Help: Attempting a Geofex Spyder

[minnow]

Hi friends,
I am in the midst of attempting to build my own power supply unit   

I am planning to use this circuit by RG which can be found at his site: http://www.geofex.com/Article_Folders/Spyder/spyder.htm


I have this transformer and bridge rectifier:



I need some help to understand a few things:
1. What would be the transformer Amp output? It says 12VA. I did a search on the net but couldn't understand the conversion from VA to A

2. I couldn't get any DB01s, but I have RB154 which has similar tech specs (http://www.rectron.com/data_sheets/rb151-157.pdf). I can't understand the pinout of the RB154 though, there's only one lead with a "+" mark. I'm guessing that the "-ve" pin should be the one directly opposite the +ve? The remaining two pins shouldn't matter since it is AC, right?   

3. How do I wire the transformer to obtain 12V (since it's got 6V-0-6V)?   

4. What should be the value for components A,B,C,D & the diode in RG's schematic?   

Is this correct?:

[JKowalski]

Hi friends,
I am in the midst of attempting to build my own power supply unit  

I am planning to use this circuit by RG which can be found at his site: http://www.geofex.com/Article_Folders/Spyder/spyder.htm


I have this transformer and bridge rectifier:



I need some help to understand a few things:
1. What would be the transformer Amp output? It says 12VA. I did a search on the net but couldn't understand the conversion from VA to A

2. I couldn't get any DB01s, but I have RB154 which has similar tech specs (http://www.rectron.com/data_sheets/rb151-157.pdf). I can't understand the pinout of the RB154 though, there's only one lead with a "+" mark. I'm guessing that the "-ve" pin should be the one directly opposite the +ve? The remaining two pins shouldn't matter since it is AC, right?  

3. How do I wire the transformer to obtain 12V (since it's got 6V-0-6V)?  

4. What should be the value for components A,B,C,D & the diode in RG's schematic?  

Is this correct?:

1. VA means Volt-Amp. It is just volts x amp. If you notice power is also in V x A - but this is not actually power in the real sense - it's like "power handling" capability. Whereas power in watts means power dissipated or absorbed, VA is power transferred. In an ideal transformer, you wouldn't lose any power.

So say you have a 12V RMS transformer. It is 24VA. That means 24VA/12V=2A.

2. You are correct in your second question, that will work, bridge rectifiers are all the same the only things you typically need to think about are current capability, reverse voltage breakdown, and (less often) voltage drop. The pinouts you worked out are correct as well, that's the only way that would make sense. If you really wanted to you can double check it with a diode function on a multimeter. AC pins are interchangeable.

3. to get 12 volts, just ignore the center tap (0V) and use only the two ends (6V). A more logical way to view center tap transformers is (in your case) 0V-6V-12V instead of 6V-0V-6V. The second way would be if you used the center tap as ground, and truthfully should be read as (-6V)-(0V)-(6V) since each end of the transformer will be inverted from each other in that configuration (this is how two diode center tap full wave rectifiers work)

4. Information is provided as to those values on the LM317 datasheet (how to calculate them) and using this but I would suggest ditching the LM317 and using a fixed regulator instead. The adjustments aren't very useful IMO.

I know R.G. says D is 200 ohms, take a close look at his site.



Yes your diagram is right.

[minnow]

Hi JKowalski,
Thank you for your reply! It's been extremely helpful!!!
I shall proceed with my attempt with extra confidence since having read your reply!!! 

Only 2 simple reasons for me wanting to use the LM317:
1. I have a lot of them lying around
2. Wanting to check out the "dying battery" feature 

Once again, thanks for your help!
A lot of my questions if not all, have been answered, and a lot of doubts reassured!
THANK YOU!!! 

[R.G.]

One other thing to think about - total output current.

The LM317L is rated for approximately 100ma. If you intend to rum a number of pedals from the output, you may need more than that. Pedals are commonly in the 20ma-50ma range, although simple pedals may be much less and some commercial digital based pedals are much more. The LM317 (no "L") is rated for 1A or more, and can run many pedals. So you need to be clear with yourself about how many pedals go on one output, if you are making more than one output.

If you are making a single 100ma output, you don't need a 12VA transformer. You only need about 1.5 to 2VA. For a multi-output transformer powering N outputs, you need N times 1.5-2VA. The 12VA rating is for a hypothetical rewind of an 8-output transformer.

