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Noobie question about using a diode to protect power to a Looper pedal

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roadtrip

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I am building a Looper that runs on 9 volts, sleeve positive. Is it easy to put in a diode to protect against reversing the power by mistake? Do I put the diode in the negative connection from the DC jack (tip), and which way does the band face?

Also, how do I select a diode? I'm looking at an inexpensive 1N4007 1A 1000V General Purpose Rectifier Diode. Will that work? If not, what am I looking for?

Thanks in advance!
 
Diodes can be configured as simply protective. That is, they short the input (adding a resistor maybe advised) when the polarity is reversed.

They can also be configured to facilitate either polarity. A bridge rectifier does that. The disadvantage is the small voltage drop you get. If your circuit needs 5V, the drop from a 9V battery can be tolerated.

What is a "Looper?" Can you provide the circuit you plan that will let us correct it?
 
A diode can go in either lead, but traditionally it goes in the positive, the silver band towards the device being supplied, in this case the looper.

What Jp was saying is that sometimes the diode is placed inside the device to be protected across the supply inversely connected, so that if you power the thing with the wrong polarity the diode conducts and effectively shorts out the supply, blowing a fuse and protecting it, only if there isnt a fuse, and there often isnt problems are going to arise.

Jp a looper is a sampler, you pressa button it samples audio, then when you next press the button it sstops ampling audio then plays what it has just sampled over and over, you can sort of record & play your own accompaniment as you go, very popular for street performers.
 
Thanks for the quick reply jpanhalt. The type of Looper I am building is for electric guitar pedal effects. It uses five Clickless Bypass Relay type circuits wired together in series to put effects in and out of the signal chain. They run at 9 volts.

The relays run at less than 1 mA, and the led that shows if the loop is engaged can be around 2 mA but I tested a 20 mA Led on one relay circuit already and it works. The relay peaks at 23 mA (not including the Led) when it switches.

My concern is that I may plug in a 2.1 mm sleeve negative 9 volts, and fry the whole thing. There is no diodes on the circuit boards.

I just wanted to block 9 volts from going to the circuits negative connection in case I use the wrong AC/DC adapter.
 
Thanks dr pepper. Ok I get what jp is saying now. That makes sense with a fuse.

Yes your right about a looper being a sampler. I think looper is also slang maybe for an effect switching system as the simple ones just loop the effects in and out of "the loop" in series.

I would have thought that the diode would work on the negative connection of the power jack as then it would block the 9volts + from reaching the circuit.
 
Your right it would work in the - line, its just the majority of that kind of thing that I have seen its in the +.

Edit: thinking about it the diode needs to go in the +, if you put it in the - then ground will be lifted by a diode drop - 0.6v, this could cause ground loop issues if you have other effects units in the system, or even just between the loop and the amp.
 
Great info thanks dr pepper. I understand how to wire the diode now. (It's obvious too, like Nigel Goodwin says. I'll try to find a diagram too, Nigel)

So If I put a diode in line with the positive lead to the circuit, is there a small voltage drop there?

Also, can I use any diode as long as it is rated for more than the 9 volts and (about) 200mA? (I'm going to assume that it's not a Zener diode that I need but a Rectifier Diode?)

I'm making an order of misc. parts that I need for my projects, and I would like to add a handfull of diodes for my various projects specifically for this safety feature. Apart from the Zener Diodes, this company offers:

1N4007 1A 1000V General Purpose Rectifier Diode - $0.09 each
1N5408 General Purpose Rectifier Diode 3A 1000V - 10-Pack - $1.90
Power Supply Rectifier Diode 6A 1000V - Used as a rectifier in higher current power supplies, also known as 6A10 - $0.48 each
1N4148 Logic Diode (DO-41 case style) - $0.04 each
1N914 Small Signal Fast Switching Diode - $0.06 each
1N4935 Fast Recovery Rectifier Diode 200V 1A DO-41 - $0.19 each
1N4937 Fast recovery Rectifier Diode 600V 1A DO-41 - $0.48 each

Will any of these be suitable? Thanks!
 
If you want to avoid the 0.7V or so drop of the diode you can use a P-MOSFET as an ideal diode to block any reverse voltage, as discussed here.
Its forward drop is just the circuit current times the MOSFET on-resistance, which can be less than a tenth of an ohm for many types.

For 9V you just need a P-MOSFET with no resistors, as shown below:
The P-MOSFET can be just about any device with voltage rating of 20V or more and an on-resistance of less than an ohm.

upload_2017-9-8_1-9-57.png
 
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Alright you guys are great!

