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Quick question, Ohm's law!

Started by Boba7, December 03, 2019, 06:55:37 PM

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Boba7

Hey guys,

I'm working on a pcb that uses a charge pump.
Some of the circuits using a charge pump can induce noise in certain conditions (like a Klon with a Strymon Timeline), but in the past I've had good luck adding filtering before and after the charge pump. It seemed to reduce, or even cancel the noise.

But anyways my question is about Ohm's law, and more specifically R31 in the schematic. I'd probably use 47r or 100r instead of 22r. But a regular 1/4w will be fine right?
The circuit around is basically a Klon (so boring, I know), so there's not much current draw

But how do you guys do the maths? Even for R14, how do you guys figure that a 1/4w will be just fine? (and I know it is fine, I've used that value a lot)

Thanks! :)


Max

Well, V=RxI (Ohm's law) and W=VxI (definition of power). You know R, if you know V then W= V^2/R gives you the maximum power dissipated on that resistor. If you know I then use W=RxI^2.

Hope this helps.

Boba7

Quote from: Max on December 03, 2019, 09:06:52 PM
Well, V=RxI (Ohm's law) and W=VxI (definition of power). You know R, if you know V then W= V^2/R gives you the maximum power dissipated on that resistor. If you know I then use W=RxI^2.

Hope this helps.

Thanks!
So if I take R31: it's 22r, and the voltage is 18v.
W= V^2/R
18^2/22= 14.7
What does that mean now?

And if I estimate that I is at maximum 0.01A (10mA is quite enough to feed a dual opamp I think)
W=RxI^2.
22x0.01^2=0.0022

So yeah, I've never studied those things, and I don't quite understand how it works!

EBK

Voltage across that resistor won't be 18V unless you short the other end to ground.  Even if you do short it, the charge pump has a maximum amount of current it can source and it has its own output impedance, both of which would limit the voltage.  Beyond that, I'm not much help, except my gut says 1/4-W is fine.  I definitely recommend that you do not try to find a 15W resistor to put there.
"There is a pestilence upon this land. Nothing is sacred. Even those who arrange and design shrubberies are under considerable economic stress in this period in history." --Roger the Shrubber

Boba7

Quote from: EBK on December 03, 2019, 10:30:20 PM
Beyond that, I'm not much help, except my gut says 1/4-W is fine.  I definitely recommend that you do not try to find a 15W resistor to put there.

Hahaha that's gonna be a huge pcb to accomodate for my much needed big resistor

Im convinced a 1/4w will work but I would love to understand better how that works!
Thanks for your message, and of course you're right about the voltage accross the resistor, I had not thought about that
:)

Max

R31 is a current limiting resistor like R14, therefore the voltage drop across it should be very small. The voltage is raised by TC1044 and the diodes-caps net. R31 can only cause a voltage drop, not a voltage rise, so if you have 18V after it, it will be a bit higher before, but it depends on the current flowing, according to Ohm's law.
If the current is zero then you have 18V before and after. V=RxI gives you the current drop, which would be 0, and thus W=0. W calculated as I said above is the power dissipated by that resistor with a certain current.
If you know the maximum current through R31, you can calculate the maximum power dissipated by R31 and therefore its minimum power rating. If it dissipates 0.1W it needs to be rated for at least 0.1W (of course with some margin, so, let's say 0.25W).
You can probably figure this out from the datasheets of the components downstream, but it can be tricky, you'd better use some simulation if you can.

Boba7

Quote from: Max on December 04, 2019, 08:12:01 AM
R31 is a current limiting resistor like R14, therefore the voltage drop across it should be very small. The voltage is raised by TC1044 and the diodes-caps net. R31 can only cause a voltage drop, not a voltage rise, so if you have 18V after it, it will be a bit higher before, but it depends on the current flowing, according to Ohm's law.
If the current is zero then you have 18V before and after. V=RxI gives you the current drop, which would be 0, and thus W=0. W calculated as I said above is the power dissipated by that resistor with a certain current.
If you know the maximum current through R31, you can calculate the maximum power dissipated by R31 and therefore its minimum power rating. If it dissipates 0.1W it needs to be rated for at least 0.1W (of course with some margin, so, let's say 0.25W).
You can probably figure this out from the datasheets of the components downstream, but it can be tricky, you'd better use some simulation if you can.

Thanks a lot, that's very helpful.
I don't have any simulation software but I may try one soon.

I'm pretty sure the current output of a TC1044s chip is 20mA max. And downstream there's only one TL072.
So if W=RxI^2
W=22x0.02^2
W=0.0088

That seems so low though, is that normal? :D

madbean

Quote from: Boba7 on December 04, 2019, 02:12:25 PM

Thanks a lot, that's very helpful.
I don't have any simulation software but I may try one soon.

I'm pretty sure the current output of a TC1044s chip is 20mA max. And downstream there's only one TL072.
So if W=RxI^2
W=22x0.02^2
W=0.0088

That seems so low though, is that normal? :D

That's simply the minimum wattage you could use. In real world practice you are going to use something that exceeds that and corresponds to the available specs from manufacturers (eg 1/8W, 1/4W).

I've found a good rule of thumb is 1/4W for 22R-100R and 1/2W for 10R.

You also want to consider the LP filter created by using a current limiting resistor and decoupling cap. For example, let's say you are using 100uF/100n combo for your PS filtering.

