# DC to DC ?

Discussion in 'The Projects Forum' started by rougie, Sep 17, 2012.

1. ### rougie Thread Starter Active Member

Dec 11, 2006
410
2
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

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2. ### KCHARROIS Member

Jun 29, 2012
292
1
Hello,

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

R1 being 3.4K
R2 being 6.6k

3. ### rougie Thread Starter Active Member

Dec 11, 2006
410
2
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????

r

4. ### t06afre AAC Fanatic!

May 11, 2009
5,939
1,222
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?

Jun 22, 2012
4,779
707
6. ### rougie Thread Starter Active Member

Dec 11, 2006
410
2
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!

r

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7. ### t06afre AAC Fanatic!

May 11, 2009
5,939
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Have you read the data-sheet here http://www.murata-ps.com/data/power/ncl/kdc_nke.pdf
It states clearly
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

8. ### rougie Thread Starter Active Member

Dec 11, 2006
410
2
Hello t06afre,

much what you have recommended.... Based

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?

r

9. ### strantor AAC Fanatic!

Oct 3, 2010
4,295
1,983
yes thats what he means

Dec 11, 2006
410
2
Thanks guys.

r

Dec 11, 2006
410
2
12. ### Bernard AAC Fanatic!

Aug 7, 2008
4,139
393
Could use a Si diode, 1N4002, between 4.08 V & #.3 module for a +_ .7 V drop & a 110 Ω bleeder.

13. ### WTP Pepper New Member

Aug 1, 2012
21
6
Because as soon as you draw current, the voltage will fall.