Thanks for the tip Nigel!

crutschow, I'm not sure yet if I need to be worried about a 0.7 volt drop with this circuit. I just found out the Clickless relay boards can run from 6 to 30 volts DC (I`m assuming they mean the whole board, in which case the Clickless circuit can run as low as 6 volts, I guess.) And I`m not really up to speed with wiring up a P-Mosfet, you say it's just like a diode? Cool. But I am going to look into where to get them, just in case. Is there a specific type I should look for?

Thanks again you guys!
 
And I`m not really up to speed with wiring up a P-Mosfet, you say it's just like a diode?
It acts like an ideal diode but it has three connections instead of two.
The P-MOSFET Drain goes to the positive input voltage supply, the Gate goes to ground, and the Source goes to the positive circuit power.
Is there a specific type I should look for?
Just one that meets the voltage and on-resistance criteria I mentioned, is easy to solder to, and is cheap.

Do you have a space issue?
Here's an inexpensive example of one that should work.
It has a 0.6Ω on-resistance so will drop 60mV @ 100mA current.
The pinout is shown in its data sheet.
 
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As you say, the 0.7V loss shouldn't make any difference. If it is a problem at all then you can switch to a schottky diode instead - the 1N5817 has 0.4V drop at 1A. You haven't filled in your location but in the states you can get a suitable one at Sparkfun for $0.15.

Mike.
 
Thanks Pommie. I'll look at Sparkfun's website.

crutschow, that's an interesting part! Does the "drain" pin also connect to the "drain" flange?

Can I ground it just by soldering the drain pin to ground and sticking it in the enclosure with velcro or double sided sticky tape? Or should I screw it to the back of the enclosure?
 
As you don't know where to stick a diode, you don't want to be confusing yourself trying to use an FET instead :D

EVERYTHING commercial uses a simple diode, and the 0.7V drop is of no consequence at all.

As we've said before, post a circuit and we'll show you exactly where and how to connect the diode.
 
Does the "drain" pin also connect to the "drain" flange?
Yes, they are connected together.
That pin is the substrate connection to the MOSFET chip, which is soldered to the flange.
Can I ground it just by soldering the drain pin to ground and sticking it in the enclosure with velcro or double sided sticky tape? Or should I screw it to the back of the enclosure?
As I stated in post #13, "The P-MOSFET Drain goes to the positive input voltage supply, the Gate goes to ground, and the Source goes to the positive circuit power."
Why do you think the drain goes to ground?

Since the MOSFET will be dissipating very little power when operating it doesn't need to be mounted to the enclosure.
Any convenient way to mount it is fine (even just using the leads).

But as noted by others, if the 0.7V drop of a diode is of no consequence in your circuit than just use a diode.
 
At the risk of being burned I'll explain how its done commercially with 0 volt drop (and this has already been mentioned).

Put a diode across the supply terminals of the looper in reverse, so the silver strip goes to the +, and the other side to -, then you put a fuse, or fusible resistor inline with the supply (before the diode).
Then if you power the device in reverse the diode will turn on and blow the fuse.
Obviously a fuse or some means of disconnecting the supply is important here.
 
ok that makes sense with a fuse, dr pepper.

crutschow, I see that I got the P-MOSFET connections wrong, but if the Drain is connected to the positive connection, and the Drain is also the Flange and I screw it to the enclosure, and the flange is then grounded, it will ground the Drain! That's why I thought the Drain goes to ground! (unless I'm getting this wrong!)

Nigel, here is a diagram of the Lace Looper from Mammoth Electronics, available in 3, 4, or 5 loops and lots of powdercoat options. Can be ordered with mechanical switches or Clickless Bypass Circuits. Comes with the Clickless circuits for that option. (Mine was around $112.00 for 5 loops and powdercoat with the clickless option. The Clickless circuits are preassembled).

Note: The Send and Return are labeled backwards according to the connections on the Clickless circuit boards.

Untitled.jpg
 
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I see that I got the P-MOSFET connections wrong, but if the Drain is connected to the positive connection, and the Drain is also the Flange and I screw it to the enclosure, and the flange is then grounded, it will ground the Drain! That's why I thought the Drain goes to ground! (unless I'm getting this wrong!)
The MOSFET flange cannot be connected to ground.
That's why you must not screw the flange to the enclosure without an electrical insulator between the flange and the enclosure along with an insulator for the screw (or use a nylon nut and bolt).
 
Obviously a fuse or some means of disconnecting the supply is important here.
That's why I like a MOSFET for reverse protection with a near zero forward voltage drop (you can get MOSFETs with only a few milliohms of forward resistance), it doesn't require a fuse.
 
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