Brown Out Circuit

Discussion in 'The Projects Forum' started by tazntex, Sep 2, 2010.

  1. tazntex

    Thread Starter Member

    Sep 29, 2008
    I am using the external brown out example for my PIC16F628A processor because I my supply voltage is 3.3v.

    Note 1: ​
    This brown-out circuit is less expensive,
    albeit less accurate. Transistor Q1 turns off
    when V
    DD is below a certain level such that:

    2: ​
    Internal Brown-out Reset should be
    disabled when using this circuit.

    3: ​
    Resistors should be adjusted for the
    characteristics of the transistor.

    DD x R1 = 0.7 V
    R1 + R2

    I don't know how to insert my schematic so I will try to explain:

    R1 and R2 form a voltage divider, R1 to 3.3v and R2 to ground, the junction in between R1 and R2 goes to the base of Q1,2N3906. Then emitter of Q1 goes to 3.3V and the collector goes to a 40K resistor(shown in the exampe) to ground. In between the collector and the 40K resistor is a juntion to connect to the MCLR of the processor. Using the above formula, I have R1, 82K, and R2, 330K which alone should use about 8uA, running a simulation of the entire Brown Out exampe using my values for R1 and R2, but no load at the collector of Q1 it draws about 7mA's. Since this is a battery powered circuit and I am trying to keep current consumption to a minimum, how can I recalculate the value of the 40K resistor that when I am pulling MCLR to ground at the collector when Q1 is off.