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In my circuit, a lm78l05 is used to provide +5V voltage to other parts. And now I need another +2V source to shift my input signal from +-2V to 0~4V as the ADC ad7682 limits the input signal range from 0~4V.

Can I just connect the lm78l05 with a simple voltage divider(maybe made by ada4841)?(as the picture showed)

If yes, any parameters I should consider carefully such as the opamp offset current/voltage or input resistance of opamp?

If no, any good suggestion to achieve the goal?

Thank you very much!

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Your idea might be good enough but suboptimal. What do you need these 2V for? Add this detail in the question so that a proper solution might be designed. \$\endgroup\$ – Vladimir Cravero Aug 15 '14 at 9:11
  • \$\begingroup\$ @VladimirCravero: I have added some detail.Do you have any suggestion? \$\endgroup\$ – billyzhao Aug 15 '14 at 9:25
  • \$\begingroup\$ well yes, you just need a reference so you can omit the op amp buffer. If you are concerned about noise and such search for an integrated voltage reference. \$\endgroup\$ – Vladimir Cravero Aug 15 '14 at 9:29
  • \$\begingroup\$ @VladimirCravero: Do you mean I should take another regulator to generate the 2V? \$\endgroup\$ – billyzhao Aug 15 '14 at 9:38
  • \$\begingroup\$ No. You can either live with the two (noisy and inaccurate) resistors or search for a voltage reference chip that is not a voltage regulator. A voltage reference provides a very stable and noise output voltage but very little current. \$\endgroup\$ – Vladimir Cravero Aug 15 '14 at 9:43
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This idea will work providing you don't need to supply more than (say) 20 mA of current at 2V to other devices because it's likely that the op-amp wouldn't be able to do this - some op-amps will supply (say) 50mA such as the AD8605 (from memory) of course.

DC offset is an issue if you want to use an op-amp so check the data sheet.

Why not find an adjustable regulator that can do the job - most are able to supply at least 150mA and can be adjusted by a resistor divider from about 1 volt upwards.

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  • \$\begingroup\$ Does any adjustable regulator provide 5V and 2V at the same time? And I think it does not need too much current. I have added some detail in my question. Take a look. Thank you very much. \$\endgroup\$ – billyzhao Aug 15 '14 at 9:27
  • \$\begingroup\$ Not that I'm aware of but try googling it. \$\endgroup\$ – Andy aka Aug 15 '14 at 9:28
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That won't work.

    Vin    Vout
    -2      6
     0      4
     2      2
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  • \$\begingroup\$ EM is correct, the gain looking into the non-inverting input in your circuit is +2.0 so you would need a 1.0V voltage to correctly offset the output. \$\endgroup\$ – Spehro Pefhany Aug 15 '14 at 18:33
  • \$\begingroup\$ Sorry, I just make a mistake. The rate should be 1/5 which means the resistors or OA1 should be 4k and 1k. \$\endgroup\$ – billyzhao Aug 16 '14 at 2:38
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This will function okay, particularly if you use good resistors, and don't need more than some % DC accuracy.

The 78L05 is not a very good regulator and its a worse reference so it's almost surely going to limit the overall DC system accuracy if not dealt with. You could measure the voltage with one channel of your ADC and deal with it digitally, but power supply noise will again limit accuracy and will likely have some unfortunate statistical correlations that limit what can be done in the digital domain.

The 78L05 has an accuracy of +/-200mV and a typical temperature coefficient of -0.65mV/K. Since the relatively high-end op-amp you're looking at has an offset of 300uV maximum (40uV typical) and a drift of typically 1uV/K the reference errors will dominate (by 250:1 for offset and drift typically), even with the 2/5 divider. The resistor tolerances will add some errors too.

In fact, I suggest you use the ADC reference 4.096V output as the primary reference source rather than your power rail and use precision resistors.

http://www.analog.com/static/imported-files/data_sheets/AD7682_7689.pdf

The particular op-amp you've chosen is not quite suitable for buffering the ADC reference output (only guaranteed to work to 4V input, not 4.096V). It also has an enormous input bias current (5.2uA maximum- enough to light up an LED) so if DC accuracy is important to you, resistances have to be kept quite low (the internal ADC reference can only supply +/-300uA and you had best keep loading to less than that). The main advantage of this op-amp is low noise, but there are many better choices if you don't need such low noise (it's only a 16-bit ADC, so without doing any calculations I suspect that is not a requirement even over the maximum 125kHz BW for that ADC).

Edit: Also see EM's answer regarding the error in the level of the offset voltage. The above comments still apply except the divider should be 1/5 from 5V or ~1/4 from 4.096V.

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  • \$\begingroup\$ Thanks for your reply. However, is the REF pin available for voltage reference as a source? It is not mentioned on the datasheet. It is only mentioned at page 21:analog.com/static/imported-files/data_sheets/AD7682_7689.pdf , that the REFIN pin could be used as a source.. \$\endgroup\$ – billyzhao Aug 16 '14 at 2:45
  • \$\begingroup\$ Yes, should be REFIN as on page 21. \$\endgroup\$ – Spehro Pefhany Aug 16 '14 at 5:04
  • \$\begingroup\$ So does it mean that I can not use the REF pin to be the output voltage reference? \$\endgroup\$ – billyzhao Aug 16 '14 at 5:31
  • \$\begingroup\$ That data sheet is really badly written. By my reading the REF pin should supply the 4.096V at +/-300uA (minus, presumably, whatever the converter needs) but you could use the REFIN if buffered as they describe on page 21(but it won't be 2.5/4.096V it will be 1.2 or 2.3V typical)! \$\endgroup\$ – Spehro Pefhany Aug 16 '14 at 11:53
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Just a suggestion... use a pair of metal-film resistors as a voltage divider between +5 and ground. MFRs are a lot quieter than carbons. R1 should be exactly 150% of R2.

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  • \$\begingroup\$ There is no R1, but if you're talking about the R2R3 divider, R3 should be exactly 25% of R2 in order to get a voltage division of 5:1. \$\endgroup\$ – EM Fields Aug 16 '14 at 19:34
  • \$\begingroup\$ In my response, the R1 and R2 referred to are the future voltage divider, R1 being above R2... in which case R1 should be exactly 150% of R2 so that the voltage drop across R1 would be 3.0V and the voltage drop across R2 would be 2.0V. The voltage at OA1's non-inverting input in the drawing shown in the OP would be 1.0V. \$\endgroup\$ – TDHofstetter Aug 16 '14 at 20:59
  • \$\begingroup\$ If what you're talking about is a voltage divider with a 3k ohm resistor - R1 - connected to E1, (+5V) - a 2k ohm resistor - R2 - connected to GND, and their free ends connected together, then the voltage at their junction - E2 - will be: E2 = E1 R2 / R1 + R2 = 5V * 2kR / 3kR + 2kR = 2V. Then, since the opamp is a non-inverting unity gain buffer, its output will also be 2V. \$\endgroup\$ – EM Fields Aug 17 '14 at 16:45
  • \$\begingroup\$ Is that not the goal, per the subject line and original question? \$\endgroup\$ – TDHofstetter Aug 17 '14 at 17:35
  • \$\begingroup\$ "Was" the goal; the rest of the world has moved on since then. \$\endgroup\$ – EM Fields Aug 17 '14 at 17:51

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