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3 Voltage levels needed (Resistor network ) ( resistor array) or Transistor arr — Parallax Forums

3 Voltage levels needed (Resistor network ) ( resistor array) or Transistor arr

metron9metron9 Posts: 1,100
edited 2006-06-26 17:19 in General Discussion
I would like to use 4 output pins to output 4 seperate voltages to drive a hexfet gate.
At full voltage 5V the gate is open, at about 1.5V it is off that leaves 3.5V
At about 4V it is still full on but starts to drop between 4 and 1.5V so 2.5 / 4 = .625 V per step
This controls the Fet that drives about 300MA ( a Luxion LED) as the voltage drops each step the LED brightness is less
(PWM is not an option here too much noise as the input on the circuit is in the 10mA range)

pin1 = 5V
pin2 = 3.375
pin3 = 2.75
pin4 = 2.125

(Voltages above are approx and would need to be dialed in to the perceived brightness level) Simply put so the LED looks like 4 different brightness levels.

What works for example is to drop the voltage to 2.75V range is use a 3.1V forward Voltage LED
I don't think I will be able to find the correct forward drop values in three different LED's If I could it would be just 3 parts added to the circuit
I will be making these in some quantitys so each solder point adds up in time as well as complexity.

discrete Resistor network could work but each one would have to have its own transistor switch (or diode) and thats 6 resistors and 3 transistors plus one more transistor to the full voltage pin.

I am wondering if there is a chip that has 4 digital inputs with 4 outputs that would solve this problem for me.

On off times are in the 50 to 100 mS range. Perhaps there are calibrated descrete diodes or a diode array chip?

Suggestions welcomed.

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Think outside the BOX!

Comments

  • Mike GreenMike Green Posts: 23,101
    edited 2006-06-26 16:14
    Easier is just a switched voltage divider. You connect a different resistor in series with each of the 4 output pins and the FET gate, then an additional resistor between the gate and ground of 4.7K. Choose the 4 resistors to divide down the voltage the way you want it Be sure to set all only one output to HIGH and the other 3 to INPUT so they don't load down the divider. To turn off the FET, set all 4 pins to LOW. The pin 1 resistor is just there for protection in case of a programming error and could be as small as 220 ohms. A 4.7K resistor from pin4 to the gate gives a voltage of about 2.5V. Since the HIGH voltage of the output pins isn't exactly 5V, you may want to use a 5K pot at each of the pins and measure the gate voltage. Measure the pot resistance and choose the closest resistor value. You can always choose the next lower value and keep a small trim-pot in series with that for fine adjustments.
  • metron9metron9 Posts: 1,100
    edited 2006-06-26 17:19
    Ahh yes I forgot about making the pins input.
    I will have to test the speed as the transition from Output to Input there will be a tristate period.
    I may be able to just drive it with the pins being input and toggeling the pullup resistors
    I dont need to pull the gate back to low as I have an RC time constant doing that
    The gate is driven through a diode. With no extra capacatince on the gate the leakage current of the diode makes a time constant of about twice the time I need so I added an .1uf and a 470k resistor for the bleed resistor.

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Think outside the BOX!
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