PSU series resistor, V=IR, and the voltage drop curve

[aion]

There's something I've been struggling to wrap my mind around something the past few days. I thought I understood how PSU current-limiting resistors work - the resistor we sometimes put in series with the power supply that forms an R-C filter with the main filter capacitor.

So: V=IR, got it. P=V²/R, got it.

Say we have a circuit with 10mA current draw. A 10R resistor dissipates 0.001 watts as it drops 0.1V.

Bump it to 100R, it dissipates 0.01 watts as it drops 1V.

Jump another decade to 1k and it's dissipating 0.1 watts... and 10V. It's a 9V circuit though. And a 1k resistor of course doesn't block anywhere near 100% of the voltage. Even a 10M resistor is going to let a few millivolts through.

So, the formula being immutable, if the resistance doesn't change then the current draw must. It isn't drawing 10mA anymore. And we'll allow that the circuit doesn't work very well at this point, and hasn't for awhile if we've gotten far enough in the starvation process that we are noticing these effects, so it's primarily just theory.

But, how do we predict the relationship of I to R as the value of R goes up, when V is also going down? Presumably the circuit will try to draw 10mA until it's unable to. But wouldn't different circuits have different capacities in this regard? Op-amps usually have a graph in the datasheet showing how much current they draw as supply voltage changes, but this spec varies from part to part.

At some point, the current that the op-amp wants isn't what it gets... so there is some relationship wherein the resistor reduces supply voltage, so the op-amp wants less current, but it also gets less than it wants due to the resistor that reduces the supply voltage.

I understand the overall load of the circuit in proportion to the CLR, and that the circuit on the other end of the resistor can for the most part be condensed into a single resistor, per Thévenin. But it seems that any formula describing the current/voltage curve as the resistor changes would have to take into account the specific active devices in the circuit, right?

I have the sense that there is something basic I'm missing and I will reveal my carefully-guarded idiocy just by asking. But you are all pretty friendly so I don't mind taking that risk in this space, especially if it helps me patch something missing in my knowledge.

[R.G.]

For a resistor, it's always V = I * R. That is, for the voltage available to it, a resistor always lets a current V/R flow. And for a current forced through it, the voltage across the resistor is always V = I * R.

It may be that you're missing that any other stuff in series with the resistor can be affecting the voltage across it or the current through it. You're kind of picking that up in your post, but maybe an example would help. If you have a >> perfect << 10V voltage source, and hook a 1K resistor across it, then the current is just I = 10V/1K = 10ma.
If you put a silicon diode in series with the 1K across the 10V source, the diode voltage increases to about 0.6V to 0.7V (depending on the type of diode, temperature, etc.) and the diode "eats up" that 0.6V or so. Now the resistor only has 10V-0.6V = 9.4V across it, and the current through the resistor has to be and must be 9.4V/1K = 9.6ma. The diode itself will allow a lot more current through, so the 1K resistor does, in fact, limit the current.

Change the diode to a 1.4V-forward-drop red LED and the LED uses up 1.4V out of the 10V, so the resistor only gets 8.6V to work with.  The resistor in this case limits the current to 8.6V/1k = 8.6ma.

If the 1K is in series with a JFET that has an Idss of 3ma, and it's gate is shorted to its source. This means the JFET won't let more than 3ma through at all. Now you have 10V across a 1K resistor and a JFET that approximates at constant current source. The current out of the 10V source is limited to the 3ma by the JFET, so the resistor only gets 3ma. The voltage across the resistor is controlled by the current at 3ma * 1K = 3V because the JFET is limiting the current, not the resistor. The resistor is starved down to 3V because the current is externally limited to 3ma.

If you change the resistor to 1 ohm but keep the 3ma JFET, then the resistor produces 3mV. If it's 10 ohms, the resistor produces 30mV.

A  resistor is only a current limiter if what it's in series with wants more current than the resistor would let through. If the circuit in series with the resistor naturally self-limits its current, the resistor then doesn't limit the current, it produces a voltage across it that can be used to sample/sense the current the circuit wants. A resistor is a current limiting resistor only if it is what is determining the current.

A resistor in series with the supply voltage in most pedals, especially if there is a BFC (Big Freakin' Capacitor) after the resistor should not be limiting the current the circuit gets at all. It's there to filter the incoming power by forming a low pass filter with the BFC. In this case, the smaller the better to not lower the voltage to the circuit, and the bigger the better for better lowpass filtering. Yep, they're contradictory requirements.

[Rob Strand]

So, the formula being immutable, if the resistance doesn't change then the current draw must. It isn't drawing 10mA anymore. And we'll allow that the circuit doesn't work very well at this point, and hasn't for awhile if we've gotten far enough in the starvation process that we are noticing these effects, so it's primarily just theory.

The formula is immutable but the key point is an effects circuit is a passive load.  A theoretical 10mA current source can model a load but it is actually an active device from a theoretical point of view.  That's where things mess with your head.  If you have a 10V source and a 10M series resistor which is feeding a current source:  VL = V - I*R = 10 - 10M*10e-3 = 10 - 100kV = -99.99kV.  In other words the formula does works.  Since the current source is an active source it can *supply* current and pull the output terminal negative.

