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Hi everyone,

I’ve been trying to build a multiple-load voltage divider that applies a negative voltage to one of the loads. I’ve followed the steps and example given in "Practical Electronics for Inventors" by Paul Scherz.

Using the circuit from the book and the specified load requirements, everything works perfectly. However, when I define my own loads and change the voltages, I can never seem to achieve the desired result using exactly the same technique described in the book.

I’m wondering if I’m missing something fundamental in the way I’m setting up or calculating the circuit.

Has anyone come across this before, or could point me in the right direction?

This is the example from the book:

This is the version I built:

The resistor making the divider should be at least 10 times smaller than the resistance of the "loads". (100 and 1000 times is even better)

As example, if you divide 170V with two one ohm resistors in series, add a load of 1 kilo-ohm in parallel to one of the two initial resistors and the voltage is still close to 170 V/ 2.

But it is definitively a WASTE of energy since a relatively large current will go through the small resistor and the resistor must be small (in comparison to the load) to be an effective voltage divider. Other alternatives are often considered because of that "energy weakness" of resistors to divide voltages (efficiently independantly of the loads), unless the to be applied loads are already large. (But large resistors are known to be "noisy"). Also consider the size of a one ohm resistor if subjected to 170V, ... that makes close to ... 29 kiloWatt !!! You could melt steel with that in less time than to say "rabbit" (or hopefully the in-house breaker will pop-out before without becoming a fire hazard).

Note that NOTHING tell to the circuit if it is R1 + R2 which is supposed to be the voltage divider, and R3 et R4 the loads, or if it is R3 + R4 which are the voltage divider and R1 et R2 the "loads".

In practice, the smaller of the two resistors in parallel will tend to be the "voltage divider", as long as they carry about the same current. Here, R1 <<< R3 and R2 <<< R4, so R1 and R2 would perform as voltage divider

... as long as R5 is really small.

Use R5 as 1 mega-ohm, and it is as if it wasn't there, so we are getting then, in reality, two independant circuits, R1 in series with R2, and a second circuit, R3 in series with R4, each of these two paths becoming independant voltage divider practically not seeing at all the other "voltage divider".

So, in summary, for a voltage divider by 2 resistors in series, the SAME current should flow through them ( or approximatively the same current), AND, important, the added load in parallel must be much greater than the resistor to whom they are in parallel and which is part of the divider.

Technical note: you can move over the resistor to see its current, voltage difference and power to dissipate.