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question for voltage reduction |
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| May29-06, 04:42 AM | #1 |
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question for voltage reduction
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 |
| May29-06, 05:48 AM | #2 |
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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 |
| Jun1-06, 01:28 AM | #3 |
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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 |
| Jun1-06, 06:07 AM | #4 |
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question for voltage reduction |
| Jun1-06, 06:13 AM | #5 |
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Recognitions:
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A voltage divider might do the trick as Hootenanny suggest. You need to select resitors [itex]R1, R2[/itex] such that the current drawn from the source, [itex]I[/itex], is appropiate to its specs (too small a series resistance will drain the source).
[tex]I=\frac{V_{in}}{R1+R2}[/tex] The output voltage will then be [tex]V_{out}=I R2[/tex] resulting in [tex]V_{out}=\frac{R2}{R1+R2}V_{in}[/tex] Note that the effective [itex]R2[/itex] resistance might be altered by connecting your module to the voltage divider circuit. |
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