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I want to use lm336z-5.0 as a reference voltage for Arduino Uno. In the data sheet I see an example of the chip being connected to a 10V source via a 5k resistor. In Absolute Maximum Ratings section I only see that the max reverse current is 15 mA.

What is the lowest and highest possible voltage at the input (I assume I will adjust the resistor to keep the current well below 15 mA)?

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The term "input voltage" is actually not exactly applied here, since LM336 is a shunt regulator. You can see it roughly as a zener diode.

Check also Figure 11 in the datasheet.

enter image description here

As long as you provide a bias current to it that is at least 0.6mA and at most 10mA (15mA absolute maximum) then the voltage across the regulator/diode will be around 5V.

So, the value of the supply voltage is not important in this application, as long as it is of course higher than the reference voltage.

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  • \$\begingroup\$ But I cannot use the built-in Arduino 5V Vcc, which is in reality around 4.6-4.8 V depending on the USB power source I use and load? \$\endgroup\$ – Eiver Jun 30 '17 at 12:35
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    \$\begingroup\$ @Eiver Oh, that of course you cannot! The supply voltage should be higher than the reference voltage. Check out the "Shunt Voltage Reference External Resistor Quick Start Calculator" from TI. \$\endgroup\$ – nickagian Jun 30 '17 at 12:57
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The LM336 type you have is a shunt voltage regulator capable of regulating between 4 volts and 6 volts.

What is the lowest and highest possible voltage at the input

Because the LM336 is a shunt regulator it means that the input voltage is also the output voltage i.e. between 4 volts and 6 volts.

I see an example of the chip being connected to a 10V source via a 5k resistor

Providing you use an input resistor that adequately limits the current to 15 mA then you will be ok. Check that at higher temperatures the 15 mA figure isn't degraded of course.

For instance you could run with an incoming voltage of 100 volts but the resistor will need to drop 95 volts and, if you want an operating current of 15 mA then the resistor value will be 6333 ohms and dissipate nearly 1.5 watts.

But you don't need to operate at 15 mA - the LM336 will work at 0.6 mA and this means a resistor value of 158 kohm and a power loss of only 57 mW.

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The data sheet for the part lists the operating current range at 0.6 to 10mA in the teaser specs section on the first page. The specification itself gives this same range in the "Conditions" column where they specify the max voltage change over operating current:

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So to answer your question you need to use a resistor to limit current to the shunt regulator to within this range if you want to achieve the specified accuracy. The size of the resistor determines the input voltage range.

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