Hello,

I have a bread board which requires two voltages. The first is 5VDC and the second one is 3.3VDC.

Please view the attachment for a simple schematic.

The problem is that when I adjust my power supply to exactly 5VDC in order to satisfy the 5Volts part of my board, the DC/DC modules shown in attachment deliver 4.27VDC instead of 3.3VDC.

Now I imagine this is because the DC/DC modules have little load on them, but this is really annoying because my components connected to the DC/DC modules really require 3.3VDC or at worst 3.5 MAX!!!

My DC/DC modules are "NKE0503SC". I also realize that the more components I connect to the DC/DC modules the more I get cloeser to the expected DC/DC module voltage of 3.3VDC


My question is, what can one do to temporarily achieve a steady 3.3VDC while building up the board?

I was thinking of putting a 50 ohm resistor between the 3.3VDC rails to temporarily attain the desired approximate 3.3VDC.

all feedback is appreciated!
r

 

Hello,

Why not use 2 resistors and form a voltage divider to give yourself 3.3V.

R1 being 3.4K
R2 being 6.6k

 

Hello KCHARROIS,

questions:

Q#1:
If,
It = 4.27/10K = 0.427ma
VR1 = 0.427ma x 3400ohms = 1.4518V

Voltage divider will be 4.2 - 1.4518 = 2.74Vdc ???

Is there a reason why you set the voltage to 2.7 instead of 3.3

Q#2)
Will your voltage divider solution be able to hold 3.3VDC at 150ma????

thanks for your help!
r

 

I think you need a bleeder resistor connected to the ground. To ensure a minimum load that will give 3.3 volt out from the converters. By the way. Why are you not using a low dropout regulator instead of those DC to DC converters. Do you need isolation?

 

Use a 3.3volt regulator instead of the dc/dc boards.

http://uk.farnell.com/texas-instruments/lp2950-33lpe3/ic-volt-reg-micrpwr-3-3v-sd-to92/dp/1755063

unless your trying to create an isolated supply.

 

Okay,

The 4.27 volts was wrong. I disconnected all my DC/DC modules and I was still reading the 4.27VDC on the secondaries of the DC/DC modules???? I realized, I had a wire connected from the 5V system to a diode which connected to the 3.3VDC rails .... LOL!!!

So still though, I read 4.08VDC instead of 4.27VDC which really isn't that trivial... but still I would prefer 3.3 exactly.

So figured out some temporary circuit!

Thanks for your help!
r

 

Have you read the data-sheet here http://www.murata-ps.com/data/power/ncl/kdc_nke.pdf
It states clearly

Minimum load
The minimum load to meet datasheet specification is 10% of the full rated load across the specified input voltage range. Lower than 10% minimum loading will result in an increase in output voltage, which may rise to typically double the specified output voltage if the output load falls to less than 5%.​

You need to have a bleeder resistor that ensure about 20% of max current draw. Max current for your device is 303mA. Se the TOLERANCE ENVELOPES figure. If you do as I say everything should work. But it is your call​

 

Hello t06afre,

I have made an adjustment circuit that I think does pretty
much what you have recommended.... Based
on the load I adjust the circuit so it corresponds
to a stead 3.3 VDC.

This is all temporary because eventually the circuit on
it's own will draw at least approximately150ma when I will
be done with it.

When you say a bleeder resistor, do you mean a resistor
from +3.3VDC to ground ensuring a load of
20% of max DC/DC current module?

Thanks for your help?
r

 

yes thats what he means

 

Thanks guys.

r

 

>Why are you not using a low dropout regulator instead of those DC to DC converters.

You mean something like this:

http://www.linear.com/products/ldo_linear_regulators?gclid=CI3mvqbBv7ICFUGo4AodL2sABw

I had these DC/DC modules hanging around so I just used them. Perhaps when the project gets a little more serious, I will go with LDO's!

thanks for your help!
r

 

Could use a Si diode, 1N4002, between 4.08 V & #.3 module for a +_ .7 V drop & a 110 Ω bleeder.

 

Hello,

Why not use 2 resistors and form a voltage divider to give yourself 3.3V.

R1 being 3.4K
R2 being 6.6k

Because as soon as you draw current, the voltage will fall.