Electronics > Beginners
Voltage Divider Power Question
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bryancostanich:
Hi folks! Here with another noob question!

In one of the electronics books I'm reading, it says that in practical situations, when building a voltage divider circuit, you want to choose the lowest possible value for the total resistance.

I'm a little confused by this. Wouldn't you actually want to use the largest possible R that still allowed for the necessary amount of current that Vout required? Otherwise, wouldn't you just be leaking potentially large amounts of current to ground and draining the voltage source? Or is that current going to be dissipated as heat if you use large resistance anyway?
Dave:
You want high resistance in terms of power dissipation and to reduce the drift that self-heating of the resistors is going to cause.

On the other hand, you want a low resistance to improve frequency response, because there is always going to be some capacitance in the circuit and the lower the impedance, the lower the time constant. Another argument for low resistance is noise. Johnson-Nyquist (thermal) noise voltage is proportional to the square root of the resistance.

So what you're really looking for is a balance between the two, so you get a tolerable power draw but still keep a decent frequency response and low noise.
Zero999:
It's a compromise. The lower the resistance values, the lower the output impedance, but the higher the current will be. The higher the resistor values, the higher the output impedance but the lower the  current will be.

The output impedance is equal to both of the resistor values in parallel: ROUT = (R1×R2)/(R1+R2)

One way round the above, is to use add a low power, unity gain op-amp to the output of the potential divider.
rstofer:
The voltage divider probably hooks up to something.  If it is an op amp, no problem, the op amp impedance is, for all practical purposes, infinite.  It doesn't load down the divider.

OTOH, if the divider hooks up to something with a lower impedance, you want about 10x the amount of load current flowing in the divider.  If the load requires 10 mA, you want 100 mA running through the divider.  That way, as the load varies, the voltage doesn't move around as much.  There's nothing magic about 10x, other numbers work but the effect of a varying load needs to be considered.

And, of course, the load impedance needs to be including in parallel one resistor of the divider.  If the load connects to ground, the load impedance is in parallel with the bottom resistor.  If it connects to Vcc, the load impedance is in parallel with the upper resistor.

Ian.M:
Potential dividers are great when you want a scaled down bias voltage or signal, but absolutely suck if you need to supply power to a load.   As Rstofer just pointed out, you need far more current flowing through the divider than the peak load current to keep the voltage anywhere near to stable.    One way to help it out if you are dealing with DC voltages and can tolerate an extra drop of about 0.7V is to hang a BJT emitter follower off the potential divider tap, so the divider 'sees' the load current divided by the transistor HFE.   Its cruder than the OPAMP buffer Hero999 mentioned, but still can be useful.   You can roughly compensate for the drop by inserting a small signal silicon diode in series with the bottom of the divider, but at the top end it still wont come closer than one diode drop to the supply voltage.
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