The transformer world works on "VA" instead of watts because they historically have to provide constant volts into varying loads, much of which is composed of motors which are heavily inductive, so the current is phase shifted compared to the voltage. But the power lines have to supply the current anyway, in phase or not. So they use volts times amps, "VA", not watts.

Notice that I do NOT recommend that you rewind a transformer, even following the instructions for the split-bobbin construction from the article. Done properly, there is little chance of electrical faults, but there is some. There is a commercial transformer at an attractive price which provides multiple outputs and is cheap enough to make rewinding your own a bit silly. Check the webervst web site. Mouser electronics also has single output 12V transformers for about $2.50 each that will do one output nicely. This makes producing an N output power supply a matter of replicating the single transformer/rectifier/filter/regulator setup N times.

Finally, if you are using one output for more than the 100ma of the single LM317L, you'll have to buck up that filter cap. See "Power Supplies Basics" at GEO.

[minnow]

Hi RG, thank you for your precious inputs.
However, since my technical knowledge is indeed limited, I am still a little blur on your reply.

Ok, I have the LM317AT, datasheet: http://www.national.com/ds/LM/LM117.pdf
It says "Guaranteed 1.5A output current". I couldn't find the 100mA rating
But it is not really an issue, because I intend just one pedal for one output.
I will be using your exact suggestion, one transformer for one output:


About what you said that I do not need a 12VA transformer for a single 100mA output...
Based on your design above, shouldn't I need to have a minimum of 12V into the bridge rectifier to achieve a 9V (variable via "dying battery") at the output?
Actually I found 2 different requirements, one from your site that the input should be 2V higher than the desired output. The other is from http://www.muzique.com/schem/lm317.htm? (Thanks JKowalski for the link!!): "Remember that the input voltage to the LM317 must be at least 1.5v greater than the output voltage. "
And from the muzique site, if I am right about the R1 & R2 values, with R1=240ohms & R2=1.5MOhms, output will be 9V.
Now, I have a 12V in, which still doesn't meet that requirement I should have about 13.5V in to the LM317.

Argh, it's getting very technical, and I still can't work out the values for A,B,C,D & the diode...
I am now contemplating on attempting the "Basic Regulator" instead...

[composition4]

Ok, I have the LM317AT, datasheet: http://www.national.com/ds/LM/LM117.pdf
It says "Guaranteed 1.5A output current". I couldn't find the 100mA rating

LM317AT is different from LM317L, note the suffix.  The L indicates TO-92 package, not TO-220.  TO-92 is much smaller, hence lower current capability.  Use the TO-220 version if you have the space.  You don't want to draw more than about 200ma anyway without a heatsink, depending on your input-output voltage difference on the regulator.

Now, I have a 12V in, which still doesn't meet that requirement  I should have about 13.5V in to the LM317.

When you rectify (convert from AC to DC) your power source, it will come out at a higher voltage.  Specifically 1.414 times the input voltage, then minus 1.4 volts.  This is because AC is measured in RMS whereas DC is peak-to-peak.  The minus 1.4 volts is because your power source passes through two diodes in the rectifier which drop the voltage.'
So technically if you have a 12V AC power source, this ends up as 12 x 1.414 - 1.4 = 15.568V after rectification.  More than enough for 12V output for the regulator.

Argh, it's getting very technical, and I still can't work out the values for A,B,C,D & the diode...

C ... I like 220 ohms. (makes the following values easier)
You want between 9V and 12V as your min/max voltages from what you said?
A ... make this 1.4k ohms.  If you can't find this value of resistor, just parallel two 2.8k resistors, or series a 1k and 390 ohm resistor.
B ... this is a pot or a trimpot.  Make it 500 ohms.  Probably just test this with a multimeter to make sure the range comes up ok as some pots can have pretty poor tolerance.

This means that minimum resistance is 1.4k series with pot turned to minimum (0 ohms) = 1.4k ohms between the ADJ pin of the regulator and GND.  This will give you 9.2V
Maximum resistance is 1.4k in series with pot turned maximum (500 ohms) = 1.9k ohms between the ADJ pin of the regulator and GND.  This will give you 12.05V
Check out the datasheet if you want to understand the calculations behind these values and behind the AMZ calculator link you gave.

D ... this is only for altering the power supply impedance, simulating a dying battery... sort of.  Sounds like you don't want this, just leave it out.  Or let me know if you do want it and we'll discuss.

