Electronics > Projects, Designs, and Technical Stuff
Voltage dividers on power supply rails
oschonrock:
--- Quote from: new299 on August 03, 2020, 01:37:57 pm ---
Are there any examples voltage dividers on rails in production designs? I don't think I've ever seen it.
--- End quote ---
Only in bad designs or if you are talking about a few uA and don't care to waste 10x of your uA in the divider and are not fussed about the exact voltage.
It's just a bad idea, if you can afford an LDO (price, space, etc) , use that...
David Hess:
--- Quote from: new299 on August 03, 2020, 01:37:57 pm ---Yes, I've seen a op-amp used to drive a supply rail in an older Lecroy oscilloscope, it makes sense to me that it could provide a reasonably stable supply (though the approach seems rare, and I'm not clear why you'd want to do it).
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Operational amplifiers used to be relatively expensive so using one that way would be considered wasteful. Their relatively high open loop output impedance also limits performance unless you add a shunt capacitor and most people do not know how to do this without causing the operational amplifier to oscillate.
But there are cases where they are exactly what is needed, like to power a circuit which has limited power supply rejection like a VCO or oscillator.
--- Quote ---But I've never seen a voltage divider used to create rails before. I'm kind of confused as to why someone would want to do this (having been presented with the design).
Are there any examples voltage dividers on rails in production designs? I don't think I've ever seen it.
--- End quote ---
There are plenty of examples where a resistor divider provide low current bias supplies for things like cascode transistor output stages which require neither precision nor high current.
Zero999:
We need to see the schematic, to make a full judgement. The output impedance of the potential dividers is 60Ohm and it seems plausible a low frequency op-amp, will work with that sort of power supply impedance, without any problems, as long as not much current is taken from the output.
To get a total power supply voltage of 5.5V, rather than 6V, change the lower resistor to 100R, on the negative supply. The reason why I'm suggesting to reduce the negative, rather than positive supply voltage, is because the works closer to the negative voltage, so reducing that, will have less impact.
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