Using voltage dividers to bias Op-Amps
Posted by Matthew
DIY Live has an interesting article up describing voltage dividers as power sources. This is an excellent way to create plus/minus voltages from a single positive voltage sources. Unfortunately, the downfall is continuous current flow (and power drain) though the divider. The following ±4.5V supply is detailed in the article. (UPDATE: Read the comments for a discussion on the possible issues with this circuit)
The mathematics behind a voltage divider are quite simple. After dividing the 9V supply into 9V, 4.5V, and 0V, the ground reference is changed to 4.5V. This leaves us with -4.5V, 0V (new ground reference), and +4.5V.
This is bad design
Even though I wish your comment was more constructive, I thank you for reminding me how desperately this article needs appending. There is a second potentially critical issue with this circuit.
First, as mentioned in the article, the circuit draws a constant current through the resistors thereby draining the battery even when the circuit is not in use. The current draw can be minimized by increasing the size of the two resistors.
Unfortunately, increasing the value of these two resistors makes the second (and more troublesome flaw) worse. If there is substantial current drawn (or fed) from 0V, the excess current effects the flow through one resistor. This offsets the voltage divider, and it no longer divides directly in half as intended. Larger values of R will result in larger deviations from the desired voltage.
Fortunately, Greg is using this circuit as a temporary supply to bias op-amps at +4.5/-4.5 volts. The current drawn by op-amps is negligible as long as R is very small compared to the op-amps input impedance and no no other current paths give or take current from the 0V node.
With minimal resources and proper usage, this circuit suffices. A good and simple solution is simply to run two 9 volt batteries in series and bias the op-amp with +9/-9 volts.
This is a bad way of doing this. My old analog prof said ‘you float the ground, you muck around’. First off, you’re gonna be hard pressed to find two matched resistors - even still, if you load the divider it will alter everything making it useless. Plus, it constantly drains the batteries. Try this - for quick and dirty, use 2 batteries and take 0V in the middle, although this will still become unbalanced during loading. The professional way would be to use the battery to drive an oscillator, then feed the waveform into an appropriate transformer with a centertap on the secondary. Use the centertap for 0V and rectify your remaining two taps for +/-. Trust me though, this thing is only useful on paper as a 1st year quiz question.
woah, sorry for all the redundancy. I’ll shut up now. :P
/me wishes he was made of money and had all those components in his desk drawer. :)