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

  • Thread starter Thread starter rollyasilo
  • Start date Start date
  • Tags Tags
    Reduction Voltage
AI Thread Summary
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.
rollyasilo
Messages
2
Reaction score
0
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
 
Physics news on Phys.org
A simpler method may be to reduce the input voltage by using a potential divider. Or would that interfere with your counting process?

~H
 
Last edited:
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.
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top