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Is this circuit possible?

Viliuks

Dec 5, 2015
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Heyo, I'm making a portable boom box and I ran into something, I got a 3pin ON-OFF-ON tumbler switch and I was wondering if this connection is possible.
I want it to be able to switch from battery pack to the barrel connector as inputs.
I cannot leave the boost converter connected because it will drain the battery on idle. (Atleast I think it will, its the xl6009 boost converter)
t6mlSnM.png

If you know any better solution for this, let me know! Thanks!
 
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dorke

Jun 20, 2015
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I think it would be better to switch the (+) and not the (-) ,
with a DPDT rocker switch,and for "added fail-safe" insert schottky diodes to prevent 12V power "collision" .

Like so:
t6mlSnM.png
 

Viliuks

Dec 5, 2015
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I think it would be better to switch the (+) and not the (-) ,
with a DPDT rocker switch,and for "added fail-safe" insert schottky diodes to prevent 12V power "collision" .

Like so:
View attachment 26815
Damn, I bought the SDPT today, didnt think I would need DPDT, Im guessing without that I would not be able to make this. If there is a way to make this work with SDPT that would be great, because the electronics store is fairly far and I cannot really go there wherever I want :/ And Im new to electronics I dont really know much about it, learning the intermediate stuff slowly :p
 

Colin Mitchell

Aug 31, 2014
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There is nothing wrong with your circuit.
It amazes me that supposed "electronics persons" create such absurdities.
 

(*steve*)

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There is no problem switching negative, UNLESS there is any external connection which makes a connection.

If the 12V supply is also used to power other equipment, the ground connection (on say amplifier inputs) may be connected to it. In this case, a cable connected between the two devices could complete the circuit.

In the worst of possible worlds, if your device requires high current, the audio cable or other wiring could be damaged by the unexpectedly high current.
 

Kiwi

Jan 28, 2013
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Your circuit is nice and simple. It will work fine.
The only suggestion I would make is to put the switch in the positive line, not the negative line.

If you look at most circuit diagrams you will see that the negative is usually drawn as the common.

Having a constant common negative makes using a voltmeter for testing or fault finding easier.
Connect the negative voltmeter probe to a good negative point, and then use the positive probe for testing.
 

Viliuks

Dec 5, 2015
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There is no problem switching negative, UNLESS there is any external connection which makes a connection.

If the 12V supply is also used to power other equipment, the ground connection (on say amplifier inputs) may be connected to it. In this case, a cable connected between the two devices could complete the circuit.

In the worst of possible worlds, if your device requires high current, the audio cable or other wiring could be damaged by the unexpectedly high current.
No other equipement is powered only the amp :p

Your circuit is nice and simple. It will work fine.
The only suggestion I would make is to put the switch in the positive line, not the negative line.

If you look at most circuit diagrams you will see that the negative is usually drawn as the common.

Having a constant common negative makes using a voltmeter for testing or fault finding easier.
Connect the negative voltmeter probe to a good negative point, and then use the positive probe for testing.
I will change that :)

Thanks everyone, will post pictures of the finished project :)

Oh but Im still worried that having the vin and vout of the boost converter connected together being a problem or it isnt?
 

(*steve*)

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Oh but Im still worried that having the vin and vout of the boost converter connected together being a problem or it isnt?

It all depends on the topology of the boost converter. If the output is isolated from the input then you can tie either output to either input (but just one of them!)

If it is not isolated, then it is likely that the -ve lead is common. Use a multimeter to read the resistance between the -ve input and -ve output. Note that a common boost topology will read the same from the +ve input to the +ve output, but it will be *slightly* higher due to the presence of an inductor.

A photo of the boost converter (or a link to a description on the web) may allow us to give you a quick answer.

It also may be possible to power the boost converter from 12V. This would make it simple to switch the +ve. Whilst it's not an ideal solution it does have some benefits, allowing you (for example) to run the unit from any power source from about 4V to 12V
 

Viliuks

Dec 5, 2015
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It all depends on the topology of the boost converter. If the output is isolated from the input then you can tie either output to either input (but just one of them!)

If it is not isolated, then it is likely that the -ve lead is common. Use a multimeter to read the resistance between the -ve input and -ve output. Note that a common boost topology will read the same from the +ve input to the +ve output, but it will be *slightly* higher due to the presence of an inductor.

A photo of the boost converter (or a link to a description on the web) may allow us to give you a quick answer.

It also may be possible to power the boost converter from 12V. This would make it simple to switch the +ve. Whilst it's not an ideal solution it does have some benefits, allowing you (for example) to run the unit from any power source from about 4V to 12V
http://m.banggood.com/XL6009-Step-U...upply-Module-Converter-Regulator-p-916222.htm
Thats what I bought.
Measured the resistance it was 0.2ohms which would mean that it has 0ohms in resistance because its my multimeters resistance when I just touch the probes together.
 

