Anybody please tell me How to convert 5 - 3.7 volt. Is there any problem If I convert it using resistors.
Thank you.
What are you going to use it for?
If you use a potential divider you need to use values that draw ten times the amount of current than the device is going to use.
That voltage is not standard so why do you need that value?
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Grumpy_Mike:
What are you going to use it for?
If you use a potential divider you need to use values that draw ten times the amount of current than the device is going to use.
That voltage is not standard so why do you need that value?
I'm going to use it for 3.7 volt and 10ma load. So that only.
Thank you.
For maximum efficiency use a switching voltage converter with adjustable output. You can either build the circuit yourself or you can purchase a ready built unit for
No, I want to create it. I don't want to ready made materials.
Thank u.
Venki:
I'm going to use it for 3.7 volt and 10ma load. So that only.
Use a series resistor of 130 ohms.
Caveat: Your load, which you are choosing not to not reveal to us - must draw exactly 10 mA at 3.7V.
Unless you tell us what you are doing it is hard to help.
Your answers so far have been very poor.
A MCP1252-adj can do the job with a minimum of external components.
If you want an even simpler circuit you can use a linear regulator like the lm117/317 but with lower efficiency.
An alternative would be using a zener regulator with a series resistor a little less than 130Ohm
nilton61:
If you want an even simpler circuit you can use a linear regulator like the lm117/317 but with lower efficiency.
I very much doubt it could save 13 mW! (1.3V drop at 10 mA).
It is micro mini car.
A simple voltage divider may be enough for you
To calculate the resistors that you need, here is the formula:
Vin * ( R2 / (R1 + R2) ) = Vout
For example, 5V to 3.3V: R1 can be 1.8K and R2 can be 3.6K
Because
5 * ( 3.6 / (1.8 + 3.6) ) = 3.3333
Try to not use low values resistors (below 1K), because they will be very hot.
Two diodes in series will drop to 3.8 volt.
Foggiest:
Two diodes in series will drop to 3.8 volt.
Is this dissipate heat. Is 6 amps diode is enough to reduce it without heat.
Thank
guix:
A simple voltage divider may be enough for you
To cal
Thank u
Venki:
Is this dissipate heat. Is 6 amps diode is enough to reduce it without heat.
No all solutions will dissipate heat.
Try to not use low values resistors (below 1K), because they will be very hot.
That is what is wrong with a resistor divider. As I said you have to have 10 times the current going down the two resistors as you want to extract at the junction to stop the load dragging down the voltage. So that dissipates 10 times power that you need to with other solutions. So for a 10mA load you should have 100mA doing down the resistors. That means R1 and R2 should total 50 Ohms.
Grumpy mike/Foggiest: can either of you break down the math more for me?
Why are you saying its 10x desired current; is that based on venkis value of desired 10mah? If so how do we know its 10x am i missing a supply mentioned?
i had a 3.7v 230mah battery feeding a heating element in a vape pen (watts not labeled) but it broke so i will replace the battery with 3x AAA battery in series, this provides 4.5v 750mah, im not understanding why the mah really matters as i thought that was just capacity or run time, more important to me is how to calculate the voltage drop based on the resistance?
How do we mathematically go about saying and solving i have 4.5v in and i added a ?ohm resistor that will cause voltage drop of ?v
Any help would be appreciated, lean on the vape pen example because id like to get it running but id rather learn this math for myself if you guys would be so kind as to aid me so stray as needed 
cowboy10068:
Any help would be appreciated, lean on the vape pen example because id like to get it running but id rather learn this math for myself if you guys would be so kind as to aid me so stray as needed
The math:
Well this is a post from 2014 you know. Doing this is know as a necro post.
Why are you saying its 10x desired current;
Because as a rule of thumb you should have about 10 times the current going down a potential divider as you try and draw out of the bottom to maintain the divider ratio. In theory this should be infinitely more but 10 times is normally enough.
Any load on the divider will be in parallel with the bottom leg and so in effect reduce its value changing the top and bottom ratio, giving you this:-
This is why potential dividers for heavy loads are very inefficient.
How do we mathematically go about saying and solving i have 4.5v in and i added a ?ohm resistor that will cause voltage drop of ?v
Voltage drop required = Voltage you have - voltage you want
If the current you want through your heater is I mA then
voltage drop/ I = resistor you need.
Note that 750mah is NOT a current 750mA is a current and 750mAh is a capacity. You need to know how much current you heater needs not what the battery will supply.
Grumpy mike/Foggiest: can either of you break down the math more for me?
Why are you saying its 10x desired current; is that based on venkis value of desired 10mah? If so how do we know its 10x am i missing a supply mentioned?
i had a 3.7v 230mah battery feeding a heating element in a vape pen (watts not labeled) but it broke so i will replace the battery with 3x AAA battery in series, this provides 4.5v 750mah, im not understanding why the mah really matters as i thought that was just capacity or run time, more important to me is how to calculate the voltage drop based on the resistance?
How do we mathematically go about saying and solving i have 4.5v in and i added a ?ohm resistor that will cause voltage drop of ?v
Any help would be appreciated, lean on the vape pen example because id like to get it running but id rather learn this math for myself if you guys would be so kind as to aid me so stray as needed
Did you mean to repeat the question in reply #15?
Or is this just a case of ignoring an answer you don't like?
If you don't understand an answer then ask about what you don't understand about it.