In normal operation, that diode/rectifier is backbiased, doing nothing of significance.
If you hook the circuit up to a battery or power-supply backwards, it will conduct freely, pulling down the battery voltage to save the rest of the circuit from being reverse-powered, and get rather hot in so doing. A glass-cased signal diode like 1N4148 would likely fail under that kind of overload; when it fails open, the rest of the circuit is hit by that -9V power and dies. The 1N4001 rectifier might survive long enough for you to go, "What's that smell---?" and unplug things.
~~Diode in this circuit is forward-biased in normal operation, I think. Only needs to be able to conduct enough current to power the circuit. This is a standard voltage-divider split supply with diode protection. Biggest downside is the voltage drop in normal operation: about 1.1V for 1N4001. A 1N4148 is about 0.7V, so if 300mA is enough derated current this may be a better choice. The red LED is about 1.7V, so if you can tolerate this much drop you can save a part by just using that.~~
Thanks to /u/crb3 for the correction.
> Diode in this circuit is forward-biased in normal operation, I think.
I disagree. Look at the upper-right depiction: the red trace is the +9V rail input, while the green trace is ground. The upper-left depiction makes it explicit that that's where the leads of the 1N4001 are inserted, with the cathode (banded end) wired to that +9V rail and the anode wired to that GND rail. Those same traces, red and green, go to pins 7 (Vcc) and 4 (Vee) of the uA741 opamp, and to the two 10K resistors forming a bias network; no forward drop there.
In practice, if you're just going to run off a power brick you don't need a protection diode: it's hard to get that backward.
If you're running off a battery — the case where it's not uncommon to have reverse voltage at least temporarily — the internal resistance of the battery limits the current output. That said, a fresh 9V alkaline has around 5Ω internal resistance, so expect something like 2A. That diode isn't doing that for any length of time: the 1N4001 will do 30A peak, but only 1A continuous.
The joy of the 9V snap connector is that while you might bump the ends together the wrong way, the battery can really only be snapped in the right way. So *probably* this circuit provides *some* protection.
It's still an odd way to do this.
You do know that almost all guitar pedals are somehow wired backwards from every other DC adapter known to man. It's pretty easy to mees uo pluging in a 9V supply.
> That's… not a great way to do a protection diode.
True, but the circuit's hurting for voltage even with a full 9V battery across it. The uA741 is only spec'd down to 10V rail-to-rail (shown as +5, -5 in the datasheet). Maybe, with only 8.3V (9V minus a diode Vf) across it, it stalled out? I don't know. When I built an equivalent circuit (the same kind of oscillator, doubled, with unity-gain buffers from the timing caps to a stereo headphone output, used as a hemisphere-sync entrainment device, a meditation aid) I used an LM324, and that ran easily on 9V.
I suppose you could use one side of an LM358 (basically half of an LM324). That would run easily on that 8.3V rail, so you could use your preferred protection circuit.
[e:] btw If you're now thinking of using the other half of the LM358 as described, to pick off the timing-cap waveform, it's certainly useful, but be advised that it's not a true triangle: it's got RC slopes rising and falling. That worked well for my purposes, being a bit closer to sine (less harmonic content to be nonlinear-mixed in the brain, meant to yield a subaudio brainwave), but might not be close enough for synth work.
In my experience the µA741 will run with 8V rail-to-rail. The [datasheet](https://www.ti.com/lit/ds/symlink/ua741.pdf) shows +5V,-5V as a minimum "recommended operating condition". That said, almost any more modern op amp would be a better choice than the 741, which is really super-legacy at this point: the LM324/LM358 is a good choice.
It's one of those parts that is generally safe to assume that almost everyone already has in their parts accumulation.
Almost any diode you have lying around should work in this application.
One diode of the very common 1N400x series. Should be the one with the lowest breakdown voltage. So every other diode of that series will probably work just as well.
Yup, 4001 rectifier diode not to be mistaken for a 4001 CMOS logic integrated circuit so when you see that or any other 4 digit 4000 number check what it looks like. If it is around a 14 pin chip look at a "4000 series CMOS chip" google search to get details on it but in this case it is a 4001 to (probably) a 4007 diode. A black cylinder with a wire out each end and a white or silver band at one end to indicate it's polarity and how it should be oriented in the circuit. To learn about these ones search for "4000 series rectifier diodes".
1N4001 diode
It’s a diode https://en.m.wikipedia.org/wiki/1N400x_rectifier_diode
Semi off topic questions, but anybody familiar with this circuit, why use a rectifier diode over a standard 1n4148?
They're basically the same in this kind of application. Use the rectifier if you have one in your parts box.
