How Can I Reduce Voltage from 1.3V-4.8V to Below 1.15V for My Microprocessor?

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To reduce the voltage from a range of 1.3V-4.8V to below 1.15V for a microprocessor, a potential divider is suggested as a simpler solution. Users have attempted using diodes and resistors, but these methods have not successfully lowered the voltage below the required threshold. The voltage divider requires careful selection of resistors to ensure the current drawn from the source is appropriate. It is important to consider how connecting the microprocessor affects the effective resistance in the circuit. A well-designed voltage divider could provide the necessary voltage reduction for proper device operation.
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i have a module with microprocessor of 1.15volt sense level. when the potential is above this value it registers high and below it registers low. however the potential from the source varies from 1.3volts to 4.8volts only.nothing less than 1.15volts so all registers are high.

anybody can help me reduced the lowest potential to be less than 1.15volts so this device will work. i used this device for pulse counting from a solid state digital output.

any idea?

rolly
 
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A simpler method may be to reduce the input voltage by using a potential divider. Or would that interfere with your counting process?

~H
 
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I have tried with diode 1N40001 to reduced the lowest and highest potential by 0.7volt but it did'nt work at all. I've tried with resistor but it reduced the higher potential only and lowest potential still the same.

anybody can help me resolved this matter.

rolly
 
rollyasilo said:
I've tried with resistor but it reduced the higher potential only and lowest potential still the same.

I thought you needed only a potential difference of 1.15 V ?
 
A voltage divider might do the trick as Hootenanny suggest. You need to select resitors R1, R2 such that the current drawn from the source, I, is appropiate to its specs (too small a series resistance will drain the source).
I=\frac{V_{in}}{R1+R2}
The output voltage will then be
V_{out}=I R2
resulting in
V_{out}=\frac{R2}{R1+R2}V_{in}

Note that the effective R2 resistance might be altered by connecting your module to the voltage divider circuit.
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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