How to use LM334 3 terminal adjustable current source electronics component by electronzap

[@cyrilsli]

Great video! Helped me a lot

[@edwingeovanimendoza5832]

Hello. Sorry for my bad English. So what happens if the input voltage (V +) is not enough to deliver the constant current to the load. Is there no voltage passing and therefore there will be no current through the load? (I simulated a constant current source with the LM317 and when it happens that the input voltage is not enough, it simply passes less current through the load but I don't know what happens in these cases with the LM334)

[@Electronzap]

Current will go down. It will go down to no current if all of the voltage drops added up are higher than the supply voltage.

[@SilentPokeman]

hello does it still need that 1v voltage input for it to work ?

[@JBattler]

Thank you. !

[@Electronzap]

Thanks for watching!

[@microscopeholder5406]

How can I use this as an electronic adjusted current source? Like for a small voice coil motor? Do I need a digital resistor?

[@Electronzap]

Anything that changes resistance will work.

[@Debraj1978]

LM334 has a certain dependency on temperature. Can you pl do an experiment by bringing a solder iron close to LM334 and check the current flowing through?

[@Electronzap]

I'll keep that idea in mind! I'll probably cover it with something instead.

[@yasirshafiullah3016]

Hi, Can you also make a nice video about ultra-low noise voltage references as well? it would be nice to see that in your channel as well

[@Electronzap]

I'll keep low noise in mind.

[@jurek7792]

TNX!

[@Electronzap]

Thank you!

[@omaral-gallab7735]

please, can you give us the circuit diagram?

[@reynolds087]

Digital multimeters can be inaccurate when measuring series current. It's better to measure the voltage across a known resistor and use ohms law to calculate the current. Also, the 6.8 milliamps is correct for 10 ohms because you have to add the bias current. The datasheet has a chart which shows the ratio between the set current and the bias current, which, in the case of a 10 ohm resistor is approximately 13:1. So: 0.0064 + ((0.064/10) / 13) = approximately 0.00689 Not perfect, but much closer to the observed output.

[@Electronzap]

Thanks for the info!