stepping down from 6.3V to 5V DC

[priya123]

Hello

I am new to hardware and my ques may seem stupid. I apologise for that.
I wanted to step down 6.3V to 5V and I did that using a 1.75 Mohm resistor.
Although when I connect it to the module ( DRM -which requires 5v) the voltage drops to around 0.3V
Is it because the resistor is consuming too much power ?
Could you kindly help.

Thank you.

 

[prakashvenugopal]

Hi,

1. Use a Voltage regulator to step down 6.3V to 5V. or
2. Place two Diodes in series to drop 1.4 Volt.

 

[eklikeroomys]

Remember that you are using a voltage divider, which consists of two series resistors, in order to obtain the 5 V. When you connect the module, you are effectively connecting its input resistance in parallel with the second resistance of your voltage divider. The parallel combination will effectively reduce the value of your second resistor. The majority of the 6.3 V will now sit across your first resistor, which is now larger than your parallel combination. This is why you are only seeing 0.3 V at the input of your module.

I would rather do the following: Connect a voltage buffer between your voltage divider and the input of the module. Or, as Prakashvenugopal said, rather use a voltage regulator.

 

[Hewitson]

The above solutions are superior to doing this, but if you wanted to use resistors you should make a voltage divider.

 

[mvs sarma]

@ Priya
as you are handling a power supply ( changing a 6.3V to 5V) you should always aim at very low internal impedance, such that it delivers the needed load current. you feeling that so much power ... is not correct, power is a multiplication of voltage and current. with Meg ohm resistor , where is the current at all.

even for the digital multimeter (with 10MOhms impedance, to read a voltage it needs some current to flow thro. p you might appreciate that you poser supply internal impedance is in order of M ohms.

you would appreciate designing or scaling power supplies with potential dividers, and the criteria for selection of components. I would rather feel that such an arrangement should be capable of supplying 10 times the current than what you need.
While using regulators, obvious that you need to use LDO types that work even at 5.6V input to give out 5V at its output.

suppose you current demand is not much, like a 74HCxx CMOS chip , then you may use 100 ohms with 5.1V, 1/4 watt zener diode to ground. filter the Zener diode by say 1uF capacitor and use it.

 

[pranam77]

@Priya
Could you let us know what will be your load connected to the said/ derived 5 volts? You need it to power some devices or you need it just as a reference source?
Cheers

 

[priya123]

Thank you for all your valuable suggestions.. I will try all the possible solutions and get back to you.

 

[Externet]

+6.3 VDC----------------|>|----------------|>|----------------+5 VDC

Use general purpose silicon diodes capable of handling twice the current consumed.

 

[Raza]

As Externet said the quickest and cost less solution is to use two general purpose Diodes in series to the load. It will drop 1.4 volts and the final circuit will get 6.3-1.4=4.9 volts which will be assumed as 5.0 volts by the circuit.