純筆記。

沒(méi)整理。沒(méi)翻譯。沒(méi)嘗試。沒(méi)時(shí)間。但有用。

本來(lái)想企圖設(shè)計(jì)設(shè)計(jì)下學(xué)期的實(shí)驗(yàn)。

看ADS1115和MCP3421等等一眾老外的差分ADC模塊價(jià)格一路瘋漲。有高手建議用AD7705/7706。但是搜索一下相關(guān)的Arduino庫(kù)，很古舊，也少有建議使用。

碰到一個(gè)without ADC的討論，其實(shí)應(yīng)該是differenciral input，沒(méi)準(zhǔn)兒能用。有衰減，但簡(jiǎn)單。先記著，有時(shí)間試試。因?yàn)椴簧婕巴鈦?lái)的電壓，所以全部都轉(zhuǎn)成Single End，一下子測(cè)負(fù)電壓的代碼就簡(jiǎn)單了。

the -1 to -3V that I'm measuring is the output of a LM337 that I want to monitor. –

Kevin Boyd Jun 21, 2010 at 6:23

related: Measure -20V to +20V voltage with pic –

davidcary

Jun 25, 2011 at 22:29

I don't have enough details to flesh out a complete answer, but if your PIC has a VREF- (negative reference voltage) pin then you can probably do this directly, without an additional inverter, through creative use of the reference voltages. –

Sparr Aug 16, 2013 at 2:04

An inverting amplifier does not need a negative rail to invert the voltage.

gschem

Try to think of your power rails as what supply your output. If you look at the circuit, all op-amp pins are tied to a voltage of 0V or higher. When your range of -1 to -3 comes in, it will show up as the exact opposite of 1 to 3 on the output. This also gives you some advantages as a buffer, as the input impedance of your pin will not affect this circuit very much (so long as Rin||Rf is large).

I agree that a simple resistor divider does the job -- just letting you know that this also works.

+1 -- this is the proper way to do it if you don't have an ADC taking in negative inputs. –

Jason S

Jun 20, 2010 at 19:41

3

No, the rails of the op-amp are the bounds for your output. Where a rail to rail op-amp will go very close to the bounds. You could, and many have, spent their life designing op-amps. There is no such thing as a perfect op-amp, but there is normally a perfect op-amp for a specific case. –

Kortuk Jun 21, 2010 at 14:03

4

On that note, in this configuration, if you surpass around -Vcc as your input your output will hit the Vcc rail. Some op-amps will not go within a volt or more of the rail, some op amps will go within 50mV. If you get a larger input signal, divide it by a larger amount, if you have a -1 to -10 V signal, divide by 2, problem solved. –

Kortuk Jun 21, 2010 at 14:04

2

@Kortuk - There is is never a perfect op-amp for any case. However, there is the best op-amp out of the available selection, when price is taken into consideration, for every project. –

Connor Wolf

Oct 19, 2010 at 4:27

3

@Fake Name, I think this is a case of a communication error. If it meets the specs you need and is affordable then you have the perfect op-amp. That would be my wording, I understand what you mean though. I accept the limitation and use perfect because I am positive. <3 –

Kortuk Oct 19, 2010 at 14:36

You could use a voltage divider, with one end hanging off the positive supply rail. Say you have one with equal resistors and a 5V power supply, this will result in a voltage between +2V and +1V for your -1 to -3V range.

+5V +

|

R

|

+-- OUT

|

R

|

IN -+

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This will require the IN terminal to sink a current of (5-Vin)/(2R), so you'll have to choose R large enough to not overwhelm the input with unwanted current. then, your ADC input will need to have at least an order of magnitude more input impedance than the R value, in order not to unduly load down the divider network. Which all may (or may not) be possible. To the data sheets! –

JustJeff Jun 20, 2010 at 13:35

3

-1: you'd better use precision resistors and a precision reference for 5V. –

Jason S

Jun 20, 2010 at 19:39

1

+1 for being the cheapest simplest way. But as Jason S says, precision will take effort, if that matters. –

DarenW Oct 15, 2010 at 20:34

3

Any noise on your positive rail will show up in your ADC. –

endolith Oct 15, 2010 at 21:37

If reference voltage is not present (e.g. MCU is not powered up) input may go deep minus in regard to the ground, damaging the input or MCU in whole. –

Anonymous Nov 6, 2020 at 7:42

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You might not even need an op-amp. Some ADC's (like the MCP3304, see datasheet: http://ww1./downloads/en/DeviceDoc/21697e.pdf have a built in differential mode, where the ADC returns the difference of two channels, which can be a negative number. If you tie one channel to ground (called pseudo-differential mode), the ADC can accept a negative input voltage on the other, and translate it into a negative number, all without needing a negative voltage.

Of course, this only applies if your ADC supports this sort of thing. Many don't have differential mode at all.