A simple Project(Beginner)

Discussion in 'General Electronics Chat' started by Kadmos, Mar 19, 2005.

  1. Kadmos

    Thread Starter Member

    Mar 15, 2005
    20
    0
    My phone at home needs 3 x1.5 Volt batteries to work.Cause they get out of juice every two weeks, i decided to use the 9V adaptor.Although the adaptor says its Voltage is 9 ,the multimer says 15.Propably cause here in greece we have 220 AC
    60hz and that was made for america.So i used some resistors and achieved to reduce the voltage at 4.49 using 2 resistors and a variable one.
    My phone works perfectly now,but i could have used many different combinations and different total resistance.
    The question is
    ok about the voltage but the current?how could i know how much current is desired?
     
  2. Firestorm

    Senior Member

    Jan 24, 2005
    353
    0
    batterys are measured in amp hours(how many amps are put out per hr)...Not sure what the specs are for a AA battery are but try looking on google and it should tell you or the battery might state it. Once you figure this out you are almost done! :)...thx l8er

    -fire
     
  3. Brandon

    Senior Member

    Dec 14, 2004
    306
    0
    Connect a meter in series with your adapter so that the current has to flow through the meter. It will give you a number then.
     
  4. Firestorm

    Senior Member

    Jan 24, 2005
    353
    0
    never thought it could be that easy brandon..thx 4 the easier method :)

    -fire
     
  5. Kadmos

    Thread Starter Member

    Mar 15, 2005
    20
    0
    Well the problem is that i dont know how much current is needed
     
  6. Firestorm

    Senior Member

    Jan 24, 2005
    353
    0
    if your phone will run off of 3 1.5v batteries, then those 3 batteries will supply enough current(needed current) to run the phone. Just hook up the 3 batteries together and measure the current using a DMM. Then you know your needed value...thx l8er

    -fire
     
  7. Kadmos

    Thread Starter Member

    Mar 15, 2005
    20
    0
    if i do that then i will get the current that is needed when the telephone is in use.After that what?Make a circuit which has the same current?
    if i do that then wouldnt the current drop more?
     
  8. Brandon

    Senior Member

    Dec 14, 2004
    306
    0
    Lets do a electronics 101 here.

    You want to power a circuit. Powering a circuit require a voltage source, not a current source. Current sources are used more for amplifiers which need current sinks or sources.

    Ideally you want your voltage source such that when you connect a load to it, (load being your phone) that the load does not 'load down' the voltage sourse. Loading down a source is when your circuit draws too much current from the source, and this is the explination why this happens.

    We'll start with an 'ideal' voltage source. An ideal source is one such that the voltage never drops on it no matter how much current it supplies and what it is connected to. This does not exhist, but its a model to make the concept easier. Inside of all voltage sources there is an internal resistance. Sometimes it just connections wires, internal connections, many times it stuff you just can not see that is part of the make up of the source. An ideal source has an internal resistance of zero.

    Now, just like you make a voltage divider out of the resistors to drop down the 15 volts, the same happens with a voltage source. You have your voltages an internal resistance, then your load resistance. If your load resistance is anywhere near your internal resistance you're going to get a huge voltage drop and your circuit is now 'loading down' the source.

    Since you used a voltage divider to get your voltage, you have a resistor which is adding to the internal resistance of your adapter. But since your phone is working alright, the next set of components that you connected your source too must have a fairly higher resistance than your divider.

    A long as your internal resistance of your voltage source is low, you will only supply as much current as your circuit is built to draw. Since its a manufactured phone and works on 4 volts or so for an extended period of time, the current draw is probably quite small, so you have no issues to worry about.

    If you wanted to make your circuit more along eletronics standards, you would use a voltage regulating component instead of the divider you built. Voltage regulators have a very low internal resistance at their output so it takes a heavy load to bog them down.

    Since you have a meter, MEASURE the current. You have to 'open' up the circuit at some point and put the meter inline with the circuit so that the current can flow through the meter. MAKE SURE YOU HAVE YOUR METER ON CURRENT MODE BEFORE you connect it into the circuit. You can blow a meter if you don't. Honestly, I doubt this would happen in your case, but its good practice to make sure you do it right.

    You would bascially take say the + terminal of the batteries connect it to the - input for the meter, then connect the + terminal of the meter back to your circuit where the + battery terminal would have connected to. Another way to see the possible max current draw is to measure the resistance of the phone. Pull the batter out, connect your meter in Ohm mode across the positive and negative terminal where the batteries plug into. Make sure NO POWER is in the circuit when you do this. You will get a resistasnce vale for the phone. Take the voltage your giving it divide it by the resistance and you will get the current draw. This may not be exact since your phone probably has a microcontroller in it and the current will change as it operates.
     
  9. Firestorm

    Senior Member

    Jan 24, 2005
    353
    0
    well said brandon,
    if you still need help just say so :)

    -fire
     
  10. Kadmos

    Thread Starter Member

    Mar 15, 2005
    20
    0
    thanks brandon
     
Loading...