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Rectifier Diode As Steering Diode

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Urahara

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Hi

Been a while! Need help to see if my schematics is correct. Am designing a 5V power supply circuit for my PIC. See attached. For the 3V to 5V battery supply, am using a MAX756 IC. Datasheet recommended putting a rectifier schottky diode at the location specified. I assume this is to prevent battery reverse polarity.

Am wondering if it is ok to place the same diode(D3) further down to also function as a steering diode (two in one).

Also, to favour the wall wart power in the steering diode setup, I have placed a small 1ohm resistor in series with D3. This will drop about 0.15V (150ma * 1ohm). This should work right?

Appreciate your comments and advice. Thanks!
 

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No, it must be located where it is in the data sheet. It prevents the output capacitor from discharging through the internal FET switch. Also note that the diode should be placed between the inductor and the output cap, not between the LX pin and the inductor as you have indicated by the arrow in your diagram.
 
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The 1 thay show is a steering diode it keeps your output from shorting your input why do what to use the 3 volts for
 
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Here is something a little easier
 

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Hi Speakerguy78

I have re-drawn the schematics. Is this what you meant?

Hi be80be,

I was thinking of using 2xAA batteries since a 9V battery isn't as lasting as the AA ones, but the simplicity of the circuit is something worth considering. In your schematics, your -5V means GND?

Thks! :D
 

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That should work. Yes the 5 volts - is gnd I should of change it. You no cel phone car chargers have DC-DC switching regulators in them I have reused them to get 12 volts and 5volts for a hard drive controller that I was making a clock out of.
 
I use 4 AA for power to circuit I drawled or a 9 volt. Have you tryed your circuit.
 
Why did you not use this in your circuit
The output voltage of the MAX757 is set by two resistors,
R1 and R2 (Figure 1), which form a voltage divider
between the output and the FB pin. The output voltage
is set by the equation:
VOUT = (VREF) [(R2 + R1) / R2]
where VREF = 1.25V.
To simplify resistor selection:
R1 = (R2) [(VOUT / VREF) - 1]
Since the input bias current at FB has a maximum
value of 100nA, large values (10k½ to 200k½) can be
used for R1 and R2 with no significant loss of accuracy.
For 1% error, the current through R1 should be at least
100 times FB’s bias current
 
Knew abt MAX757 but guess I was lazy :eek:

Since I wanted something just below 5V so that the steering diodes work in favour of the wall-wart power when it is used, I thought a small resistor in series would do the trick. Saved me the calculation and another resistor. :D

Do you think my logic is flawed?
 
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