I have seen in some analog circuits that if we need a 2.5 V supply there is way to do it that generate a 5V square pulse with duty cycle
1/2 and pass it through a low pass filter to get a DC 2.5V.I was thinking of another way of generating 2.5V. Use a voltage divider and then use an opamp buffer.
Which of the above 2 ways is a recommended way of generating 2.5V DC?
I just need 2.5V to be at one of the inputs of the opamp. I will be using the virtual ground property of opamp getting 2.5V at the other input of the opamp also. I dont need practically any current to be drawn by the load because after opamp theren will be only a schmitt trigger(comparator). So that you know this is part of an water overflow detector. Still you think the regulator or voltage divider are good ideas for this.
This method has problems with additional voltage drop through the low pass filter when the load on the 2.5 volt output uses current.
As long as the load doesn't upset the opamp (less current than its output can deliver without current limiting, and capacitance to ground that does not make the buffer unstable) this can work fine. You can also add a capacitor from the divider output to ground to clean up any high frequency noise that was present on the 5 volt supply. The 2.5 volt output will be less accurate than the 5 volt supply (must include divider tolerance and drift and opamp offset voltage).
I prefer the second for most low current needs. There are also fixed and adjustable regulators that are made for this sort of thing.
For example:
These may improve on the voltage accuracy of the 5 volt supply.
If the load current is more than a few tens of milliamps, you may want to look into a switching buck regulator, because of its higher efficiency.
A simple two resistor divider may be all you need. If only an opamp input is connected to it (no current load except the opamp bias current) a pair of 100k resistors may be stiff enough.
normally a circuit like that is used to generate - voltages from a
thats what she said!
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