The short version is: I've been trying to do this with a voltage divider, but it's just not working.

Part A is an EHX LPB-1 Booster clone circuit: I measure ~0.365mA max going through it.*

**Interestingly, the official specs list 1mA of current draw.*

Part B is a karaoke microphone w/ echo: I measure 108.5mA max going through it.†

*†varies based on volume of input source, delay time, feedback amount, etc. The minimum I read was ~99.6mA.*

Skipping ahead a bit, it's now come to my understanding that, according to this video, a voltage divider generally shouldn't be used if ≥10mA needs to be supplied.

So what are my options?

For context (though it may just confuse readers more than it helps) here's my math & notes:

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As I understand it, 4.5v could be attained simply by putting two equal resistors in series between 9v and GND, then attaching a lead to where the two resistors meet: makes sense.

One of the earliest problems I came across was, the higher the resistor values, the lower the supplied current: okay, so that just means lower resistor values, right?

Well, I initially came up with a value of 42Ω for the resistors (R = V/I = 4.5v/0.1085A = 42Ω): this seemed suspiciously low to me and, predictably, it didn't work. There was obviously

*some*current being drawn for Part B, but not enough. At the same time, I didn't want to risk lowering it any more and frying the circuit (the max voltage for the primary IC in Part B is ~5.5-6.0v).

I tried simplifying my calculations with the Rthevenin equation from the first video and came up with 21Ω:

Rth = (Rtop * Rbottom)/(Rtop+Rbottom) = (42 * 42)/(42+42) = 1764/84 = 21Ω

Meanwhile, I couldn't get a solid read on the cumulative resistance (Rload) of Part B's circuitry. Again, it's to my understanding that DMM measurements shouldn't be trusted when calculating Rload, so I went with the last method highlighted in this article and came up with 45Ω, like so:

Runknown = Rload = the cumulative resistance of Part B's circuitry

Rknown = 38.7Ω

Vsource = 4.65v (in this case, I'm omitting Part A entirely and only powering Part B)

Measured voltage across Rknown w/ Rload in series = 3.97v

Current of Rknown:

I = V/R = 3.97v/38.7Ω = 0.10258A = 102.58mA

Resistance of Rload:

R = V/I = 4.65v/0.10258A = 45Ω

So then, plugging this back into the Rthevenin calculations I got:

Rth = 21Ω

Rload = 45Ω

Rth + Rload = 66Ω

Vload = Vin * ((Rload)/(Rload + Rth) = 4.5 * (45/66) = 3.068v

Plug that into Ohm's law:

I = V/R = 3.068v/45Ω = 0.06818A = 68.18mA

WTF this doesn't make sense: the only way I'd even get close to 100mA is if Rth = 0, and if Rth = 0, then that would effectively mean 9v straight to Part B, which would fry the thing.

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As you can imagine, I've spent enough time on this already: is there a simpler solution I could try?

I've got a few standard op-amp chips on hand (e.g. TL072): would

*they*help?

I mean, I guess I could just do 6xAA's with a center-tap, but it'd be nice to be able to power this thing through a standard 9v wall wart [like a guitar pedal].

One more bit of context: the only reason I'm adding Part A at all is because the output signal on Part B is so low that I need a way to boost it. Part A actually comes

*after*Part B in the chain.