On Tue, 7 Jun 2016 15:20:30 -0700, you wrote: >Good explanation. The other benefit from using an opamp is the low impedance >it provides to the ADC Sample&Hold circuit. As you know, when the circuit is >in sample mode, it is charging and internal capacitance which means the >current spikes for a short time and that will cause noise and measurement >inaccuracy if a simple resistor divider is used. The opamp on the other hand >is a low impedance source and can supply the current required by the ADC >Sample&Hold circuit.
True, and if there's a buffer amplifier, this can possibly be set for low or high impedance (can in the XMEGA... ARM? no idea). Didn't occur to me, and on second thought, if a beginner can understand that one, perhaps they're not a beginner? Nice comment, though. Harvey > >Regards, >John > > > > >> On Jun 7, 2016, at 2:54 PM, Harvey White <[email protected]> wrote: >> >> On Tue, 7 Jun 2016 13:03:29 -0700 (PDT), you wrote: >> >>> Hi, >>> >>> chiming in here because I'm about to build my first circuit that uses ADCs >>> on BBB... >>> >>>> My standard advice >>>>>> would be to run the analog voltage through a non-inverting op amp >>>>>> configured as a gain stage. You run the op-amp (and have to pick one >>>>>> that does rail to rail and also runs from 1.8 volts) from the 1.8 volt >>>>>> supply. >>>>>> >>>>> Yes, that's what I do. There are quite a few very low power op-amps >>>>> suitable for running from the 1.8 volt rail on the BBB. If OP is >>>>> interested I can look up the device, it's from TI if I remember. >>>>> >>>> >>> In my somewhat amateurish approach to this I was planning to use a DC-DC >>> converter to provide 1.8 VCC for my sensors. I'm still learning about >>> op-amps and anything more advanced than a transistor, so I wonder whether >>> there are any advantages to using an op-amp compared to providing 1.8 V >>> from a switching DC-DC converter? >> >> Let's see. Firstly, the goal is to make sure that the voltage to the >> analog terminal does not go negative at all, and does not go more >> positive than 1.8 volts, ever. >> >> So firstly, let's feed all the sensors from 1.8 volts. Now this >> works, and can work well. The question is how to get the 1.8 volts. A >> switching converter, while efficient, generates noise, is relatively >> complex, and is moderately large. An easier way to get 1.8 volts is >> to use a simple 3 terminal regulator. There are some very small ones >> available. They generate little electrical noise. A typical current >> limit is about 100 ma, which should power a number of sensors. >> >> Now, failing that we can (and often we cannot) power the sensors from >> only 1.8 volts, we still need a method of reducing the input voltage. >> >> Some will say to put a resistor in series, thinking that it's the >> current you get with an overvoltage. This is not necessarily true, >> often it is simply the voltage itself, and the damage is done by the >> way that the processor is constructed. >> >> Others would say that the best way is to add two diodes (back biased) >> to 1.8 volts and ground, and then add a protective resistor. The >> problem with this is that the diodes must conduct before they limit, >> and that means that your voltage range is -.7 to 2.5 volts to the pin, >> which violates the limits. >> >> Another approach would be a zener diode. One problem is that the >> zener needs roughly 0.001 amp to work, and that the sensor must supply >> that voltage. Another is that if you are trying to drop 12 volts down >> to 1.8, you can, but by a fixed amount. Go to 13 volts and you still >> drop the 13 down to 2.8, and exceed the chip limits. Zeners are >> standard diodes in reverse, so you have no protection below -0.7 volts >> at the input. >> >> Yet another way would be to use a voltage divider. This reduces the >> input voltage by a fixed amount, and can be a decent solution IF and >> ONLY IF the voltage does not go negative (and can NEVER go negative), >> and the maximum voltage you can ever see is reduced down to 1.8 >> volts). As a specific cure, it can work, but as a general cure, it >> isn't recommended because you can still go over 1.8 volts and less >> than 0. For those who consider it important, the impedance of the >> divider can also be an issue, too little and nothing drives it, too >> high and the chip characteristics override the divider resistors. >> >> Putting an op amp in allows several possibilities. Op amps can be >> inverting or non-inverting. By running the op-amp from 1.8 volts and >> ground, the maximum output voltage that the op-amp can put out is >> limited to these values. You will need a chip that outputs rail to >> rail (ground to positive power supply voltage). >> >> Inverting: the voltage has to be offset to keep the opamp from going >> negative (and it will try, even when powered from 1.8 volts and >> ground). This also inverts the voltage so that as the voltage at the >> input increases, the output decreases. >> >> Non-inverting: the voltage increases and the output increases. Non >> inverting unity gain (buffer) may have output voltage limits, so that >> must be checked. It will have an extremely high impedance, though. If >> gain is needed, then the non-inverting configuration with gain can be >> used. >> >> I'd personally recommend the op-amp for the most general solution, and >> the resistive divider ONLY if the application can be guaranteed to >> never exceed the voltage limits of the chip under any reasonable >> conditions. >> >> Level converters are digital devices, and not suitable for use on >> analog circuits. >> >> Good design is frequently paranoid design. >> >> Harvey >> >> >> >>> >>> I would be very grateful if one of the electronically literate participants >>> to this discussion would share their insight with a newbie :) >> >> -- >> For more options, visit http://beagleboard.org/discuss >> --- >> You received this message because you are subscribed to the Google Groups >> "BeagleBoard" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to [email protected]. >> To view this discussion on the web visit >> https://groups.google.com/d/msgid/beagleboard/c4felb90298at6aj8j9fdgk8ub7cr8cs8p%404ax.com. >> For more options, visit https://groups.google.com/d/optout. -- For more options, visit http://beagleboard.org/discuss --- You received this message because you are subscribed to the Google Groups "BeagleBoard" group. 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