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How to lower 5v to 1.6v and 0.8v?

  1. May 24, 2009 #1
    Hey guys, ive been stuck on this for the past few hours and im about to lose it, what i am using is a usb interface that outputs 5v on each pin and when i activate one pin i need it to output 1.6v and when i activate ithe other i need it to output 0.8v, i attempted this using a voltage divider and got around 0.83v which was great but i couldnt get 1.6v at the same time, any help with this would be fantastic.

    Thanks Scott
  2. jcsd
  3. May 24, 2009 #2
    How mich current do you need in each case? Can you use a resistor divider, or do you need to regulate it?
  4. May 24, 2009 #3
    i dont belive the current matters, its actually for a analog stick, 0 volts is backwards, 0.8 is centered and 1.6v is forwards, i need to be able to change between the three, i cant figure it out. Thanks for the reply
  5. May 24, 2009 #4
    If current doesn't matter you could use a 1k resistor in series with a 1k pot and adjust the pot for whatever voltage you need. It would give you a range between 0 and 2.5 volts.
  6. May 24, 2009 #5
    Hey thanks for that, it has to be automatic though, so when i select pin one in hyperterminal it outputs 1.6v and pin two outputs 0.8v although they are both 5v by defualt
  7. May 24, 2009 #6
    So everything has to go to one input on your analog stick?
  8. May 24, 2009 #7
    If I'm understanding you correctly you can do it with two diodes, one 1k resistor, and one 1k pot. Connect the anode of one of the diodes to one of the USB outputs. Connect the anode of the other diode to the other USB output. Connect the 1k resistor between the cathodes of the two diodes. Connect one end of the pot to ground. Connect the other end to the cathode of the diode which is connected to the USB output you want to use for the 1.6 volt. Connect the middle tap of the pot to your analog stick.

    You can adjust the pot for 0.8 volts or 1.6 volts, depending on which USB output you have activated. You only have to adjust the pot once. Having both USB outputs off will give you 0 volts.

    Attached Files:

    Last edited: May 25, 2009
  9. May 25, 2009 #8
    Hey TurtleMeister thanks alot for the reply, yes both outputs must go to the same connection im sorry if im being silly here or mabye not explaining it correctly, i dont understand how the circuit you gave me would change between 0.8v and 1.6v, with only adjusting once, wouldnt i have to adjust it each time between voltages? sorry if im being silly, i assumed i would need difffrent valued resistors or something. Thanks again
  10. May 25, 2009 #9
    No, you only have to adjust once. The reason it works is that 0.8 is half the value of 1.6. And since the resistor and the pot have the same value, when you switch to the 0.8 USB output it will cut the current in half. So whatever voltage you have the pot set at, it will be halved also. Just use ohms law and you will see that it works. I've already calculated that you will need to set the pot at 320 ohms (as measured from tap to ground). With the 1.6 output active you will have a total resistance of 1k, so the current will be 5/1000=.005, and the voltage at the tap will be .005*320=1.6. When you swith to the 0.8 output you will have a total resistance of 2k, so the current will be 5/2000=.0025, and the voltage at the tap will be .0025*320=0.8.

    Of course you could also use a 680 ohm resistor in series with a 320 ohm resistor in place of the pot and get the same results. I should also mention that this circuit assumes that the current draw from your analog stick will be negligible and that your USB interface can source 5ma. And of course the USB interface ground and the analog ground must be connected together.
  11. May 25, 2009 #10
    ahh thats excellent, its kind of over my head though, should really stick to programming lol, i belive i have a few 680ohm and 320 ohm resistors althought i am unsure how to connect them, do i attatch the 680ohm to one 5v output and the 320ohm to the other 5v output and connect them both to the anloug stick?, the analoug stick wont be a problem with current draw, it was at first so i just removed it and am attatching directly onto the y axis point on the board, thanks again i appreciate your help.
  12. May 25, 2009 #11
    On second thought, I would not recommend using the resistors in place of the pot. The reason is that I did not account for the voltage drop across the diodes, which would be somewhere around 0.7 volts and may vary depending on the diode. Because of this I cannot predict exactly what value resistors you would need to get the voltages you want. However, this would not be a problem using the pot because you can adjust it using a voltmeter. Make sure both diodes are of the same type.
  13. May 25, 2009 #12
    ahh thats great, i wont be able to get to an electronics store for a few days and id love to get this complteted today, just now i am using a 3300 ohm resistor and a 660 ohm resistor which is giving me 0.8v i assumed if i connected both 5v pins anf activated one it would be 5v and give me 0.8v and if i activated both pins it would be 10v and give me 1.6v although the pins are both active and 5v each when i combine them im only getting 5v, im sure thats really basic stuff but it seems wrong lol
  14. May 25, 2009 #13
    Yeah, connecting the two 5v together will still only give you 5v. By the way, it's not a good idea to connect two outputs together like that without using diodes. In some cases it could damage the outputs when one is high (5v) and the other is low (0v). That's the reason I used the diodes in my circuit. However, the interface could already have the diodes internally, but I had no way of knowing that. To be safe you should use the diodes.
  15. May 25, 2009 #14
    ahh ok, thanks for all your help, guess ill have to go up to maplin :P
  16. May 25, 2009 #15
    i found a few POT's although i dont have diodes, i figured id give it a bash anyway, i made a voltage divider using a 430 ohm resistor and a 680 ohm i then connected the POT, one output gave me 0.8v and the other was 1.8v which is great, the 0.8v has to be as close as but the 1.6v is a minumum up to about 1.8-2v, but when i connnected the same setup again to do the other axis the values all changed, i belive its because they are using the same ground perhaps?
  17. May 25, 2009 #16
    Sorry, I'm not following that. You can change the resistance values in my circuit as long as you keep the pot value and the resistor value the same. 2k resistor and 2k pot, 5k resistor and 5k pot etc... However, I would recommend staying with the 1k values because if there is any appreciable current sink from your analog stick the lower values will work better.
  18. May 25, 2009 #17
    sorry i explained it poorly, what my ideal solution would be is to create a voltage divider on each pin one that lowers 5v to 0.8v and one that lowers it to 1.6v, i can do that and it works great, i enable pin one and get 0.8v, i enable pin two and get 1.6v when they are seperate although because they are going to the same place the y axis, it seems that the resistance changes once i join them together although only one pin is active at a time, do you know any way around this?

    Thanks again

    Edit: ahhh i was just looking up the parts u suggested getting and relised that a diode stops current from flowing in a certian dirrection, i was able to pull four from an old control pad and it works fantasticlly, thanks soo much for your help.
    Last edited: May 25, 2009
  19. May 25, 2009 #18
    Congratulations. Glad you got it working.
  20. May 26, 2009 #19
    Hey TurtleMeister sorry to bring this up again, the control pad was freaking out when i applied 1.9v that must be too high a voltage this this type of controller, i tried the setup you suggested today and i cant get it to work for some reason, i have a diode on each usb port, a 1k resistor between them and a POT connected to one of the outputs,when i set output one to 0.8v using the POT it works great, although when i change to out put two it dosent double to 1.6v it goes up to 0.92v, sorry for the trouble.
  21. May 26, 2009 #20
    Are you using a 1k pot? If all components are correct then the most likely cause of this is that your analog device is sinking too much current for this circuit. You said in your earlier post that the current did not matter so I designed the circuit based on that. If the device requires current then I will need more information about the analog device.

    You can use your voltmeter to determine if the analog device is sinking current or not. Check the voltage with the analog device connected, and then check it with the analog device disconnected. If the voltage changes then it is sinking current.
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