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WallWart conversion...

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roughshawd

Jul 13, 2020
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I have a VHS camera that requires a 9.6v 1.2amp (1200ma?) Negative ring psu wall wart. I don't find a 9.6 with 1.2 among my collection, only a 12v 2.0. can I covert this to 9.6 with an inline resistance?
 

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AnalogKid

Jun 10, 2015
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Yes, but not well.

The rating on the wallwart is its max continuous output current. Your device will draw less than this, and depending on the wallwart's internal circuit, the output voltage may vary inversely with it. The main issue is that the device current will not be constant. For example, it will draw more current when the motors are running.

With a simple resistor in series, the voltage seen by the device will wander all over the place as the drawn current varies. a better way is with a constant-voltage drop between the ww output and the device. The most simple way is a series string of power rectifier diodes, something rated for 50 V and 3 A. At 1 A, a 1N5400 will drop somewhere around 0.8 V each so three in series is a good place to start. You can test the resulting voltage with a resistive load before connecting it to the camera. You don't have to draw the full 1.2 A to see if the the approach is working.

The 9.6 V rating probably is not super-critical. A few tenths of a volt over or under shouldn't matter.

ak
 

Delta Prime

Jul 29, 2020
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(got me by 2 minutes) If that's all you have then yes. I would advise against it.If anything I would use rectifying diodes in series for cheap and quick solution each diode would drop approximately
.7Volts
1.2amp (1200ma?)
That is correct.
That's okay that your power supplies two amps no harm no foul . As long as the current is more than what your device requires as far as current.
 
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Delta Prime

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If all you have are resistors then here you go! The value of the resistance I've given you for R1 and R2 are arbitrary but in general the lower the resistance the better. Vs=Vin=12volts , R2 16 Ohm's , R1 4 Ohm's
Vout is 9.6 volts
Vout=Vs×R2/(R1+R2)

Like this!
photo_1686178140009.png
How many watts do your resistors need to be...well you know the current and you know the voltage v times I equals power in Watt the largest resistor will dissipate the most power.
 
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roughshawd

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You guys are so much fun, you're never gonna be rid of me!!! That second one.... That's what I was expecting, and I have heard of "the better way", using only diodes... But as far as I know the only way to change voltage is with a voltage divider circuit. I

I did a test of this theory when I was young, and the diode got so hot I burned my fingers. Please try the other method AnalogKid it is all about the heat...
 

Delta Prime

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Hello!!! I just hooked you up and I also saved your house from burning down remember me!

So are you good with the calculations for your power dissipation how many watts your resistors need to be?

"You guys are so much fun, you're never gonna be rid of me!"
:p

Zz
 
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Harald Kapp

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A voltage divider is next to useless in this application.
Your load (camera) will be parallel to R2 as it is connected to Vout. Therefore the current into the camera will develop a voltage drop across R1 which is not only unspecified but worse it will change with changes in the current drawn by the camera.
You have (at least) 3 options:
  1. use a 9.6 V wall wart with a rated current 1.2 A or higher. It doesn't have to be exactly 1.2 A. A 2 A supply, for example, will work equally well. Important is the correct voltage 9.6 V. The current is controlled by the camera.
  2. use an off the shelf step down module to convert the 12 V from the wall wart you have down to 9.6 V. Here's an example of such an (adjustable) module.
  3. use an adjustable linear regulator e.g. an LM317. While easy to use a linear regulator dissipates much more power than a step down regulator. In this case the dissipated power will be (12 V - 9.6 V) × 1.2 A = 2.9 W. Not the best solution.
 

crutschow

May 7, 2021
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I did a test of this theory when I was young, and the diode got so hot I burned my fingers.
No matter how you drop the voltage (unless you use a buck regulator) there will be the same amount of power (which generates heat) that has to be dissipated, whether it's in a linear regulator, resistors or diodes.
In this case three series diodes is the easiest solution.
Use diodes rated for at least 2-3A.
The will dissipate about 0.8W each @ 1A, so you will just have to deal with that heat.
 
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AnalogKid

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But as far as I know the only way to change voltage is with a voltage divider circuit.

There are many ways to decrease the value of a voltage source, and they all fall into two categories: linear and switching. Both dissipate heat, one more than the other.

