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Any Electrical Boffins - Questions about Bucks, Fuse etc

SniffTheGlove

By SniffTheGlove
in DIY Astronomer

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SniffTheGlove

SniffTheGlove

Posted
Posted

Hello,

I am reworking my custom power pack as I had a few issues with it and would like to ask some questions and hopefully an Electrical Boffin can reply.

1) I am using a LM2596 Buck StepDown (off ebay) to step down from 12v to 7.5 to power my Canon 1100D. I want to add a fuse to the powered item (Canon 1100D). Where would the fuse go to provide the best protection?

example 1:       12V Battery ------ 2A fuse ------ LM2596 ------ 5V Canon 1100D

or

example 2:       12V Battery ------ LM2596 ------ 2A fuse ------ 5V Canon 1100D

or

example 3:       12V Battery------ 2A fuse ------LM2596 ------ 2A fuse ------ 5V Canon 1100D

2) I want to add an Analogue Ammeter to my setup. Previously I been told to add the ammeter between the Battery Negative Terminal and the Negative power return, however I have see a few youtube video where they are saying it goes between the Battery Positive Terminal and the Positive power lead into the circuit. Which is the correct way. I will be adding an Ammeter onto the mount to monitor the amps (as I have been recommended) and I also want to put on to show the entire amps my custom power box is using in total.

3) If I have the following simple circuit how can I get an LED to come on when there is power to indicate that the circuit is live.

example 12V Battery -----SWITCH ----- LM2596 ----- LED  ----- Powered On Item

Previoulsy when I have tried to put an LED into a circuit there is not enough volts/amps to power the item, how do I get round this?

Thanks

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Tinker1947

Tinker1947

Posted
Posted

One of these circuits might help...

https://www.google.co.uk/search?q=analog+amp+meter+circuit&espv=2&biw=1920&bih=1099&tbm=isch&imgil=E0PtKd0HtpaU-M%253A%253B_HnJjWBaEzpHlM%253Bhttp%25253A%25252F%25252Fwww.allaboutcircuits.com%25252Fworksheets%25252Fmeters2.html&source=iu&pf=m&fir=E0PtKd0HtpaU-M%253A%252C_HnJjWBaEzpHlM%252C_&usg=__1f3drzJ-SOpsOqqUp8l7dE5u9Y0%3D&ved=0CEsQyjc&ei=BgeoVJ2JBJPuaOmXgqAL#facrc=_&imgdii=_&imgrc=E0PtKd0HtpaU-M%253A%3B_HnJjWBaEzpHlM%3Bhttp%253A%252F%252Fsub.allaboutcircuits.com%252Fimages%252Fquiz%252F01794x02.png%3Bhttp%253A%252F%252Fwww.allaboutcircuits.com%252Fworksheets%252Fmeters2.html%3B324%3B273

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SniffTheGlove

SniffTheGlove

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Thanks Tinker, but I am not an electrronic person so these circuit diagrams are quite above my understanding.

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NEM

NEM

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For the best protection use example 3.  The thing to remember about fuses is that they protect the power side of the circuit from faults that happen on the load side.  For example, in your example 2 circuit the fuse will protect the battery and the LM2596 if a fault develops in the Canon 1100D; but it will not protect the battery if a fault develops in the LM2596 itself.  If the battery is a car or leisure battery you should always have a fuse directly after it as a short circuit could cause a fire.

About the ammeter, wiring it into either positive or negative will work as the same current flows in each line.  However, I'd put it into the positive, because circuit voltages are measured from the battery -ve and there will be a small voltage drop across the meter, so the voltages in the circuit will measure slightly higher than they actually are.  In reality it doesn't matter, use what ever's easiest.

From your diagram it looks as though you're putting the LED in series with the load.  The LED should go in series with a current limiting resistor and that combination should go in parallel with the load.  The LED should light even if the load is not connected. The value of the limiting resistor depends on the suppply voltage, the LED and how bright you want it, start at 1kOhm and work downwards,  There are LED resistor calculators available on the web, the Google.

Noel

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pete_l

pete_l

Posted
Posted

Hello,

I am reworking my custom power pack as I had a few issues with it and would like to ask some questions and hopefully an Electrical Boffin can reply.

1) I am using a LM2596 Buck StepDown (off ebay) to step down from 12v to 7.5 to power my Canon 1100D. I want to add a fuse to the powered item (Canon 1100D). Where would the fuse go to provide the best protection?

It's my understanding that the LM2596 has built-in current limiting.

As far as adding an Ammeter, I'd place this on the 7.5V supply to the camera. That way you are getting direct information

about the current the camera is drawing. If you place it before the SMPS, the current drawn will vary with the supply (battery) voltage - even if the camera is drawing a constant current.

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SniffTheGlove

SniffTheGlove

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For the best protection use example 3.  The thing to remember about fuses is that they protect the power side of the circuit from faults that happen on the load side.  For example, in your example 2 circuit the fuse will protect the battery and the LM2596 if a fault develops in the Canon 1100D; but it will not protect the battery if a fault develops in the LM2596 itself.  If the battery is a car or leisure battery you should always have a fuse directly after it as a short circuit could cause a fire.

About the ammeter, wiring it into either positive or negative will work as the same current flows in each line.  However, I'd put it into the positive, because circuit voltages are measured from the battery -ve and there will be a small voltage drop across the meter, so the voltages in the circuit will measure slightly higher than they actually are.  In reality it doesn't matter, use what ever's easiest.

From your diagram it looks as though you're putting the LED in series with the load.  The LED should go in series with a current limiting resistor and that combination should go in parallel with the load.  The LED should light even if the load is not connected. The value of the limiting resistor depends on the suppply voltage, the LED and how bright you want it, start at 1kOhm and work downwards,  There are LED resistor calculators available on the web, the Google.

Noel

Thanks NEM,

With regards to the LED, do you means something like this....

                                     ---- LED ----Resistor ----- Negative Terminal  (Or should the Resistor come before the LED

12v ------ SWITCH   ---|

                                     ---- Load ---- Negative Terminal

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Davey-T

Davey-T

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Is there any reason you want to use an analogue ammeter ? I have a red LCD digital one on my power supply so it's visible in the dark (fitted in the negative lead)

Dave

Formula for LED resistor R=(Vs-Vf) / If

Vs=supply voltage

Vf=forward voltage drop across the LED

If=forward current through the LED

Example: If =20ma and Vf=2.5v to operate from 12volt supply R=(12-2.5) 0.02 =475ohm resistor nearest commonly available resistor 470ohm

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NEM

NEM

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Posted

With regards to the LED, do you means something like this....

                                     ---- LED ----Resistor ----- Negative Terminal  (Or should the Resistor come before the LED

12v ------ SWITCH   ---|

                                     ---- Load ---- Negative Terminal

Yes, that looks correct, the output of the switch goes to both the LED and the load. The order of the LED and resistor doesn't matter.

As Davey-T says, a digital meter is easier to read in the dark and I think (from what I've read on this forum) they do have to be in the negative lead.

Noel

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