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Topic: 12V Demux (Read 1 time) previous topic - next topic

Jul 16, 2012, 11:36 am Last Edit: Jul 16, 2012, 11:40 am by Enter_Kratos Reason: 1
Hey guys,

I'm fairly new to this sort of thing; only really played around with some servos, LEDs and Pots.
What I'm looking for is either a digital pot that can output voltages between 0 and 12volts which through my searching doesn't appear to be available or
a demux which can output up to 12volts so as to fit resistors of various strengths on the different outputs to the same effect.

Does anyone who has more of a grounding in this sort of thing have any suggestions?

EDIT:
Just found this:
http://uk.farnell.com/analog-devices/ad5293bruz-20/ic-dig-pot-1024pos-20k-14tssop/dp/1827274

Would this be able to be used for this purpose?

Cheers,

Kratos

Magician

Depends on how accurate voltage you need, better to use DAC (digital analog converter) than scale output with OPA. 8 or 12 bits available, interfacing circuitry with drawings usually printed in data-sheets.
Or you can even create one using resistors in R-2R ladder configuration.

Grumpy_Mike

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Would this be able to be used for this purpose?

Yes but you have to have a 12V power supply to drive it and also give it a logic supply of 5V.
However it depends on what you want to do with the output voltage once it is output. It is a 20K pot so you are not going to be able to actually power anything from this signal.

Thanks for the quick response guys.

Maybe if I explain my intended use it will make things a bit easier.
It's a bit of a pointless project but I have an electric train set which has a control box with a pot which allows you to control how fast the train goes. This control box has a max output of 12volts to the track. I wanted to hook something up between the control box and the track which will let me basically do what the analogue pot is doing but being able to program different timings to start off with with intent to build on top of this. I have an Arduino Duemilanove.

My idea was to use another chip which can pass through this max 12v and have various resistors on the different outputs to give me different steps. By controlling the chip with the Arduino I could change the speeds programmatically by telling it to output to whichever output has the strength resistor I wanted.

I hope I'm making sense. It may be a really silly thing but I'm just playing haha

Thanks

Grumpy_Mike

It makes sense but you will need an amplifier on the end of the pot in order to drive the required current into the train set. You can't do it from just the pot alone.

So what's happening to the 12 volts that's coming from the control box? I thought I could use the same current that's already being used.

Grumpy_Mike

The 12V out of your control box and applied to the track, is a voltage with a very low output impedance. That means it has the capacity to supply a large current. The 12V you get from a digital pot has a high output impedance, that means it can't supply very much in the way of current. Therefore you need a current amplifier to convert the high impedance voltage into a low impedance one. You might get away with a simple emitter follower circuit but you might have to go for a darlington pair. This has the disadvantage of dropping about 1.5V so you should not use a 12V supply but a 13.5V one if you can.

However, it is possible that the pot in your controller is a modern one with electronics after the pot. In this case you could replace the pot with the digital one.
You can tell this if you open it up. If the pot consists of loops of wire with a wiper passing over them it is the old fashioned type. If it goes into a bunch of electronics it is the new type.

Ok now I'm really confused. Why do I need to amplify the current if what's coming out of the control box is already powerful enough to run the train?

Grumpy_Mike

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Why do I need to amplify the current if what's coming out of the control box is already powerful enough to run the train?

You need to amplify the current from the digital pot in order to drive your train.

Do you understand the relationship between voltage current and source impedance?
Source impedance is like a series resistor in line with your voltage generator and limits the amount of current a particular voltage can deliver.

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You need to amplify the current from the digital pot in order to drive your train.

Is this because the digital pot is 'using' the original current which previously was powerful enough to run the train but now isn't?

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Do you understand the relationship between voltage current and source impedance?

Not really  :~ Everything I looked at regarding impedance seems to be about audio.

How have you come to the conclusion that I'll need to amplify?

Sorry if I'm being really dense. I feel a bit out of my depth. I think I'm learning though

Grumpy_Mike

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How have you come to the conclusion that I'll need to amplify?

Because that digital pot is 20K. So suppose it is half way down, between the 12V power and your train you have a 10K series resistance. Lets suppose you train had zero resistance, it doesn't but it is very small compared to 10K.
The maximum current your train could draw would be given by ohms law I = E/R where I is the current and E is the voltage and R is the resistance.
So plugging in the numbers
12 / 10000 = 0.0012 amps or 1.2mA
That is not enough current to drive a motor. It is more likely that your motor takes between 100mA and 600mA, so there is not enough current behind the voltage to power your system. This is formalised by saying that your voltage source has a 10K impedance. You see impedance mentioned in audio but it applies to anything sourcing a voltage.

Your train controller is probably not a pot but a variable resistance in series with the track. This is likely to be a very low resistance and will limit the current to the train's motor.

Ever thought why you can short out the contacts on a dry cell battery and nothing much happens, then you try it with a car battery and you get sparks and metal melting? That is all to do with the source impedance of the battery. For a car battery it is very low and so allows a large current to flow. For a dry cell it is very high and that limits the maximum amount of current you can draw from it even with a short on the output.

Thanks for giving such an in-depth explanation.
I really don't think I have what it takes for this kind of thing.
Although this seems to be a very simple thing for you to understand, I just don't think I'm cut out for it.
It's a real shame as I am interested in electronics but I think I'll just stick to coding.

Thanks for all your help, it's been a real eye opener.

Grumpy_Mike

Well give up if you must but I would recommend that you byte of a bit of a smaller lump to chew on. Get a feel for voltage, current and resistance. Play about with pots, LEDs and resistors. It will click and then you will wonder what was so difficult in the first place. It's just getting there in the first place takes a bit of doing but it will come if you don't try too hard.

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