Shop OBEX P1 Docs P2 Docs Learn Events
2.5V on demo board — Parallax Forums

2.5V on demo board

n.borreron.borrero Posts: 16
edited 2008-09-08 05:24 in Propeller 1
quick question...

i need 2.5V for the ADXRS300 gyro for the precision reference pin. how can i do that? whenever i find voltage dividers, it always has them going from +V and through resistors into -V. Since there's no -V on the board, i'm not sure what i'm supposed to do. any thoughts? roll.gif

Comments

  • HarleyHarley Posts: 997
    edited 2008-09-08 00:59
    Just connect it to ground. Let that be your most negative voltage for the divider. Scale from there.

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Harley Shanko
  • Mike GreenMike Green Posts: 23,101
    edited 2008-09-08 01:13
    The easiest voltage divider is to put two identical resistors in series between +5V and ground and use the center for +2.5V. Two 470 Ohm resistors would give you a divider current of about 5mA which is probably more than enough. Use an Ohmmeter or multimeter to match the resistors as well as you can.
  • n.borreron.borrero Posts: 16
    edited 2008-09-08 01:43
    ok, thanks. i was just worried about whether or not i could just use ground. i'm not much of a circuits guy. but i got it workin now. thanks guys.
  • n.borreron.borrero Posts: 16
    edited 2008-09-08 03:25
    for this same gyro, it says that it needs a 2.5V reference. it also says the "Load Drive to Ground" is 200 uA. the power supply is 5.0V with a max "Quiescent Supply Current" of 8.0 mA. does this mean that i'm also supposed to make sure that the 2.5V is has a current of <8.0 mA also? or does the sensor somehow regulate the current? i was able to get it below that value by just adding a 330 Ohm resistor before the sensor but i don't know if i'm supposed to do this.

    i've also been messin with an ADXL330 triple access accelerometer and i haven't done anything about regulating the current going into this. it's been working but i don't know if it'll break. this sensor is supposed to run on 320 uA. the board outputs at about 0.5 A so i guess i'm way over.
  • Mike GreenMike Green Posts: 23,101
    edited 2008-09-08 03:51
    Quiescent Supply Current means that when the regulator is "quiescent" with the load (the Propeller, the +3.3V regulator) essentially not there, it (the regulator) will draw that minimum amount of current.

    The 200uA is the amount of current the gyro will draw from the reference voltage supply. With any voltage divider, the output voltage is only what you expect if the current drawn is a small fraction of the total current drawn by the voltage divider. You can use Ohm's Law to figure out the actual output voltage given any particular output current. With the ratio I suggested (5mA : 200uA or 25 : 1), the output voltage will be pretty close to 1/2 the input voltage.
  • n.borreron.borrero Posts: 16
    edited 2008-09-08 04:15
    ok, so when i hook up a DMM, n it reads the current as 5mA that means that that's the maximum current that can be drawn from there, right? so then if you hook up a sensor that only needs something like 500uA, the 5mA won't be going to it since it doesn't need that much? is that more or less how it works?
  • Paul BakerPaul Baker Posts: 6,351
    edited 2008-09-08 05:24
    Typically when a part asks for a reference, accurate readings require a voltage reference chip. It is used for biasing analog cicuitry, and if you use 5% resistor divider, expect to see up to a 5% error in your readings. If you use a resistor divider, you should use 1% or better.

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Paul Baker
    Propeller Applications Engineer

    Parallax, Inc.
Sign In or Register to comment.