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MC33063A Vout Switching

Samuel Alessi
Prodigy 20 points
Other Parts Discussed in Thread: MC33063A

Objective: Switch Vout between two constant voltages that are toggled by a relay/photocell combo.

Input voltage: 18-24VDC (varies, cuts off if dropped below 18VDC)

Output voltage: TBD, approx. 7VDC and 9VDC

The challenge we're faced with is how to switch the voltage from one to the other via the photocell. As of right now, the relay outputs the input voltage through output A with output B switched off, unless it detects light at a certain level, then it outputs the input voltage through output B, switching off output A.

Right now our design uses a LM7812 to take the fluctuating input voltage and output a constant 12VDC and feeds it into the relay, which outputs 12VDC through either output A or output B. Both outputs are fed into separate LM317T IC's with a 5k Ohm pot to fine tune the output voltage. 

Would we be able to use one MC33063A to take the place of both LM317T's, the LM7812 or all three?

  • 0
    TI__Mastermind 24975 points
    Hello Samuel,

    If I understand what you are doing, then I believe you may be able to achieve exactly what you want with a single MC33063A,

    You would be using the MC33063A in a buck configuration as shown in section 9.2.3 . The detailed design procedure on the following page helps to design what component values you would need depending on the output voltage, loading conditions, ripple, and so forth.

    If you want to set the output voltage to either 7V or 9V, your Vout = 1.25*(1+R2/R1), so you could set the two voltages based on two separate R1 values. Lets Fix R2 to some value, lets say 4.7k. Based on this value we can calculated the different R1 resistors that we would want to "switch in" to the circuit.
    A quick calculation based on the equation in the paragraph above gives the following:
    Vout = 9V --> R1 ~=760 Ohm
    Vout = 7V --> R1 ~=1kOhm

    Since your photocell works in such a way that either the output A or the output B is activated, then you could just have each of these resistors connected through an NPN transistor to ground with output A activating one and output B activating the other.

    You could also just use a "baseline" resistor to set the initial output voltage. Meaning you could have the 1kOhm always connected to ground and you could just use one of the outputs from your photocell to switch in a resistor that will decrease the R1 total resistance down to 760Ohm. This extra resistor would take on the value of ~3.16kOhm. This method would probably be preferred since it will at least ensure that the output voltage is 7V at all times, and it will only increase when the 3.16kOhm resistor is switched in. This way you also only have to use one of the outputs of the photocell.

    Does this make sense?
    Let me know if you have additional questions and I would be happy to help.

    Best,
    Michael
  • 0 in reply to Michael J Schultis
    Prodigy 20 points
    That sounds like it will work. Thanks for the help!