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5V -> 3.3V needed!!!!!!!1

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d_a_l_3

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Hi I need converter but without lmxxxx chips and other IC

Pure diodes, capacitors, resistors

If someone have please post here...

thanks
 

hi

i have found this

but not working
 

If the above circuit didn't work then chances are you are pulling too much current at your load. If you have a high load current requirement then using a series-pass transistor like Eugene mentioned is the easy way out.
 

d_a_l_3 said:
hi

i have found this

but not working

Which solution depends on how much your load current is.

With this circuit, you can get maximum current of 17mA ((5V-3.3V)/100 ohm) for both load and zener. A 400-500mW zener usually needs about 10mA for maintaining its voltage reference, so you now have only 7mA for your load.

The circuit, comes from the link given by EugenE, is useless here since the common series transistor needs a drop out voltage of about 3V to operate in the linear region.

So, if you can, buy the voltage regulator as people called LDO (low voltage drop-out), which type, depends on your current need and local part vendor.

The another solution for you maybe a series of 2-3 silicon diodes (1N4001, or 1N5401, ... which one is up to your current) that is used for the surplus 1.7V. You have to make a experiment to decide 2 or 3 diodes is good enough. Connect this chain in series with the load.

nguyennam
 

Use a NPN transistor,

connect base to neg input with ~4V zener,
resistor between base and collector,
collector to pos input,
output is between emitter and neg input.

rate transistor for your current requirements, might have to modify 100 ohm resistor appropriately for your load.
 

Cheap and easy way to do it:

Three diodes in series with the 5volts.

2 of them Si, and the other Ge.

The 2 Si diodes will consume 1.4 volts and the Ge one will consume 0.3 volts, that will give you a consumption of 1.7 volts

5 volts - 1.7 volts = 3.3 volts.

Make sure you have the right polarity
 

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