A passive circuit cannot supply current.  It has a VI characteristic which levels off at V=0 at I=0.  We might think of a pedal as 10V and 10mA being a 10mA current source but it's not an accurate model.  You could model the effects pedal as a resistive load instead of a current source load,  RL = 10V/10mA = 1k.  The main advantage here in getting around your problem is as the voltage across the pedal drops and the predicted pedal load current drops.  So a 100 ohm series resistor will pull current IL = V / (Rs + RL) = 10/(100+1k) = 9.09mA and the voltage across the pedal will be 10-100*9.09m = 9.09V.  The point here is that lower voltage at the pedal causes it to pull less than 10mA.  And as the series resistor gets larger and larger the voltage across the pedal get lower.  For example a 10M series resistor:  IL = V / (Rs + RL) = 10 / (10M + 1k) = 0.9999uA, and the voltage across the pedal is 1k*0.9999uA = 0.9999mV  which is more what you would expect.

A real pedal circuit is a more complicated load than either a current source or a resisistor.  Even if the pedal acts like a passive current source, like RG's JFET example, at some point the supply voltage will be too low for JFET circuit to operate correctly.  It will start passing less than 10mA.  In fact as the supply voltage drops to zero the supply current will approach zero.
Here's a good example a Fuzz Face:

[PRR]

All that R.G and Rob said.

> 10V. It's a 9V circuit though.

Draw the WHOLE circuit. (I've said that before.)

Define (or assume) what is in the "pedal".

Some simple assumptions:

Pedal is a resistor. Say a few hundred or a few kOhms. Now you have a simple resistor divider and should know what to do.

Pedal is a constant current. Transistor(s) or even a hot filament lamp, sorta. I*R=V.

If I or R is not pretty constant (as you say, it can "curve"), you would like a graph or chart or equation of how it varies. This will be recursive: estimate a load point, look-up the actual value, re-figure until error is negligible (and you can negligible a lot in stage-pedals).

[Ben N]

Huh? Isn't it P=V²/R?

[Rob Strand]

Huh? Isn't it P=V²/R?

Yes.

P = V^2 / R
P = I^2 R

[antonis]

No P= V*I..??
(the simplest form - no fraction no exponent)

[ElectricDruid]

No P= V*I..??
(the simplest form - no fraction no exponent)

I'm with you, Antonis. I like that simple form best too.

[aion]

Thanks, this is incredibly helpful. Glad to hear that it's at least not as basic as I feared! But I'm starting to get it now.

A  resistor is only a current limiter if what it's in series with wants more current than the resistor would let through. If the circuit in series with the resistor naturally self-limits its current, the resistor then doesn't limit the current, it produces a voltage across it that can be used to sample/sense the current the circuit wants. A resistor is a current limiting resistor only if it is what is determining the current.

This is where it clicked. I would now describe the concept I was trying to understand as the point where the resistor takes over.

A real pedal circuit is a more complicated load than either a current source or a resisistor.  Even if the pedal acts like a passive current source, like RG's JFET example, at some point the supply voltage will be too low for JFET circuit to operate correctly.  It will start passing less than 10mA.  In fact as the supply voltage drops to zero the supply current will approach zero.

And this (plus the LTSpice proof) was another click. I had done a spice model of an OCD circuit and was changing the CLR and watching how the voltage and current draw changed. The power dissipation peaked when the supply voltage dropped 58.9% of the voltage (a resistor value just over 5.1k), and from there the dissipation went down as the resistor increased.

So again... without the specific words to describe it that directly, I knew it wasn't as simple as a current source or a resistor, though both ideas get you most of the way there as long as you stay inside the window of practicality.

Huh? Isn't it P=V²/R?

Yep, I missed the slash... I'm going to edit the first post now for future generations, but will leave the acknowledge here that I screwed it up!

I am also going to move the question about the 10R resistor to a different thread... in my head it was related, but it is really a different concept. All the responses have been to the first part of the question, so it'll be a clean break.

[Rob Strand]

And this (plus the LTSpice proof) was another click. I had done a spice model of an OCD circuit and was changing the CLR and watching how the voltage and current draw changed. The power dissipation peaked when the supply voltage dropped 58.9% of the voltage (a resistor value just over 5.1k), and from there the dissipation went down as the resistor increased.

So again... without the specific words to describe it that directly, I knew it wasn't as simple as a current source or a resistor, though both ideas get you most of the way there as long as you stay inside the window of practicality.

FYI, opamp models, in fact many IC's, often don't model current draw and hence the VI behaviour of the power rails very well.   A large majority of opamps use a macro model, not a transistor level model.  For general use macro models can end up being better but the thing that gets lost is current draw and VI behaviour of the power rails.   Transistor level models are quite rate, and often aren't accurate anyway because they don't model each of the 30+ transistors inside the IC correctly - something I've looked at in detail.

[PRR]

current draw and VI behaviour of the power rails.   Transistor level models are quite rate rare .... the 30+ transistors inside the IC

Yeah but: all "normal" opamps, 28 of those 30+ transistors work near "class A", solo or with a partner (long tail pair). At small-audio level, the glaring exception is push-pull output stages with <10k load.

But IF he has that hundred-uFd cap, even this current shift is slowed.

And what I am hearing in Kevin's posts is a desire to fix anecdotal reports from customers. Some reports might be explained as customer carelessness or stupidity (but not in public!). I've heard amazing things from clients.

[Rob Strand]

Yeah but: all "normal" opamps, 28 of those 30+ transistors work near "class A", solo or with a partner (long tail pair). At small-audio level, the glaring exception is push-pull output stages with <10k load.

You can have class A and still have the bias current (and supply current) varying with the supply voltage.

FWIW the bias varies with supply voltage  more on the older opamps than the more modern ones.   However, the bias regulation circuit on modern opamps will poop out at some low voltage and then the bias could start changing quicker than older opamps.