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Support Forum » DC adapter with LCD Array Kit only outputs 1V

June 02, 2014
by lailai
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I have an issue with the DC adapter that comes with the lcd array - it only outputs 1V (tested with a multimeter). It's supposed to output 9V, 300mA.

I am using it in Australia, so I have to put it through a plug converter, maybe that's why?

The other adapter I have outputs 9V (and it actually outputs 9V), but 1.5A which causes issues with the microcontroller.

Am I screwed?

June 02, 2014
by sask55
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Iailal

The amperage value listed on a power supply is an indication of the maximum current that can be drawn from that supply while the supply is still capable of maintaining the voltage level listed. The 9V 300mA supply should in produce ~ 9V while it is producing up to 300 mA of current. The actual current is determined by the load on the supply. Ohms law (I=R/V) can be applied to determine the relationship of current and voltage for resistive loads.

In theory a 1.5A 9V DC supply should be capable of producing 5 times as much current as the 300mA supply without overheating or the voltage dropping off. The micro itself requires very little current; it presents itself as a high resistance load and therefore either supply should work if they are actually producing ~9V DC. In practice I have found there is a lot of variation in voltage produced by some so called DC wall wart type supplies. Perhaps your 1.5A supply is producing a poor quality very uneven voltage. It may be possible to use it with the addition of a capacitor. It seams that the smaller supply or possibly the converter is not working.

There are a number of threads on this forum that go into the use of power supplies in some detail. It has been a common issue. One link of many about this toppic.

June 03, 2014
by lailai
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Thanks sask55, I believe the issue is with the poor quality. However, I have encountered a more serious issue - powering the unit with the USB, the voltage regulator overheats enough to smell bad, have bubbles appear on the metal, and melt the breadboard (unfortunately it took me a couple of minutes to realize).

I've measured that only 5 volts is coming in.

The leftmost pin of the 7805C is connected to +5V of the USB cable, the middle pin is connected to ground, and the right pin connected to the + rail on the breadboard. The ground of the USB is connected to the - of the rails.

How can this setup be frying my breadboard?!

Thank you

June 03, 2014
by JKITSON
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Lailai

I believe if you are using the USB as a power source do not use the 7805 as they require +7.5 volts or greater to regulate properly. With out seeing a photo of your breadboard it will be hard to help more. I personally don't like to use the USB as a power source for the MCU. To much chance of damage to my PC. I use a small 12 volt battery with all my testing.

Jim

June 03, 2014
by lailai
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Hi Jim,

Ah, however if I am only passing 5V that shouldn't end up with the voltage regulator frying itself and the board? Or is that expected?

This is for my led array, of which a DC battery wouldn't be enough.

https://i.imgur.com/zakQDak.jpg

(Sorry for not using blue wires to make it more clear, can't ind them)

June 04, 2014
by sask55
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Lailai

I also was wandering if the regulator heating up is result of a low input voltage. It is not what I would expect to happen. Heating up of the regulator seams likes such an unlikely result of low input voltage that I decided to test it myself Just to satisfy my curiosity.

I set up a test on a board with an adjustable supply feeding a 7805 regulator. I placed a larger size (to handle the expected output current)) 12 ohm resistor across the output. If the output voltage was maintained at 5 Volts then using I=V/R the output current would be 5/ 12 = .41 amps. This is a considerably larger load then can be expected from the micro, but it happened to be a 2 watt resistor I have at hand. That is about the value of the power that will be dissipated at the load resistor (0.4A X 5 volts = 2.0 watts). From the Atmega data sheet the absolute max rating for current draw of the micro is 200 mA. So the regulator was providing more then twice as much current as the micro could safely handle if there was a situation where there was a lot of load on the micro’s output pins. .

My findings

With the input voltage at 10 Volts the regulator was outputting 5 volts. The regulator and especially the load resistor did heat somewhat. This is to be expected.

With the input voltage at 5 Volts the regulator was outputting about 3.4 volts. It was clear by touching the regulator that it was not getting as warmer as it was when regulating an input voltage of 10 Volts. .

This may be much more information then you are interested to hearing about.

I think it is reasonable to say there is something else going on in your setup.

Perhaps if you could fix the link to your picture and someone could spot the problem.

Darryl

June 04, 2014
by lailai
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Hi Darryl,

Thank you for doing the testing, okay so something else seems to be going on.. bit surprising as I don't see anything unexpected with my multimeter.

I re-uploaded the image on minus, does this display?

test

June 04, 2014
by JKITSON
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Lailai

It appears to me you possibily have the + (red wire) on the output pin of the 7805.

Jim

June 04, 2014
by Ralphxyz
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lailai, does everything work correctly with the red USB wire directly to your + rail.

If so I'd move on!! Of course there is "potential" problems using USB power as Jim said.

Someday get a 9 volt battery and see if the regulator still works.

June 04, 2014
by sask55
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I don’t see anything obviously out of place. I can’t see clearly , make sure the metal case on the crystal is not in contact with the leads on the small power capacitor or other wires.

I would be a little nervous about connecting a board that appears to be drawing so much current directly to my USB power wire. As Jim as mentioned this could possibly result in damaging your USB adapter or the USB supply on the computer.

As Ralph has said, if you have a 9 volt battery I would isolate the regulator and check if it is still working. Are you getting 5 Volts at the output pin? If so then I would consider using that battery as a power source for the testing. It may not last long but should work to confirm a reasonable load is connected to the regulator. If the setup works using the battery then I would consider removing the regulator and connecting the USB power to the +5 rail. If not I would try to isolate the problem by systematically disconnecting and reconnecting wires on the board.

June 04, 2014
by JKITSON
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Lailai

I checked the datasheets & looked at a working Nerdkits board I have. If the lettering of the 7805 chip is facing up as I think your picture shows then you have the input and output reversed. Just my view of your board..

Jim

June 04, 2014
by sask55
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It is somewhat strange how such a clear image of a relatively simple connection could be so uncertain. After all there is only two possible ways that the regulator could be placed in that spot. I have checked and rechecked the data sheet and working boards ,I can’t see what Jim is seeing in regards to the orientation of the 7805 regulator it appears to be correct to my eye.

In any event I don’t think it is THE issue here but I would consider moving the regulator on the board. It appears that the connection did get very warm. The heat may have affected the spring characteristics of the conductors in that set of holes on your board. I think moving even one column to the left would insure a better connection. A high resistance connection would be a source of heat.

Let us know how things are going.

June 07, 2014
by lailai
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I've tried moving it to another place, but it still does not work. With a 9V battery, the voltage regulator still heats up. I think it might be a defective voltage regulator.

The input and outputs are not reversed..

I'm powering the unit with a USB directly now.

June 08, 2014
by mongo
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If the 7805 is wired correctly and is getting hot, it is likely a bad regulator.

If the output is shorted, or severely overloaded, it will also cause the 7805 to get hot but since you mentioned the circuit to be working off the USB power, it pretty much means it's a bad 7805.

July 03, 2014
by BobaMosfet
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lailai-

To answer your first, original question. Putting a meter across a regulator isn't adequate. It's a question of impedance. That regulator will only deliver the expected potential at a required current level, based on the expected load (a range in this case). Without the expected load, E and I will behave weirdly in an attempt to satisfy the physics involved.

BM

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