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.
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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