This is a pic of my multimeter.
I am trying to measure current drain.
I hooked up meter between the positive of the device, but not sure if it's right.
I was wondering what the Red Output port is used for ?
Thanks.
It used to mean that there is a internal series decoupling capacitor at that input. A way to measure AC and decouple the DC bias.
On a sunny day (Sun, 8 Dec 2013 05:25:03 -0800 (PST)) it happened Andy K wrote in :
It seems to be a 50 uA meter, so on the DC voltage scales that should be it.
It says in the display area: DC 20 kOhm / Volt -> makes 50 uA on DC ranges, figures. AC 9 kOhm / Volt -> makes 111 uA on AC ranges, rectification etc.. figures.
Maybe for continuity beeper? No idea.
As you can see, the AC voltage settings will not block any DC that is present. The "Output" input has a series capacitor.
Best regards, Spehro Pefhany
-- "it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
-- Between the positive of the device and what?
THey are trying to get you to buy another meter? :)
But really, the idea is for you to be able to measure DC with one meter and then set another meter beside it and connect the output to the input of the next meter along with the common and measure the AC ripple on that DC offset at the same time. Years ago I had one that would do this for you but it was scaled. The output was connected to a divider in the meter so that it would only generate a low level like 1 Volt at full scale. There was a knob on there so you could correct the DC reading on the meter when you connected the load. The idea was to first measure something with nothing attached to the output, get a mark on the meter, then connect the next measuring device on the output and adjust the knob to bring it back to the correct value on the meter.
Mine had a switch on it to select DC or AC coupled output. I used to use it for doing current measurements and watch ripples via the scope but you can also use it to watch via the scope anything the meter sends to the analog gage.
Things have changed over the years!
Jamie
Strange. My old unit output jack connected to a fix divider and you were required to use the load correction knob. This used to drive me crazy because at times when I wasn't using this I'd forget to put the knob back in the zero/off position. So I modified the meter with an insulated spring switch behind the jack to disable this feature when the plug wasn't in place.
A little dremel work and some glue go a long ways.
Jamie
Multi answers to one question. ??
No, I won't buy another meter. :-)
pg
s right.
That meter max's out at 0.25A, that's a pretty small current, better do a r eality check before you hook it up. The OUTPUT jack is where you plug the " OUTPUT" of an audio circuit whose AC amplitude you want to measure, and set the scale to ACV, COM goes to the circuit common.
Is this right? The f'ing crooks at Sears selling this thing for nearly $50 everyone else is pricing at $15??? Whatever it takes to pay their CEO $100M annual I guess...useless bunch of nobodies.
Give a picture we CAN look at!
For currents in the ranges indicated by the switch,use the black COMMON and the current connector; IGNORE the "output" connector. For currents much lower than those switch ranges,use voltage settings and look up (from the manual) the meter resistance for the range being used. A 10Meg handheld DVM can give full-scale sensitivity of 20nA on the
200mV scale. Naturally,if you use this trick, you should know what that extra 10Megs in series may do to the circuit operation.
This thing is a 50uA FS meter, whatever the scale, and that's DC. AC is variable because they shunt the coil with a capacitor. Dunno why they confuse the issue with that 20kOhm/Volt crap.
You should learn about meters, how they are made,their schematics, uses and limitations.
A VOM,IF it has any capacitors at all,has only _ONE_. 20Kohm/Volt is mathematically derived from the meter resistance of 50uA. Use Ohm's law equation in standard form: voltage E expressed in Volts equals current I expressed in Amperes times resistance R expressed in Ohms. All else is simple mathematics. Always keep the UNITS.
You should learn about meters, how they are made,their schematics, uses and limitations.
A VOM,IF it has any capacitors at all,has only _ONE_. 20Kohm/Volt is mathematically derived from the meter resistance of 50uA. Use Ohm's law equation in standard form: voltage E expressed in Volts equals current I expressed in Amperes times resistance R expressed in Ohms. All else is simple mathematics. Always keep the UNITS.
This one has ONE, see SP's post with link to schematic.
A.
Really now, care to explain how you use the ohms/volt for anything? And you r equation is pedantic. Ohms/Volt is Amperes^-1,(inverse amps?) but that's apparently too confusing for some people. Then inverse (50uA)^-1 is 20kOhm/ Volt. For DC measurements, the meter loads the circuit with full scale DCV /50uA Ohms.
Note that VOMs, IF they have one capacitor it is NEVER "across" the meter.
My multi meter has at least two caps along with a .5 amp fuse and spare.
It's seems to me that the switch that I have does not fully open the circuit like it should.
Because if one leg of power is broken, there can be no current leakage.
It has switch positions on both sides of the OFF position for chime mode and alarm mode.
Andy
50uA.
lts
your equation is pedantic. Ohms/Volt is Amperes^-1,(inverse amps?) but tha t's apparently too confusing for some people. Then inverse (50uA)^-1 is 20k Ohm/Volt. For DC measurements, the meter loads the circuit with full scale DCV /50uA Ohms.
er.
You mean the meter movement, but this one most certainly does. Why do you t hink the ohms per volt for ACV is on the order of half its spec for DCV?
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