A question regarding complications with resistor networks.

[msuffidy]

I just just reviewing something I did earlier and noticed there may be a complication. The objective was to use a resistor network to change the voltage of a DC adapter to supply a lower voltage device. So what I did had the sort of logic as this. Say you had 2 resistors across a battery in series. At any time if the 2 resistors are the same value in between them you would have 4.5 Volts. So you can change the values the same 2 resistors to alter the current draw of the circuit. Now if you place a load on the mid point with what becomes a parallel circuit, the resistors add in a reciprocal way. So my logic was just to find the value of, in this case, R2 and 121 ohms such then when in a parallel circuit they equalled the 100 ohms I made R1. That is R2 was 576 ohms in this case. But I didn't think of the voltage drop across the LED that you usual subtract from V. This would only apply to one branch.

So I was wondering how these 'voltage drops' are considered generally, like can a motor coil be considered a voltage drop, or a resistor? Anyway just commenting on this sort of problem as I see it as complications with tapping resistor network voltage supplies would be good.

[TimFox]

Consider the Thévenin equivalent of R1 and R2:  the ratio of the resistors gives you the open circuit (no-load) voltage across R2, and the parallel combination (R1 x R2) / (R1 + R2) gives you the source resistance driving the load from that open-circuit voltage.
In general, resistive dividers are quite inefficient when the load requires more than nominal current.

[Benta]

This is where you need to start studying.
The keywords are "Kirchhoff's Current Laws" and "Kirchhoff's Voltage Laws" (KCL and KVL).

But an approach could be to replace the LED forward voltage with a voltage source with the same voltage to emulate the LED voltage drop. This will only work for a steady-state analysis with the 9 V battery always in place.

But Tim's absolutely right: all three resistors could be replaced with one.

[msuffidy]

But Tim's absolutely right: all three resistors could be replaced with one.

Probably, but the 121 one is stuck to the thing I am powering.

[TimFox]

So, just put the remaining resistance required in the series circuit between the battery and the 121 ohm resistor.
For example, if you have a 9 V battery, and you want 10 mA through the LED, and the LED has a 2 V forward voltage at that current (read the datasheet for real value), you have a total of (9 - 2) = 7 V across the resistors, of which 1.21 V is across the 121 ohm resistor, leaving 5.79 V for the extra resistor, which would be 5.79/0.01 = 579 ohms.

[msuffidy]

Yes I guess that would work. There is a bit of an issue in that I don't actually know the value of the laser diode LED voltage drop, but know the 121 ohms and it was fed by a 4.5 volt supply. If I knew that value I could figure out the original amperage of the LED and select a resistor to supply that amperage with the 121 from 9 volt supply. I am avoiding the question about calculating parallel resistors with different branch voltage drops though.

[TimFox]

So, if it worked from a +4.5 V supply with a 121 ohm resistor, and you don't know the current, you will have to either measure it or guess.
Now, obviously the current must be less than 4.5 V/121 = 37 mA.
If the laser voltage is 2.0 V, then the present current would be 2.5 V/121 = 21 mA.
I would guess 20 mA, so the extra resistor would be (9.0 - 4.5 V) / (20 mA) = 225 .
You're not going to get a better answer with the resistor divider, unless you make the resistors so small as to pull too much current from the battery.

[msuffidy]

Thanks for the responses.

At 20ma for the laser diode and 2V forward drop. I get the total resistance required to be 350 ohms (7v / 0.020A). With the 121 there, you need 229. Which seems the same.