Maker Pro
Maker Pro

first kit, slight problem :-(

maclean

Jul 25, 2012
7
Joined
Jul 25, 2012
Messages
7
Hello everyone!

I'm new and doing a few little kit builds as I have a plan to build a kit synth once I've done some practise on cheapo kits. I've never soldered before this first little kit and although it was good fun putting it together, there is a slight problem with it!

The kit in question is a little 3.3v/5v power supply: http://proto-pic.co.uk/breadboard-power-supply-5v-3-3v/

Unfortunitly there seems to be a slight problem with it.... When I turn the switch to 3.3v and measure the output, its bang on 3.33v on my multimeter, however when I switch the switch to 5v the multimeter reads 3.60v when I'm expecting 5.00v.. Does anyone have a clue what might be wrong with it???

I plugged in a power pack with a 6v DC output (is this output not sufficient do you think?) and there is a schematic for the power supply here http://dlnmh9ip6v2uc.cloudfront.net/datasheets/Kits/Breadboard Power Supply v10.pdf

Im still working my way through 'electronics for dummies' so the schematic doesn't make 100% sense to me yet. Can any one on here shed any light on what might be up?

Thanks a lot!
 

davenn

Moderator
Sep 5, 2009
14,254
Joined
Sep 5, 2009
Messages
14,254
Hello everyone!

................... I plugged in a power pack with a 6v DC output (is this output not sufficient do you think?) and there is a schematic for the power supply here http://dlnmh9ip6v2uc.cloudfront.net/datasheets/Kits/Breadboard Power Supply v10.pdf

Thanks a lot!

Hi there
Welcome to the forums :)
yes that is the problem :) you will need a 9V PSU. the LM317 regulator chip needs at least 2V more on the input than the highest voltage its regulating to
with only 6V plugpac you are already loosing ~ 0.7V through that series diode on the input so the regulator is only seeing ~ 5.2V on its input

so easy fix get a 9V DC plugpack :)

cheers
Dave
 

Harald Kapp

Moderator
Moderator
Nov 17, 2011
13,700
Joined
Nov 17, 2011
Messages
13,700
In addition to Dave's answer, here's a short explanation of the schematic to help you get a grasp of it:

- D1 is for polarity protection. If passes current only if teh voltage on the switch S1 is positive with respect to GND.

- C1 is a filter capacitor to smooth out the ripple coming from the external power source. It supplies power to the regulator during the periods where the input voltage sags (e.g. due to a sudden increas in load).

- LM317 is a voltage regulator (as you have found out by now:) ) It works on the principle that the voltage on the ADJ pin is regulated to (Vout-Vadj)=1.25 V (details see datasheet http://www.ee.buffalo.edu/courses/elab/LM117.pdf).
R1, R2 and R3 make up a voltage divider. Assume S2 is close, then R3 is shorted and the voltage on ADJ will be Vadj=Vout*R2/(R1+R2) and Vout is regulated such that (Vout-Vadj=1.25 V), therefore Vout-Vout*R2/(R1+R2)=1.25 V which leads to Vout=1.25 V*(1+R2/R1)

- If S2 is open, replace R2 by (R2+R3) in the above equations. This way the switch S2 can set up the regulator for different output voltages.

- C2 and C3 again are filter capacitors. C2 has a rather high capacity and filters mainly low frequency distortions whereas C3 has a lower capacity but due to the different construction (probably ceramic or foil) has a lower impedance and is therefore much better suited to filter out high frequency distortions.

- R4 limits the current through LED2. An LED has a nonlinear voltage-current characteristic and will quickly draw a high current and subsequently burn out when connected directly to a voltage source. Don't do that. By inserting the series resistor a voltage drop across the resistor will develop. This voltage drop becomes bigger if the current rises. More voltage drop across the resistor means less voltage on the LED, thus less current. Want to know more? There is a tutorial in this forum on how to use LEDs.

Harald
 
Last edited:

maclean

Jul 25, 2012
7
Joined
Jul 25, 2012
Messages
7
Thanks for those cracking responses dave and harald.

Thing is - there was no instructions with it, other than the website saying you can strip down ANY DC power plug and fire it on.

Thanks for that detailed step by step description harald. Any ideas why it needs so many capacitors to smooth out the voltage if its already being supplied with DC current?

Im only on chapter three about resistors and ohms law, hopefully its all going to make complete sense with a bit more practise.

Thanks again guys.
 

Harald Kapp

Moderator
Moderator
Nov 17, 2011
13,700
Joined
Nov 17, 2011
Messages
13,700
DC is mostly not DC :)

Ideally a DC source would deliver a very constant voltage without ripple. A battery comes near (if the load is only light and not changing).
But a real source has some internal resistance causing the voltage to vary with current.

If the DC source is a conventional one, there is a transformer that generates a low voltage AC from the mains. This low voltage AC is a pulsating voltage that needs to be smoothed (see here: http://en.wikipedia.org/wiki/Diode_bridge).
If the DC source is a switched mode supply, the principle is the same, but frequencies are much higher (again see here for details http://en.wikipedia.org/wiki/Smps).

For the time being just accept the requirement for smoothing (filtering) and go along with your studies. I wish you success.

Harald
 

maclean

Jul 25, 2012
7
Joined
Jul 25, 2012
Messages
7
Thanks harald.

One more question (I promise!).

Is this 'filtering' simillar to a low pass filter on an analog synthesizer do you know? As I have used alot of synths in audio production, low pass filtering audio is something I've done alot of without quite understanding whats going on!

I know digital synths use code etc to control the shape of the filter, however for analog synths - is a low pass filter (in its most basic form) a capacitor smoothing out the high frequecy ripple in the electricity signal leaving only larger smoother waves?

Perhaps im thinking too far ahead - either way - thanks again :)
 
Last edited:

Harald Kapp

Moderator
Moderator
Nov 17, 2011
13,700
Joined
Nov 17, 2011
Messages
13,700
Well, kind of.
High frequency noise is filtered.
However, in a power supply circuit you commonly use rather big electrolytic capacitors to buffer the voltage (keep it as straight as possible) whereas in an audio circuit you generally only want to have the AC component (even if low pass filtered).

Btw: I had to correct the math for calculating the LM317s output voltage. Go back to that post an re-read the section on the LM317.

Harald

Some nit picking:
Only a truly DC coupled amplifier could make use of a true low pass. Many, if not most, amplifiers are AC coupled to minimize problems with DC offset. But an AC coupled amplifier doesn't let pass DC (that's the idea), so a low pass is in effect a bandpass with a very low upper corner frequency (the designed low-pass corner frequency) and a very, very low lower corner frequency (defined by the AC-coupling of the amplifier.
 

BobK

Jan 5, 2010
7,682
Joined
Jan 5, 2010
Messages
7,682
Yes, a smoothing capacitor is a low pass filter. Ideally it will pass only 0 frequency, i.e. DC.

On a synth, filters are used to shape the audio waveforms by suppressing different parts of the audio spectrum, so the purpose is a little different. For instance a triangle or sawtooth waveform is often used on an analog synth, but this would sound very harsh without a low pass filter to get rid of some of the high frequency components of the waveform.


Bob
 
Top