22R current limiting resistor:
f=1/(2*pi*R*C) = 1/(6.28*0.000022*100) = 72.4Hz for the 100uF cap
and, 72.4kHz for the 100n cap.
That's a decent amount of extra filtering on what may already be a well regulated and filtered 9v supply.

47R resistor:
33.9Hz and 33.9kHz resp.

So, as the resistance goes up, the freq. knee goes down obviously. And, as the resistance goes up the voltage drop across that resistor increases. The key is to find the right balance between proper filtering, minimal voltage drop and use the appropriate wattage so you part doesn't heat up as current is drawn through it. IOW, you don't need to re-invent the wheel every time you want to add some current limiting on your power. Just use this guideline:

  • High current draw requires higher wattage CLR.
  • Low current draw means you can use smaller wattage CLR and generally a smaller value.
  • Don't use a CLR too high as to cause too much of a voltage drop across it.


Also, you can look at using an inductor in place of a resistor. With those, you get a 2nd order LPF for free! Using the cap values I mentioned above a 1mH inductor gets you 2nd order LPF at 504Hz and 15.9kHz resp. So, a big improvement over just a resistor.

Boba7

Quote from: madbean on December 04, 2019, 02:50:28 PM

That's simply the minimum wattage you could use. In real world practice you are going to use something that exceeds that and corresponds to the available specs from manufacturers (eg 1/8W, 1/4W).

I've found a good rule of thumb is 1/4W for 22R-100R and 1/2W for 10R.

You also want to consider the LP filter created by using a current limiting resistor and decoupling cap. For example, let's say you are using 100uF/100n combo for your PS filtering.

22R current limiting resistor:
f=1/(2*pi*R*C) = 1/(6.28*0.000022*100) = 72.4Hz for the 100uF cap
and, 72.4kHz for the 100n cap.
That's a decent amount of extra filtering on what may already be a well regulated and filtered 9v supply.

47R resistor:
33.9Hz and 33.9kHz resp.

So, as the resistance goes up, the freq. knee goes down obviously. And, as the resistance goes up the voltage drop across that resistor increases. The key is to find the right balance between proper filtering, minimal voltage drop and use the appropriate wattage so you part doesn't heat up as current is drawn through it. IOW, you don't need to re-invent the wheel every time you want to add some current limiting on your power. Just use this guideline:

  • High current draw requires higher wattage CLR.
  • Low current draw means you can use smaller wattage CLR and generally a smaller value.
  • Don't use a CLR too high as to cause too much of a voltage drop across it.


Also, you can look at using an inductor in place of a resistor. With those, you get a 2nd order LPF for free! Using the cap values I mentioned above a 1mH inductor gets you 2nd order LPF at 504Hz and 15.9kHz resp. So, a big improvement over just a resistor.

Thanks a lot!

Yes an inductor might be even better, I've never used those and have yet to understand what they do/how they work!
I always chose the 22r/47u/100n combination by default as it was your choice in the Fatpants Jr.

Thanks again guys, its clearer in my mind now

Boba7

By the way, how do you do the maths for an inductor/cap filter? 

EBK

Quote from: Boba7 on December 04, 2019, 08:05:27 PM
By the way, how do you do the maths for an inductor/cap filter?
The math gets "fun".

You do the math just like with resistors, but you substitute the impedance of the inductor or cap, which is a complex number and varies with frequency.

Impedance of a capacitor is 1/(jωC), and
impedance of an inductor is jωL, where:

j is sqrt(-1),

ω is frequency in radians per second (equal to 2πf),

C is capacitance in farads, and

L is inductance in henries.

I can keep explaining if you are still interested.
"There is a pestilence upon this land. Nothing is sacred. Even those who arrange and design shrubberies are under considerable economic stress in this period in history." --Roger the Shrubber

Boba7

Quote from: EBK on December 04, 2019, 08:23:08 PM

The math gets "fun".

You do the math just like with resistors, but you substitute the impedance of the inductor or cap, which is a complex number and varies with frequency.

Impedance of a capacitor is 1/(jωC), and
impedance of an inductor is jωL, where:

j is sqrt(-1),

ω is frequency in radians per second (equal to 2πf),

C is capacitance in farads, and

L is inductance in henries.

I can keep explaining if you are still interested.

That's very kind of you, but I'm afraid I'll have to pass, it is way beyond my mathematical abilities right now! I'll stick with my 47r/47u/100n filtering for now haha
Thanks a lot for the answer!

EBK

There are several simpler approaches to the math, and very often we can make some time-saving assumptions.

Didn't mean to scare you with the meaty generalized version. 
"There is a pestilence upon this land. Nothing is sacred. Even those who arrange and design shrubberies are under considerable economic stress in this period in history." --Roger the Shrubber

Boba7

Quote from: EBK on December 04, 2019, 09:22:42 PM
There are several simpler approaches to the math, and very often we can make some time-saving assumptions.

Didn't mean to scare you with the meaty generalized version.

Well the I'm definitely interested in learning more yes! Feel free to write more if you have the time and willingness, Ill try to grab as much as I can

If inductors are more efficient at filtering, why are they not used more often? First time I saw one was in a VFE pedal, and I cant seem to recall any other brand using them!

Boba7

Quote from: Boba7 on December 04, 2019, 09:44:28 PM
If inductors are more efficient at filtering, why are they not used more often? First time I saw one was in a VFE pedal, and I cant seem to recall any other brand using them!
Beginning of an answer: https://www.diystompboxes.com/smfforum/index.php?topic=91002.0