Build Your Own Arcade Controls Forum
Main => Main Forum => Topic started by: lettuce on July 15, 2007, 07:58:55 am
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What type of resistor will i need if i want to get a 12v current down to a 5v? I want to connect my credit board up to my I-PAC, but the credit requires a 12v supply, so i need to add a resistor to the wire that is connected to the I-PAC and the pulse input on the Credit Board, to get the 12v down to a 5v so it doesnt fry the I-PAC
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I'd use a voltage regulator, not a resistor.
They are quite cheap at radioshack and other electronics stores.
:cheers:
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You need to invert the signal as well as isolate 12V from 5V. The ipac needs a ground to indicate a contact closure.
Use an NPN transistor and a 1000 ohm 1/4 watt resistor to make the following.
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I only have a very basic skill in wiring and soldering. Can you use simpleton terms by any chance?
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Doh, missed the pulse bit, tnx :cheers:
You need to invert the signal as well as isolate 12V from 5V. The ipac needs a ground to indicate a contact closure.
Use an NPN transistor and a 1000 ohm 1/4 watt resistor to make the following.
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Those terms were about as basic as they get. If you don't understand them you should probably not wire the circuit yourself. Get someone to assist who has a basic electronics knowledge.
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This is always a tricky subject because there are so many ways in which coin mechs and credit boards work. Most mechs and boards send a 12 volt signal and then drop to ground momentarily when a coin is inserted.
Also, most have an "open collector" type of signal which is not actually tied directly to 12 volts when there is no coin, but is allowed to float to 12 volts. This means its safe to connect to the I-PAC with a zener diode added for safety to prevent the voltage rising above 5. You will need a 5 volt (or 5.1 volt) zener diode.
There is a diagram on www.ultimarc.com/controldiags.html showing this.
The problem is, we don't know what your credit board sends exactly, but this is the most likely scenario.
Andy
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You will have to verify what type of signal your coin unit provides. If it is as Andy says then you will not need to invert the signal as the circuit I provided does. As stated previously you need to have some electronic knowledge to sucessfully interface your coin unit.
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So its the Coin Signal wire that goes to the Coin input on the I-PAC then? And the Coin Signal and Ground wires need to be connected together with a 5.1v zener diode, right?
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Thats correct, assuming its the type of signal I described. The diode needs to be the correct way around, see the diagram on www.ultimarc.com/controldiags.html
Andy
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Well i took my multi-meter to the credit board and checked the coin pulse terminal when i inserted a coin, it was sitting at about 1.5 volts, and the drops slightly when inserting a credit, so i guess im ok without a resistor/zener diode
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You need to invert the signal as well as isolate 12V from 5V. The ipac needs a ground to indicate a contact closure.
Use an NPN transistor and a 1000 ohm 1/4 watt resistor to make the following.
It doesn't sound as if you are going this direction, but for reference (perhaps someone else is remembering this circuit for some use), the circuit from BobA needs another resistor to limit the base current into that transistor. So the 12V in would go through a current limiting resistor (say 10K ohms, 1/4W) then to the base of the transistor. Connecting the 12V input directly to the transistor will allow the transistor to draw too much current, thus damaging it.
Rick
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Thanks for pointing that out rockin_rick. I had just used an opto isolator circuit then figured you didn't need the LED part of it and forgot to put the resistor into the base. For isolation the opto is probably the best because there is not direct connection even with component breakdown.
I think I would still advise Lettuce to use the zener on the input to the ipac as Andy recommended. 1.5V falling to ground when tripped does not like a proper output for a credit board.
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But if the voltage on the Coin pulse terminal doesnt exceed 5 volts do i really need to?, i would probably worry if it was near 5 volts but its not even half that. I had to set the multi-meter to 2000m on the DCV setting for it to even register a voltage, before that i had it set at 20 and didnt think the credit board was working as there was no reading.
The last credit board i had (Shefras Green Board) i used on an I-PAC for over 2 years without a problem, i think that was a really old credit board as is this one i think, so maybe its the older credit boards that drop down voltage on a coin insert?
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It sounds like you have an open-collector signal which has no pull-up. So this is fine to connect to the I-PAC directly, and the internal pull-up resistor in the I-PAC will pull it to 5 volts with no coin.
Andy
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What Andy says. He is the expert on the ipac.