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After reading for about 3 hours about the different problems and situations regarding very low and very high resistor values (source resistance / feedback resistance) with op-amps theoretically I understood all the mentioned facts.

But converting this knowledge into practically constructing a real schematic is still a problem to me. The datasheet of the op-amp used (AD4522 doesn't really go into explicit detail (or at least not into sufficiently transferred details for beginners).

On page 30 "USE OF LARGE SOURCE RESISTANCE" they go into detail with explicit values, but this is for a Unity Gain Follower circuit and not really a non-inverting amplifier circuit.

I tried to transfer this answer (How to choose resistor values in op-amps) to my case, but still struggling a lot.

My current thoughts are:

  • Since "the basic op-amp load is the feedback resistor Rf" (Or does it only apply to inverting configurations?): @page6 "paramater@50V Vsupply" it mentions "Continuous Output Current" as 14mA. For safety reasons, let's assume continuous 10mA is okayish (am I right?). Therefore: R_f should be:
  • R_f = V_output,max /I_cont. = 45V/10mA = 4,5 kΩ.
  • Therefore with a gain of 37,5 => R_s = 123 Ω.

Am I right or is the calculation of the resistor values based on something completely different?

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2 Answers 2

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The op-amp load in the case of the non-inverting amplifier is the sum of the two resistors.

In general you want to stay well away from drawing the maximum load from the op-amp in most cases at low frequencies (and you've picked a 3MHz op-amp optimized for DC specifications so I assume a low frequency application- kHz). At high frequencies the impedance of stray capacitance starts to affect the feedback.

If you actually operated that op-amp at 50V single supply then you'd have thermal issues before current limit issues. For example, the RM8 package has a thermal resistance junction to ambient (under some particular mounting conditions) of 194°C/W. So if we want to limit heating to, say, 20°C we can't dissipate more than about 100mW. Worst case will be with 25V out, so no more than 4mA, which implies a load resistance of no less than 6.25K.

In fact we might well want to use less loading than that by an order of magnitude at least. Unnecessary thermal gradients on the chip can cause other issues (such as increased distortion), and the resistor values are plenty reasonable at 10x that.

We might use something like 49.9K Rf and 2.74K Rs, or perhaps 100K and 5.48K for an amplifier operating at DC to kHz.

The maximum values practical are similarly fuzzy. Usually you want to be somewhere in between (in a log sense) so maybe 10x or 50x the minimum and 1/10 or 1/50 the maximum.

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For Non-inverting gain 1+Rf/Rs=37.5 then Rf/Rs=36.5 and the current values are safe, but the R values may be scaled up by orders of magnitude until input currents create an input voltage offset then the Vin+ ought to have a series R= Rs//Rf to Vin+.

The other factor is for RRIO, the output impedance is low for unity gain but will reduce the error between output and each rail, so they often specify 10K load min.

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  • \$\begingroup\$ Yes, I didn't wrote it down correctly, the calculated resistor values are the minimum values and can be higher by orders of magnitude as you mentioned. So where would I end up roughly? Something like minimum values *3 => Rf=13.1 k Ohm and Rs = 360 Ohm? \$\endgroup\$
    – Dakalaom
    Commented Aug 7, 2021 at 14:15
  • \$\begingroup\$ Correct……………….. \$\endgroup\$ Commented Aug 7, 2021 at 14:37

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