Thread for random DIY-related questions

[crochambeau]

Could be an IR LED or sensor, similar to those used in old TV remote controls. Semiconductors do react to light, and the saw waveform could be a result of a diode blocking conduction in one direction. I would expect it to be rather inefficient as an LDR, but gain corrects many shortcomings.

One of those automated component testers (like the Peak or many of the mega328 based knockoffs) can answer this sort of question at the push of a button. I have only speculation to offer.

[FAP]

Talked to one of my [vastly more knowledgeable] coworkers and they basically came to the same conclusions you did: they even tested it with one of those mega328-type devices (though results were inconclusive).

I’m leaning towards photodiode, but without a way to emit infrared to test it, I can’t rule out IR sensor, either. I’m also totally clueless about IR to begin with so perhaps there’s another way to test it…?

I kind of like these compared to traditional LDRs: they’re far less sensitive to light (ambient or otherwise) though I can totally see how this would be a problem in many applications.

[crochambeau]

Any standard/classic remote control can quickly determine if the parts are sensitive to infra-red, just aim and shoot while the circuit is under operation. Could be a super cool application.

[FAP]

CONFIRMED they are, in fact, IR sensors. Thanks!

Using these in an anti-parallel pair in lieu of a proper LDR yields some intriguing results indeed. The TV remote interference is an obvious plus, but I was kind of surprised at how similar the waveforms of a proper LDR vs. an IR sensor anti-parallel pair could look at times. See attached screenshots: the first one compares both methods with a static frequency (not exactly the same frequency for both, but close enough), while the second highlights an example drop in frequency from high to low (again, the control in these experiments was anything but consistent, so analyze these screenshots with that in mind).
EDIT: the source signal is a square wave oscillator running from a range of about 30Hz–3kHz (via a 40106 chip).

The one major drawback (IMO) is a proper LDR has a more "natural" sounding transition from high to low frequencies: in practice, said transition sounds like a nice, smooth bass drop. In contrast, the same transition in an IR pair lacks that same impact. Obviously an IR pair necessitates twice the number of components as well, though this isn't an issue for me since I have a bag of like 50 of them.

Having just one IR sensor simply gives half of a waveform, though certain conditions (e.g. exposure to visible light) can pull it closer to zero; up close, it looks more akin to a series of pulses than it does a sawtooth wave.

I just keep on tripping into new rabbit holes, each one deeper than the last...

[FAP]

I'm trying to add a control to an LFO that'll allow me to adjust how much light from the LED will "bleed" into the low state of a square wave function (and eventually the triangle wave, if possible).
For context, this is the circuit I'm using:

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And here's a visual aid of what I'm trying to achieve:

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As you can see, the sine wave output already has a trim that can set how "deep" the waveform swings; rotating the trim fully counter-clockwise essentially yields a solid light (LED is always on), while rotating to the other side gradually decreases the brightness of the low state until the low state is simply off.

I tried to replicate the same trim circuit for the square and triangle waves (near U1b-pin 5 and U1a-pin 2, respectively), but it didn't work: this makes sense given how the op amps are cascaded/daisy-chained together, as adding a similar trim circuit earlier in the "chain" would affect every op-amp in it thereafter.

Also, the circuit in the bottom right adjusts the brightness of the high state: what I'm trying to do is essentially the inverse of that.

[FAP]

Bumping for previous comment^

[crochambeau]

I'd have to build one in order to fully comprehend what is going on and what will work and not work within that topology. I'm not doing that. Here's the method I would start with in trying to achieve your aims, no idea if it will actually work within the context (but I have employed the concept to other things, so I know I'm not completely talking out my ass):

Build a variable DC node, limited to whatever range fits your design best.

That node is now "MINIMUM" (a virtual "ground"), reference that to drive side of the LED/light source and you can adjust it so the illumination just barely comes on or whatever, because the LED/light source itself is referenced to an actual ground or lower reference point on the back side.

With me so far?

Then simply output the LFO waveform such that the variable DC is the LFO reference point, and you've got your wiggle on top of whatever baseline you wish.

Again, just speculation on my part...

[FAP]

Okay, there's no way this question doesn't have a simple answer, but I sure as hell ain't finding it: how do I measure the amount of watts I need for a resistor, when I know the voltage, current and resistance?

Here's the known values I'm working with:

V = 9vDC
I = 2A
R = 10kΩ

I try plugging it in to ohm's law for solve for power but get different results each time:

P = V * I = 9 * 2 = 18w
P = I^2 * R = 2^2 * 10k = 4 * 10k = 40kw
P = V^2 / R = 9^2 / 10k = 18/10k = 0.0018w

What the hell am I missing?