With *any* linear technique, the power dissipated in the voltage reducing circuit will be, worst case, 2.88 watts. If this is dissipated in a single 10 W resistor, the part will get warm or hot, but your fingerprints will be safe. And, as above, the voltage stability with a varying load will be - none. With the linear methods, it all comes down to package size. 2.88 W in a 3 W resistor will be very hot; in a longer, fatter 25 W resistor it will be warm enough to know its working. The diode method will dissipate the same amount of heat but give much better load regulation. If they get too hot for your comfort, use fatter diodes. Really, that's the answer. A 6A05 or 6A10 diode in an R-6 package will run considerably cooler than a 1N5400.

A linear method not yet mentioned is a small circuit with one transistor and two resistors.

Switching regulators are a much more complex solution to grow from scratch, but modules are available on ebay for $1. Efficiency is around 80%, so the heat that has to be dissipated in the regulator components is less than 1 W.


ak
 

roughshawd

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I strung a bunch of resistors together in a linear fashion on the ground line. I achieved various results, more smaller value resistors 9.7v, less higher value reisistors 9.5v I think I will keep changing until I have the 9.6v!!! Lots of resistor options...
Because I am not dealing with an ideal circuit, but only a source, ~ maybe I am barking up the wrong tree. A 12v source is 2.5v to large for the device, thats not alot of over volt, and it delivers plenty of power, that whopping 2A is .8A much, but according to hoyle and everybody else, the device won't use any but what it needs. Maybe I should just use the 12v source, and accept the consequenses.....
Back to Ohms law again! I will run some numbers and get right back!!!
 

Delta Prime

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I strung a bunch of resistors together in a linear fashion on the ground line
The people who responded to your request are what they call experts(I know for a fact they are experts as well) they have your best interests at heart. They could not" resist" themselves. Someone had to say it!
Try googling multiple resistors voltage divider calculators. I have provided you an equation using Ohm's law just plug in the values of the resistors you have on hand . Easy peasy lemon squeezy.
 

Delta Prime

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I specified 4 Ohm's R1(as stated before the values are arbitrary use only as an example; picked out of blue sky!)
If calculations are done to provide a current of 2 amps the unit only requires 1.2 amps per specifications but in practice VHS camera motors peak voltage, current requirements and things do decline with age will be the final deciding factor . Changes will occur but the gentleman specifically requested resistors so I have given him an example at his request.
 
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AnalogKid

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Maybe I should just use the 12v source, and accept the consequenses.....
I wouldn't.

Please post a clear image of the power connector area on the camcorder. Also, somewhere on the camera body, maybe on the bottom, there should be a sticker, plate, or molded-in text defining the input power requirements and model number. With a clear image of that we might be able to track down a schematic. This will tell us exactly what the power input section can handle.

ak
 
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roughshawd

Jul 13, 2020
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yep the numbers work!!! OK so when you have a 9.6v power requirement, and you have a 12v source you need to reduce the volts by 2.4v... So put a 2.5v led in the line, and you have used 2.5v, but the source is a dedicated 12v dc, and the requirement is not 9.5vdc it is 9.6vdc so the optional 2.5v light bulb ain't getting it! Because the supply is DC, it is going to require a linear reduction to effect the proper change, as a parallel reduction would change the current equally. Where the 12v circuit requires a reduction by a ratio of 1.25 then the resistance should probably be >9.35, but I am stomping a +3m chain just to slow the power to the 9.5v state, I didn't stop at 9.7 because of the size and type of the source power. I would think that as I tested the wall wart the power climbed from 6vdc to 12vdc before leveling off at 12.1vdc. The current is plenty I would think, for my application, but where the power requirement is less that the source, the source should require a shunt of the power that is not needed, and it probably should be isolated from the original PSU. using a bridge network and or schottkey diodes to keep the current from avalanching. Reducing the 12v to 9.6v is superfluous with voltage division. In this application 1.25v must be shunted with no more than a .8a loss. For the PSU to remain stable, an ideal circuit might be applied which maintains the 12v, at 2A, then reduces the power, and drives the device.
 

roughshawd

Jul 13, 2020
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Ha ha!!! Or should I say HO Ho...
I got the 9.61vdc, slightly jittery from 9.57vdc to 9.61, but the voltage is there. Now which line is the divider to go in, the positive(like all the tutorials say) or in the ground(where I tested it). Or.... Should I redux the tesxtx with the resistor chain in the power line, rather than the ground just to be sure???
In basic electronics I heard something about this but missed what it was....
 

roughshawd

Jul 13, 2020
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Here's the chain16863503734682028048534637785443.jpgand here's the connector plug, and adapter...
 

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