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An operational amplifier has a high-impedance input. That means that its current consumption would be really small, so, we can say it can measure voltage without affecting the circuit in terms of consumption. I have got a design like this:

15 V - LDO - 5 V - VOLTAGE DIVIDER (for Vref) - OPAMP (BUFFER)

And then, this one:

OPAMP (COMPARATOR) (previous 5 V to power it, and Vref to compare) - OUTPUT OF THE SYSTEM

So, I have a input that I am going to compare to Vref and set a logic 1 when the threshold is reached. This design is not mine.

The comparator already has a high-impedance input (it is an op-amp), so why would I need another op-amp as a buffer before it? Notice that I have got:

Reference value --> Buffer --> Comparators (more than one)

For me, it would be the same if I delete the buffer. What do I have to do to check this? Check the input bias current of the comparator? Multiply the input bias current by teçhe number of comparators reading Vref and check if it is lower than my LDO output current? I am a bit lost in terms of high-impedance and the need for buffers.

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The input bias current per comparator is 80 pA. I have got 14, so the result would be a total of 1.12 nA. The output current of the LDO is 100 mA. The divider is formed by 300Kohm and 200Kohm

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    \$\begingroup\$ @tobalt Done. Do I have to check currents? \$\endgroup\$ Commented Apr 4, 2023 at 7:56
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    \$\begingroup\$ "n operational amplifier has a high-impedance input. That means that its current consumption would be really small ... " - we know what you mean, but this statement is incorrect. The opamp's current consumption, and its input current are NOT the same thing. Terms matter, and you mean input current, not current consumption (which is usually used to talk about the supply current drawn by the stage as a whole). \$\endgroup\$
    – danmcb
    Commented Apr 4, 2023 at 8:21
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    \$\begingroup\$ @danmcb You are right, I get it. Thanks. \$\endgroup\$ Commented Apr 4, 2023 at 8:51

2 Answers 2

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If you omit the buffer, the input bias current of the comparators will flow through the voltage divider (R1, R2) and cause an additional voltage drop.

It is up to you to determine if this additional voltage error is tolerable or not. It can be mitigated by reducing the divider impedance, but that will draw more current from the LDO.

For most comparators, and e.g. 10 kΩ divider impedance, this is absolutely fine for everything other than the highest precision tasks.

On a sidenote, powering the buffer from 15 V and feeding its output without protection into a 5 V IC is dangerous. If the buffer outputs a high voltage (because of some transient or malfunction), it can fry the 5 V component.

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  • \$\begingroup\$ The input bias current per comparator is 80 pA. I have got 14, so the result would be a total of 1.12 nA. The output current of the LDO is 100 mA. The divider is formed by 300Kohm and 200Kohm. If the current of the comparators flow through the voltage divider, the voltage would drop 1.12nA * 500Kohm = 560 uV. No? \$\endgroup\$ Commented Apr 4, 2023 at 8:47
  • \$\begingroup\$ In addition to that, you have said "if you omit the buffer, the input bias current of the comparators will flow throught the voltage divider". The same as the input bias current of the buffer would flow throught the divider, no? Which is the different? One versus fourteen opamps? \$\endgroup\$ Commented Apr 4, 2023 at 8:49
  • \$\begingroup\$ @DevelopingElectronics 'yes' to both comments. Your divider resistances don't make sense: How do you divide 5V to 2.5V with unmatched resistors ? But in general, yes the voltage error is rather small. Depending on the sign of the input bias current, it will flow only through R1 or R2 though. \$\endgroup\$
    – tobalt
    Commented Apr 4, 2023 at 9:06
  • \$\begingroup\$ Sorry, Its a mistake. I need 2V not 2.5V. Anyways, and without matter the sense of the current, it a voltage drop of uV. I can handle that. I will delete the buffer. \$\endgroup\$ Commented Apr 4, 2023 at 9:23
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For sure, you need to check if the combined input current of the comparators can change the voltage of the voltage divider. This is an obvious thing to do.

There could be another reason to use a buffer. Comparators can give a kick-back to the input. Whenever the output toggles, it can cause a glitch on the input.

If you are going to remove the buffer, check if the comparators toggling can affect the LDO or cause false triggerring of the other comparators due to such input glitches.

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  • \$\begingroup\$ I have edited the question to add the values. Regarding the last part: which parameter do i have to check in the datasheet? \$\endgroup\$ Commented Apr 4, 2023 at 8:49
  • \$\begingroup\$ @DevelopingElectronics, I don't think such information is available in the datasheet. Atleast I am not aware. Will have to try it out on the board and check if there is any performance impact. Why do you have 14 comparators? How much is the difference between the threshold of the comparators? \$\endgroup\$
    – sai
    Commented Apr 4, 2023 at 9:04
  • \$\begingroup\$ Any toggling current feeding from the comparators to the voltage divider is easily taken care of by a small capacitor after the divider, e.g. 1 nF. \$\endgroup\$
    – tobalt
    Commented Apr 4, 2023 at 9:09
  • \$\begingroup\$ @sai I have got 14 inputs. All of them with the same Vref. \$\endgroup\$ Commented Apr 4, 2023 at 9:14
  • \$\begingroup\$ @tobalt so, adding a ceramic capacitor in parallel to the Rload of the dividir would eliminate the necessity of a buffer, right? \$\endgroup\$ Commented Apr 4, 2023 at 9:16

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