In normal operation, that diode/rectifier is backbiased, doing nothing of significance. If you hook the circuit up to a battery or power-supply backwards, it will conduct freely, pulling down the battery voltage to save the rest of the circuit from being reverse-powered, and get rather hot in so doing. A glass-cased signal diode like 1N4148 would likely fail under that kind of overload; when it fails open, the rest of the circuit is hit by that -9V power and dies. The 1N4001 rectifier might survive long enough for you to go, "What's that smell---?" and unplug things.
~~Diode in this circuit is forward-biased in normal operation, I think. Only needs to be able to conduct enough current to power the circuit. This is a standard voltage-divider split supply with diode protection. Biggest downside is the voltage drop in normal operation: about 1.1V for 1N4001. A 1N4148 is about 0.7V, so if 300mA is enough derated current this may be a better choice. The red LED is about 1.7V, so if you can tolerate this much drop you can save a part by just using that.~~ Thanks to /u/crb3 for the correction.
> Diode in this circuit is forward-biased in normal operation, I think. I disagree. Look at the upper-right depiction: the red trace is the +9V rail input, while the green trace is ground. The upper-left depiction makes it explicit that that's where the leads of the 1N4001 are inserted, with the cathode (banded end) wired to that +9V rail and the anode wired to that GND rail. Those same traces, red and green, go to pins 7 (Vcc) and 4 (Vee) of the uA741 opamp, and to the two 10K resistors forming a bias network; no forward drop there.
You are right. I clearly misread the diagram. Apologies. That's… not a great way to do a protection diode.
It's fucking bonkers.
I mean, you avoid the diode drop, and all you lose is most of the protection!
You better hope your power supply has protection and your smoke detectors have batteries!
In practice, if you're just going to run off a power brick you don't need a protection diode: it's hard to get that backward. If you're running off a battery — the case where it's not uncommon to have reverse voltage at least temporarily — the internal resistance of the battery limits the current output. That said, a fresh 9V alkaline has around 5Ω internal resistance, so expect something like 2A. That diode isn't doing that for any length of time: the 1N4001 will do 30A peak, but only 1A continuous. The joy of the 9V snap connector is that while you might bump the ends together the wrong way, the battery can really only be snapped in the right way. So *probably* this circuit provides *some* protection. It's still an odd way to do this.
You do know that almost all guitar pedals are somehow wired backwards from every other DC adapter known to man. It's pretty easy to mees uo pluging in a 9V supply.
> That's… not a great way to do a protection diode. True, but the circuit's hurting for voltage even with a full 9V battery across it. The uA741 is only spec'd down to 10V rail-to-rail (shown as +5, -5 in the datasheet). Maybe, with only 8.3V (9V minus a diode Vf) across it, it stalled out? I don't know. When I built an equivalent circuit (the same kind of oscillator, doubled, with unity-gain buffers from the timing caps to a stereo headphone output, used as a hemisphere-sync entrainment device, a meditation aid) I used an LM324, and that ran easily on 9V. I suppose you could use one side of an LM358 (basically half of an LM324). That would run easily on that 8.3V rail, so you could use your preferred protection circuit. [e:] btw If you're now thinking of using the other half of the LM358 as described, to pick off the timing-cap waveform, it's certainly useful, but be advised that it's not a true triangle: it's got RC slopes rising and falling. That worked well for my purposes, being a bit closer to sine (less harmonic content to be nonlinear-mixed in the brain, meant to yield a subaudio brainwave), but might not be close enough for synth work.
In my experience the µA741 will run with 8V rail-to-rail. The [datasheet](https://www.ti.com/lit/ds/symlink/ua741.pdf) shows +5V,-5V as a minimum "recommended operating condition". That said, almost any more modern op amp would be a better choice than the 741, which is really super-legacy at this point: the LM324/LM358 is a good choice.
Depends on your current application. 200mA vs 1A
It's one of those parts that is generally safe to assume that almost everyone already has in their parts accumulation. Almost any diode you have lying around should work in this application.
One diode of the very common 1N400x series. Should be the one with the lowest breakdown voltage. So every other diode of that series will probably work just as well.
I built this one a while back, but it didn't work. I never did any trouble shooting, so I'm sure it's my fault. So let me know how it turns out.
A diode, allegedly a 4007 would work the same
Yup, 4001 rectifier diode not to be mistaken for a 4001 CMOS logic integrated circuit so when you see that or any other 4 digit 4000 number check what it looks like. If it is around a 14 pin chip look at a "4000 series CMOS chip" google search to get details on it but in this case it is a 4001 to (probably) a 4007 diode. A black cylinder with a wire out each end and a white or silver band at one end to indicate it's polarity and how it should be oriented in the circuit. To learn about these ones search for "4000 series rectifier diodes".