EDIT:
On a similar note, I've tried two different multimeters, both in perfect working condition, and still cannot for the life of me measure any current. I place the meter in series with the circuit (an LED and a resistor: the circuit works fine (LED turns on) so current is definitely passing, it's just not being read at all. I've always had this problem trying to read current: again, what am I doing wrong? It's so frustrating!

[crochambeau]

Okay, there's no way this question doesn't have a simple answer, but I sure as hell ain't finding it: how do I measure the amount of watts I need for a resistor, when I know the voltage, current and resistance?

Here's the known values I'm working with:

V = 9vDC
I = 2A
R = 10kΩ

I try plugging it in to ohm's law for solve for power but get different results each time:

P = V * I = 9 * 2 = 18w
P = I^2 * R = 2^2 * 10k = 4 * 10k = 40kw
P = V^2 / R = 9^2 / 10k = 18/10k = 0.0018w

What the hell am I missing?

EDIT:
On a similar note, I've tried two different multimeters, both in perfect working condition, and still cannot for the life of me measure any current. I place the meter in series with the circuit (an LED and a resistor: the circuit works fine (LED turns on) so current is definitely passing, it's just not being read at all. I've always had this problem trying to read current: again, what am I doing wrong? It's so frustrating!

Where are you getting 2 amps?

Are you referring to a 2 amp power supply?

If so, that is a power supply that is capable of delivering 2 amps into a circuit. It does not force a circuit to eat 2 amps.

Ohms law: (V= voltage, I= current, R=resistance)

V=IR or I = V÷R or R= V÷I

So, let's take the two reasonable numbers. 9 volts and 10K. 9÷10000 = 0.0009 amps, or a little less than 1mA

Use THAT current to calculate dissipation.

To measure current directly the set-up you describe is correct, so long as the meter is set up to read current. Place the meter IN LINE/series connection with the power feeding the circuit under test, select current read functionality - which typically requires your leads to be plugged into specific jacks, and then directly measure.

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To correctly read the ~1 mA on topic here, I would want to insert my red lead into the jack to the immediate left of the black lead, which remains in place. Orientation of black lead into circuit will determine if the digital meter read out indicates a positive or negative current (whereas a mechanical movement might just deflect harder into the stop if oriented "backwards"). A user manual for your particular meter will indicate the correct orientation of current wire and common wire with relation to supplied power and circuit.

Or you can use a calculator: https://www.rapidtables.com/calc/electr ... lator.html

Note, it instructs you to enter TWO values, and my theory is that by entering three you introduced an element of chaos.

[FAP]

Are you referring to a 2 amp power supply?
If so, that is a power supply that is capable of delivering 2 amps into a circuit. It does not force a circuit to eat 2 amps.

This alone explained a lot: thank you.
And wouldn't you know it? My meter works now, too! I guess I just had the range set too high? I did readings of a few of my pedals—newer ones, older ones, digital ones, analog ones, brand name, boutique, DIY, etc.—and none of them cracked the 100mA threshold. A cursory google search claims most pedals to be in the ballpark of 80mA, though in some special cases they can draw upwards of 500mA (e.g. Eventide pedals, though to be fair I have none of my own to confirm this).

I'll cut straight to the chase: I'm trying to figure out why the pot on the power starve I built not too long ago seemingly cooked itself to death one day. That's why I was asking about power dissipation: I want to know how much power could've been going across that pot to have made it go all smokey the bear on me, and if possible, find a pot with a more robust power rating.

My theory is the pot burned itself out because there was no load (in this case, a pedal). I did briefly try reading current with only a pot in the test circuit and, while I can't remember exactly what I did, it did light up like a Christmas tree. Does that make any sense? Could I actually destroy a power starve by simply leaving it plugged in without a load?

The circuit I've been using is attached below. If lugs 1 & 2 were connected in this configuration (full-CCW rotation), shouldn't that then mean a direct short to GND?

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Now I also found a variant of this circuit where there's a resistor between lug 1 & GND (instead of straight to GND): this apparently limits the minimum voltage to ~1.8v (given a max of 9.0v). Would this also not, then, prevent the pot from burning? If the lowest possible resistance between +9v and GND is, for example, 2.2kΩ (see attached, below), then max power dissipation = V^2 / R = 9^2 / 2.2kΩ = 81 / 2,200 ≈ 0.037w. So then at that point, would I even need to worry about appropriate power ratings since the most power that could be dissipated across the entire starve circuit (assuming no load) is well under a fraction of a watt?

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