What's the formula for this?

[Nigel Goodwin]

Thanks Nigel I stand corrected. Does anyone care to explain why ?

Apply ohms law and plot it out

Or just do two values - nothing connected, and the full load (minimum of five times the potential divider source impedance. See how much it changes, then compare to a simple series resistor with the same test.

 

[MrAl]

Hi,


I was going to add that the circuit sometimes depends on the load.
The load is usually taken to be either a resistor to the positive supply,
to ground, or a load current to either supply. There are simple formulas
to quickly calculate either configuration if you are interested.

The resistive divider voltage is calculated as:
Vout=Vin*Rdown/(Rup+Rdown) [Equation 1]
where
Rup is the upper resistor and Rdown is the lower resistor.
This is with no load.

If the load is a resistor to positive supply, then before calculating put Rload
in parallel with the upper (Rup) resistor with:
Rp=Rload*Rup/(Rup+Rload)
and call that the new Rup and calculate Vout with Equation 1.

If the load is a resistor to ground, then before calculating put Rload in
parallel with the lower (Rdown) resistor with:
Rp=Rload*Rdown/(Rload+Rdown)
and call that the new Rdown and calculate Vout with Equation 1.

If the load is a current to ground, then calculate the resistive divider voltage Vout with Equation 1
above and subtract from that the product of the current times the parallel resistance
Rp=Rup*Rdown/(Rup+Rdown).

If the load is a current to the positive supply, then calculate the resistive divider voltage Vout
with Equation 1 and add to that the product of the current times the parallel resistance
Rp=Rup*Rdown/(Rup+Rdown).