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Author Topic: Basic Ohms law problem  (Read 1868 times)
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Need some help guys & gals. And try to hold back the laughter.

 What I have is a basic Ohms law problem.

Ok, I made a small project as a Joke for a co-worker. It is basically a Larson scanner  incased in acrylic.

The issue I am having is figuring out the supply voltage. You see as I was making it I was using my Switching Power supply set at 5v. I used  5 Volts because of the atTiny85 and I was planning on using a 5v wall wart that I have.

 Everything was working great on until I went to use the 5v 550mA wall wart.  Witch in reality puts out 6.24vdc and makes my scanner blink crazy.

 Now I know I could of used and 5v regulator. But it’s to late. So I was wondering if I could use a resistor to lower the voltage. Is so can someone show  me how to figure it out.

Thanks   
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What is the nominal current at 5v? 5v devices do not generally tolerate >5.5v too well.
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The resistor would have to precisely match your current draw to get the right voltage drop. Given your predicament the easiest route might be to just burn the extra power with a 5v zener diode to lower the voltage.
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The resistor would have to precisely match your current draw to get the right voltage drop. Given your predicament the easiest route might be to just burn the extra power with a 5v zener diode to lower the voltage.

That sounds cool, but over my head. Can you explain.

Thanks
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What is the nominal current at 5v? 5v devices do not generally tolerate >5.5v too well.

How do I figure that out?

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Or throw a 7805 in there....

(We learn everyday... I had to Google "Larson scanner", now I know what it is....)
« Last Edit: February 01, 2013, 11:44:07 pm by JimboZA » Logged

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Or throw a 7805 in there....

(We learn everyday... I had to Google "Larson scanner", now I know what it is....)

A 7805 needs 2 to 3 volts differential in order to start regulating (i.e. 7 to 8 volts in minimum). It won't work for this application.

There's no reason why a 5 volt board won't work at 5.5.... but if it's really necessary to throw away that 1/2 volt, why not just use a series 1N4001 diode and drop 0.7v off the end of the wall wart?

The real reason that the OP's board is going nuts is not the extra 1/2 volt, but probably something else like maybe lousy filtering in the wall wart (AC ripple?).
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Or throw a 7805 in there....

(We learn everyday... I had to Google "Larson scanner", now I know what it is....)

A 7805 needs 2 to 3 volts differential in order to start regulating (i.e. 7 to 8 volts in minimum). It won't work for this application.

There's no reason why a 5 volt board won't work at 5.5.... but if it's really necessary to throw away that 1/2 volt, why not just use a series 1N4001 diode and drop 0.7v off the end of the wall wart?

The real reason that the OP's board is going nuts is not the extra 1/2 volt, but probably something else like maybe lousy filtering in the wall wart (AC ripple?).
He said the wall wart is putting out over 6v not 5.5.
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There are other low dropout regulators that will work with a 6.24V input. 7805 is not the only choice.

http://www.digikey.com/product-detail/en/MC33269DT-5.0G/MC33269DT-5.0GOS-ND/1479179
Vout = 5V with Vin  >=6.1V

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He said the wall wart is putting out over 6v not 5.5.

Right you are. I gotta get to sleep... I'm seeing things. Where on earth did I get 5.5 from???
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He said the wall wart is putting out over 6v not 5.5.

Right you are. I gotta get to sleep... I'm seeing things. Where on earth did I get 5.5 from???
Probably from my post. I mentioned 5.5v as the maximum that TTL circuits will behave normally.
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Before going too far, most wall warts have a significant internal thevenin resistance and put out a higher voltage with no load (read current).

Be sure the wall wart's voltage is measured with a reasonable load on it.

Just sayin'...
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The lesson here is that a wall-wart is unlikely to be suitable for powering digital electronics unless
used with a proper voltage regulator, or unless it is already "fully voltage regulated" internally - the
good news is that mains _USB_ wall warts are _required_ to be fully regulated to 5V - this is what you
should have gone for I think.

A resistor won't solve this problem since the load is not constant - most useful devices with a microcontroller
are varying loads - a voltage regulator adjusts itself to deal with a varying load, a resistor cannot.

If it were a simple fixed load a resistor could be used.
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Now I know I could of used and 5v regulator. But it’s to late. So I was wondering if I could use a resistor to lower the voltage. Is so can someone show  me how to figure it out.

A diode would be better. Silicon diodes drop the voltage by about 0.7V when it passes through them. Two of them in series would drop it by 1.4V, etc.

Pretty much any diode will do, it doesn't have to be a Zener or anything special. If you want a part number, try a 1N4001.

Transistors are also diodes, a fully open BJT like a 2N2222 will also drop the voltage by about 0.7V.
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The lesson here is that a wall-wart is unlikely to be suitable for powering digital electronics unless
used with a proper voltage regulator, or unless it is already "fully voltage regulated" internally - the
good news is that mains _USB_ wall warts are _required_ to be fully regulated to 5V - this is what you
should have gone for I think.



And a very important lesson it was. And i am glad I learned it now and not after making a real project. smiley
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