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I am designing a circuit which will amplify a few millivolt input (~+/-10mV full range) to a few volts (~0-8V output, centered at 4V). I have spec'd out an instrumentation amplifier with a voltage reference to give the desired center voltage, and now am designing the voltage reference part.

The option I am considering is the AS431 shunt voltage reference. I like the simplicity, but I am worried about output impedance as the INA126 specifies(ish) that ~8ohm output impedance will reduce the CMRR to ~80dB, and to prevent the AS431 from getting fried, I would prefer to limit current to the AS431 to ~10mA.

Reference voltage regulator circuit (taken from the datasheet):

enter image description here

Vin=8V, thus the desired R3 is 4V/10mA = 400 ohms (assuming R1+R2 >> R3). If the output impedance is 400 ohms, the CMRR is completely toast. However, being a voltage reference I don't know if the output impedance is necessarily the same as the input impedance. Even playing it fast and loose by drawing ~100mA through the AS431 requires ~40ohm shunt resistor.

Would I have to add something like a voltage follower to the output to get the desired low-impedance reference the INA126 needs? Or is there something I'm not quite understanding about how shunt regulators/voltage references work?

If I do need the voltage follower after the voltage reference circuit, why would I even bother using the shunt reference in the first place? I could just as easily feed the voltage-divided reference directly into the voltage follower to get a low-impedance reference output.

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I think uou have some basic misconceptions involved here. Please post the section of your circuit which shows how this voltage reference circuit connects to the instrumentation amplifier. You are probably making this more complex than necessary, and perhaps non-functional as well. –  FiddyOhm Feb 1 at 15:42

2 Answers 2

up vote 2 down vote accepted

The output impedance isn't the series 400 ohm feeding the device, it is the device in parallel with the series resistor. The device is likely to have an output impedance below 1 ohm. Having said that, at high frequencies it'll get worse so I would recommend placing a 100nF cap in parallel to ensure at HF the net output impedance seen by your IA remains low across all frequencies of interest.

The data sheet tells you that the dynamic impedance is typically 0.15 ohms but you should still put a capacitor in parallel because the impedance will rise for significantly higher frequencies.

EDIT

Here's a screen shot of what I mean: -

enter image description here

Above 1MHz the output impedance is about 10 ohm so it's pretty good but what happens at 10MHz+ is just guesswork. 100nF has an impedance of 1.6 ohms at 1MHz and obviously gets lower as frequency rises. A slight change of mind would be to use 1uF because then the composite impedance of device and capacitor should never really rise above 1 ohm for any frequency up to well over 10MHz.

Remember, that it's not just "your" signal that you might be fighting common mode problems with but, broadband noise at your input.

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As Andy says, the dynamic impedance of the 431 for currents 1mA~100mA is maximum 0.5 ohm at \$ \le \$1kHz , typical 0.15 ohm in series with 40K (in the INA126). It won't reduce the CMRR noticably.

enter image description here

You'll get better PSRR by keeping the series resistor relatively high value, a few mA should be fine.

I also agree with Andy's comment on adding parallel capacitance. Be sure to avoid the so-called "tunnel of death" instability:

enter image description here

You can see that 22nF or 27nF would be an "unwise" choice.

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