(*steve*)

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OK, the easiest thing for you to do is to connect the -ve lead of your battery and -ve from the input jack to the -ve input of this regulator, then use a switch to select between the positive of the battery or the positive of your input jack to be connected to the positive input of this regulator. Then connect the amplifier to the output of the regulator.
 

dorke

Jun 20, 2015
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No other equipement is powered only the amp :p


I will change that :)

Thanks everyone, will post pictures of the finished project :)

Oh but Im still worried that having the vin and vout of the boost converter connected together being a problem or it isnt?

Let me repeat a free advice(you probably know it,but anways )
"better safe than sorry".
And "Don't be penny wise and dollar stupid"...I think it is from Hop;)

The difference in price between a SPDT and a DPDT is very small,
2 schottky diodes will cost you 1$ or less.

Use the circuit in post #2 !
It takes care of all your requirements and concerns.
 

Viliuks

Dec 5, 2015
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OK, the easiest thing for you to do is to connect the -ve lead of your battery and -ve from the input jack to the -ve input of this regulator, then use a switch to select between the positive of the battery or the positive of your input jack to be connected to the positive input of this regulator. Then connect the amplifier to the output of the regulator.
Well it cannot step down the voltage, if someone plugs in say 20v into the barrel connector and the boost converter is set to 12v it would blow up, wouldnt it?
I might just add another switch to turn on and off the boost converter and the other switch to switch the power supplies
 

(*steve*)

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Well it cannot step down the voltage, if someone plugs in say 20v into the barrel connector and the boost converter is set to 12v it would blow up, wouldnt it?

No, it would just let the 20V straight through.

The 20V may not be very good for the amplifier, but that's another question.
 

Viliuks

Dec 5, 2015
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No, it would just let the 20V straight through.

The 20V may not be very good for the amplifier, but that's another question.
Aplifier can handle from 8v to 25v so thats no issue.
Damn, I tested it and it lets the 20v go through even if the boost converter is set to 12v.
This is neat, it will make my whole life easier :)

But I'm still wondering how does it let it go through without damaging other components?
 
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AnalogKid

Jun 10, 2015
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Use the circuit in post #2 It takes care of all your requirements and concerns.
So does the circuit in post #1. Unless he has the only low-cost toggle switch on the planet that is make-before-break, the inherent air gap between the two circuits is more than adequate for 12 V isolation, is less complicated to wire, doesn't dissipate any heat as diodes would, etc. *Technically*, this means I'm agreeing with Colin. How did the world come to this?

I agree with everyone else about moving the switch to the + wiring and hard-wiring all commons/grounds together.

ak
 

Viliuks

Dec 5, 2015
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So does the circuit in post #1. Unless he has the only low-cost toggle switch on the planet that is make-before-break, the inherent air gap between the two circuits is more than adequate for 12 V isolation, is less complicated to wire, doesn't dissipate any heat as diodes would, etc. *Technically*, this means I'm agreeing with Colin. How did the world come to this?

I agree with everyone else about moving the switch to the + wiring and hard-wiring all commons/grounds together.

ak
I'm going to move the switch and stuff, its just in the planning stage, I got all the components, but wanted to just see if "technically" everything will work. Because I dont want to damage any of my components :p
 

(*steve*)

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This would be my recommendation

poweralt.png
 

(*steve*)

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But I'm still wondering how does it let it go through without damaging other components?

The maximum input voltage is likely to be around 35V Check the voltage rating for the input capacitor. If it's lower than 35V, then that will be setting the limit (the chip has a 35V max from memory)
 
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(*steve*)

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If your question was "how does it work" rather than "what's the max voltage", then the answer is that a boost regulator like this has the input connected to the output via an inductor and a diode. When the input voltage does not need boosting, the current just passes through the inductor and the diode (leading to a small voltage drop). If the voltage requires boosting a transistor connected at the junction of the inductor and the diode is used to cause the inductor to periodically store energy and release it through the diode. It does this using a very counter-intuitive method :)

When used this way the regulator works like you do when riding down hill on a bike without brakes. You don't need to pedal to maintain speed, but there's also nothing stopping you from going faster than you want to.

So, when the input to a boost regulator is higher than the setting for the boosted output, it just "coasts", but it can't reduce the input voltage either.
 

Viliuks

Dec 5, 2015
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If your question was "how does it work" rather than "what's the max voltage", then the answer is that a boost regulator like this has the input connected to the output via an inductor and a diode. When the input voltage does not need boosting, the current just passes through the inductor and the diode (leading to a small voltage drop). If the voltage requires boosting a transistor connected at the junction of the inductor and the diode is used to cause the inductor to periodically store energy and release it through the diode. It does this using a very counter-intuitive method :)

When used this way the regulator works like you do when riding down hill on a bike without brakes. You don't need to pedal to maintain speed, but there's also nothing stopping you from going faster than you want to.

So, when the input to a boost regulator is higher than the setting for the boosted output, it just "coasts", but it can't reduce the input voltage either.
Thank you very much for all the help, I really appreciate it! :)

Now to make the dream come true and put it together. :p
 
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