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So i have a constant current source , datasheet here: http://www.ti.com/lit/ds/symlink/lm134.pdf. My setup is this : 9 volt + side goes into V+ or the source , which in turn is connected to the R pin through a 34 ohm resistor. V- of the source is connected to the battery - through my multimeter using crocodile clamps. I would expect the reading to be 2mA for the current but i'm getting a flat 0. The resistor checks out to work ok. My multimeter may be crappy but the amp reading on the battery itself works good. The only thing i can think of is the source, which i have no idea how to check to see weather it's faulty or not. Any suggestions?

Edit: After switching the rin pin connection from V+ to V-, here is what the circuit looks like, with 0 amps still being what i read.

schematic

simulate this circuit – Schematic created using CircuitLab

What i'm trying to figure out is how to test the LM by itself. Even more confusing is the fact that i have 2 of these , so the possibility of both being defective seems low. I also triedusing a 10 ohm resistance in case my multimeter is crap, with the same results, and linking the multimeter black before the R1, which predictibly yielded no results. It seems that whatever i do, there's no current flowing through. The battery has the proper voltage and amp rating when checked by itself. Please help me figure this out. Thank you guys in advance

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    \$\begingroup\$ Look at figure 13 of the data sheet. Rin is connected to V-, not V+. Please use the schematic function to edit your post and show how you have connected this. \$\endgroup\$
    – WhatRoughBeast
    Dec 16, 2016 at 19:39
  • \$\begingroup\$ I edited the original question to include the requests you mentioned \$\endgroup\$
    – omu_negru
    Dec 16, 2016 at 22:51
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    \$\begingroup\$ Many multimeters are fuse protected on the current ranges, it is possible that you blown it with your various experiments. The voltage ranges would probably still work. You could try putting a resistor such as 1k instead of the multimeter then measuring the voltage across the resistor with the meter on a voltage range. \$\endgroup\$
    – Kevin White
    Dec 16, 2016 at 23:28
  • \$\begingroup\$ @omu_negru Please look at the RoughBeast's schematic in his answer. This is the way people who know what they're doing draw schematics, positive to the top, negative to the bottom, and for more complex diagrams, signals going from left to right. !daer ot reisae ti sekam tI. \$\endgroup\$
    – Neil_UK
    Dec 17, 2016 at 7:46

2 Answers 2

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enter image description here Voltage in R1 is kept at 64 mV/298K. (From datasheet)

Thus, at room temperature (~298 K), you'll get 64 mV at R1:

I = 64 mV / 33 Ohm = ~ 2 mA

So you should see 2mA drawn from your battery.

As a side note, using a bad multimeter in ampermeter mode is not a good idea. You have way too high series resistance. Instead of that, just measure the voltage on R1 and see if it's actually ~64mV.

If you really really need to use the multimeter in ampermeter mode, - Take multimeter in ampermeter mode - red wire to Vbat "+" terminal - Black wire to "V+" of LM 334Z - of course delete the current wiring between the to shown in the picture

edit: As pointed out in comments, the multimeter is in voltmeter mode (shown in picture).

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    \$\begingroup\$ Remember - he intends to use the current input of his meter. Doing it your way will effectively short out the resistor - unless, of course, you make sure to tell him to use the voltage function. \$\endgroup\$
    – WhatRoughBeast
    Dec 16, 2016 at 23:31
  • \$\begingroup\$ So i ended up using this as the accepted answer , mainly after opening my multimeter and finding that i only have one fuse for a maximum of 250mA, and that's blown. The voltage did indeed read 64mV, so by Ohm's law i should be drawing 2mA \$\endgroup\$
    – omu_negru
    Dec 17, 2016 at 1:36
  • \$\begingroup\$ @omu_negru - Yes, but only as long as the 334 is not supplying current through its V- lead. This is almost certainly true, but you should be aware that it is an assumption rather than a demonstrated fact. \$\endgroup\$
    – WhatRoughBeast
    Dec 17, 2016 at 4:07
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Ack! No.

schematic

simulate this circuit – Schematic created using CircuitLab

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