops the current draw will increase.

2) Most batteries are less efficient at higher currents, so you need to look up the data for the batteries that you are using. According to the data sheets at eveready for their 2500mah cells the discharge at 500ma would still come out to about 4.5 hours of run time down to less than 1.2 volts. Most of that time would be at around 1.2 volts, so you should make your calculations with the higher current at 1.2.

A typical alkaline cell, such as the energizer X91 is rated at 3,135 mAh, but that's at a draw of only 25 ma and running down to .8 volts. Jump up to 500ma (750 milliwatt) and you get only 100 minutes or about 800mah. The voltage would drop below 1.2 at about 30 minutes, so the current would climb quickly, draining it even quicker.

The Eveready Energizer L91 and L92 have capacities similar to the NiMH.

We live in good times. /ubbthreads/images/graemlins/smile.gif

Daniel