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I want to use an LM4041 shunt reference voltage to generate 1.225V, as demanded in the AD7705's datasheet for a 3V supply voltage. However, to calculate the required impedance before the shunt I need to know the maximum current drawn by the load (in this case the AD7705). This is not given in the datasheet, though. If I assume it to be much smaller than the minimum current drawn by the shunt (45µA) and neglect it I get a value of:

R = ( V_supply,min - V_ref ) / ( I_ref,min + I_load,max )
  = (3.3V - 1.225V) / 0.000045A
  = 46111 Ohm

However, with a 46k and even with a 30k resistor I measure an output voltage in the range of 3.25V so what would be a better assumption or how is the regular procedure for dimensioning shunt voltage references?

Regards,
Dominik

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2 Answers 2

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You can start to realize a value by looking at figure 13 in the data sheet. I'm aware this is for 5V operation but it will give you a reasonable idea of things....

It shows a REF192 connected to the ref pins. Now if you take a look at the REF 192 data sheet you will see that it has a load regulation spec of between 5 and 20 ppm/mA. From this you can calculate the effective output resistance of the REF 192.

So the REF 192 produces 2.5 volts and 20ppm of that is 50 uV i.e. 50uV per mA or 50 milli ohms.

Now you can probably be a bit worse than this so take a read of the section about the ref input on page 23 - it tells you that you need to drive the ref input like you would drive an ADC input to obtain accuracy. It informs you that the dynamic load is 5 kohm. However, figure 13 shows 10.1 uF (two caps) across this input so that would nominally take care of this dynamic impedance change. However you could make an argument (when capacitance is zero) that to avoid anything more than a 1 bit error you need an output impedance that is 5k/65536 = 76 milli ohms.

For a bit more info you could look at the graph of REF192 output impedance versus frequency (figure 11). It shows it peaks at about 2.5 ohms when a 1 uF is used. Hence, this is why a 10 uF is used across the REF input on the AD7705/6.

So, armed with this info, take a look at the LM4041 to see how it might perform. If you look at figure 5 in its data sheet I think you will conclude it is possibly unsuitable. It appears to have an output impedance of about 0.4 ohms i.e. about 5 or 10 times higher than the REF192.

This was by no means a thorough check on things and therefore, the main focus of this answer is to provide a mechanism for working out stuff that is implied rather than definitively stated.

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  • \$\begingroup\$ Thanks for your answer, it gave me a good comprehensive insight on the involved mechanisms. However, shouldn't the LM4041 still produce 1.225V after all? If the impedance of the AD7705 is 5kOhms that would equal a current drawn of 0.245mA @ 1.225V, so this is the value I was initially searching for. With this maximum load current I can get the maximum impedance of the shunt resistor, which would be 6 kOhm, minimum being 240 Ohm. This shows me that my used 46 kOhm resistor was way too big. Still, the LM4041 doesn't work properly. Well, I think I will have some more research to do. \$\endgroup\$
    – Domme
    Commented Jul 15, 2016 at 14:43
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    \$\begingroup\$ And the problem is found... quite unobvious, but the part pins differ from the datasheet! \$\endgroup\$
    – Domme
    Commented Jul 15, 2016 at 17:24
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I just found the quite yet unexpected solution: the LM4041's datasheet is lying about the pin order. It is mirrored!

I am using an original TI part from www.conrad.de and it has a different pin order than given in the data sheet from the shop or other sources.

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