If i want to power the sensor using a digital pin, since the lipo battery voltage (and hence Arduino digital pin voltade) can be higher than the sensor supplied voltage, what value resistors should i use for a voltage divider so the sensor is not damaged?
If i want to power the sensor using a digital pin, since the lipo battery voltage (and hence Arduino digital pin voltade) can be higher than the sensor supplied voltage, what value resistors should i use for a voltage divider so the sensor is not damaged?
Thanks for any help
You really shouldn't power your Uno only on 3.7V.
Input Voltage (recommended) 7-12V
Input Voltage (limits) 6-20V => minimum of 6V on external power supply
If supply range is 1.4V to 3.6VDC why don't you use the 3.3V output of arduino ? :~ :~
can you post a schematic of the power side of your set-up ?
are you currently powering it with a battery voltage that varies ?
almost any set of diodes or resistors will always be in play. may not be horrible as the device can live at lower voltages. In other words, if your power comes in at 4.7v and you drop to 3.3 [or 1.4 volt drop through the diodes ], Then if your power supply drops to 3.7, the diode version will drop the voltage by 1.4v to 2.3. (fixed voltage drop) the diodes drop the voltage, not a percent.
On the other hand, the resistors selected to drop from 4.7 to 3.3 are dropping the voltage by 70% If the voltage drops to 3.7, then the voltage will be about 2.59 or 70%
the voltage divider consumes power all the time. putting a diode between the power supply and the device just consumes some voltage as it is being fed to the device.
what are your battery requirements?
[corrected punctuation to make it read more clearer]
dave-in-nj:
In other words, if your power comes in at 4.7v and you drop to 3.3 or 1.4 volts, then your power supply drops to 3.7
Sorry, if i use a 3.3V voltage regulator on a 9V battery then my battery will supply 8 volts and not 9 ?
You re suggesting that if i use a multimeter to check the voltage on the battery it will be 8V and not 9V ?
dave-in-nj:
In other words, if your power comes in at 4.7v and you drop to 3.3 or 1.4 volts, then your power supply drops to 3.7
Sorry, if i use a 3.3V voltage regulator on a 9V battery then my battery will supply 8 volts and not 9 ?
You re suggesting that if i use a multimeter to check the voltage on the battery it will be 8V and not 9V ?
maybe I was a bit confusing......
if your battery is outputting 9 volts, your multimeter should read 9 volts when connected to the battery.
if the battery drops to 8 volts, then your multimeter should read 8 volts when connected to the battery.
if you use a diode that drops 0.7 volts. then after the diode, the voltage would be 8.3 volts, with the 9 volt output of the battery.
if the battery drops to 8 volts at the battery, then after the diode, the reading will be 7.3 volts.
the voltage drop of the diode will always be 0.7 volts.
dave-in-nj:
In other words, if your power comes in at 4.7v and you drop to 3.3 or 1.4 volts, then your power supply drops to 3.7
Sorry, if i use a 3.3V voltage regulator on a 9V battery then my battery will supply 8 volts and not 9 ?
You re suggesting that if i use a multimeter to check the voltage on the battery it will be 8V and not 9V ?
Discussion has been about using diodes or a voltage divider, not voltage regulators. For a voltage regulator, you need to read the data sheet for that device to get the values.
of course, voltages would be after the device, not at the battery.
Is there any disadvantage of using a silicon diode over a resistor voltage divider to reduce the voltage to the temperature sensor?
Yes much so... the resistor divider you create to power the arduino will sag with load much so... the diode will not.
Can you please clarify this. I am unsure what you mean.
The lipo battery voltage will vary between 3.5-4.2V. The temperature sensor requires 1.4-3.6V. With the diode as shown, the voltage at the sensor should always be in the required range, if i assume a voltage drop of 0.7V across it. Please correct me if i am wrong.
Be careful with approximations! The voltage drop of a semiconductor diode is an exponential function of the current and temperature, and the approximation of 0.6 - 0.7 V is only good for currents on the order 1-10 milliampere (for a typical small diode like the 1N4148).
The SparkFun temperature sensor draws only 10 microamperes in the active mode and according to the model for the 1N4148 diode built in to LTSpice, the voltage drop will be only about 0.37 V at 10 uA.
If you put a 1K to 5K resistor in parallel with the temperature sensor to increase the total current draw, the diode will then drop around 0.7 V.