You said "& the diode" ? Do you mean the diode  connecting in/out of the regulator?  This is just to prevent the output voltage being greater than the input voltage of the regulator when power is cut, due to the cap holding the output pin higher.  Output voltage > Input voltage on 3 terminal regulators = bad.


Let me know if that makes it more clear or if you have any other questions

Jonathan

[minnow]

Composition4,
Thank you for your reply!!

Yeap, I'll be using the TO-220 version, LM317AT.

Thank you for the values.
For parts A & C I think I can get around getting those values that you suggested (A = 1.4Kohms, C=220ohms)
For B, I will be using a trimpot, 500ohms as suggested
Ok, I understand the need for the diode after your explanation!
A 1N4001 diode should be able to handle the job, right?

For D, yes, I am interested to try the "dying battery" feature.
From RG's site, it should be 200ohms (as pointed out by JKowalski, I missed it the first time).
I do not think I would be able to find any pot under 500ohm value readily available.
I do however, have a 1K pot in access. I believe I may use a 250ohms resistor in parallel with the 1K pot to get the 200ohm value, right?
And the correct way of achieving it would be to take the resistor and connect it between lugs 1 and 3 of the 1K pot, right?

Ok, here's RG's schematic again, with values in it this time.
From RG's reply, I should look at the output capacitor value. I really do not get it, even after going through the Power Supply Basics page a few times
Also, please correct me if I've got the pot lugs wrong  



Thanks again ya. Really appreciate your help and your patience

[composition4]

No problems

1n4001 - 1n4007 is good

Yep the 250 ohm resistor in parallel on lugs 1 and 3 will work - but the taper will be a bit bunched up one end I think - so half way on the pot won't be half of your adjustment range for "battery deadness".

I don't really like putting that much DC current through a pot to be honest, in my experience pots don't really like a lot of DC current.  I'd probably prefer to use a 3 - 6 pole rotary switch, and switch in a few different resistors with the option of straight though with no resistor for no extra impedance on the power supply line.

Jonathan

[R.G.]

About what you said that I do not need a 12VA transformer for a single 100mA output...
Based on your design above, shouldn't I need to have a minimum of 12V into the bridge rectifier to achieve a 9V (variable via "dying battery") at the output?
Actually I found 2 different requirements, one from your site that the input should be 2V higher than the desired output. The other is from ...: "Remember that the input voltage to the LM317 must be at least 1.5v greater than the output voltage. "

My comment was about the LM317L particularly. I'd have to go refer to the datasheet for the LM317 (no "L"). It's possible they're both correct.

If you want 9.00000V out, let's say the real "overhead" number is 2.0V; you need 11V minimum to keep the regulator from falling out of regulation. AC is converted to DC by diodes by filling the filter cap to nearly the peak of the AC sine wave. For sine waves, the peak of the sine wave is 1.414 times the RMS value, and transformers are always rated in RMS voltage. So your "12V" output is really 12Vrms, and the peak is 12*1.414 = 16.98V. The diode bridges lose you two diodes' worth of voltage, about 1.2 to 1.4V from this, so you really get 15.558V peak at the filter cap. This is the voltage after the cap is just charged. The capacitor voltage runs down supplying the load current until you get the next charging peak, 1/120 second later on the next half cycle of AC power line. If you're using a 220uF cap, and supplying 100mA, the voltage sags by V = I*dt/C = 0.1A*8.6mS/220uF =3.91V, so the voltage runs down from 15.558 to 11.65V, and you have 0.65V left over the 2.0V you need to keep the regulator regulating. It would still work OK if you turned the regulator up to 9.65V.

There are some optimistic simplifications in that analysis. Probably the biggest one is that "12V" transformers are almost never just exactly 12V. Especially for smaller ones, they're wound with a higher open circuit voltage, maybe 14vac, so they can sag down to 12V out because of winding resistance at full load. So your no-load voltage on the filter cap will be higher, probably 18-20V under no load, and sag down to 11.6 under load. You can also help things by reducing the capacitor run-down. If you double the capacitor, you halve the run-down. So a 470uF cap in stead of a 220uF will run down V = 0.1*.0086/470uF = 1.83V of ripple down from the peak instead of 3.9V. I put in a 220 to keep the circuit as small as possible.

And from the muzique site, if I am right about the R1 & R2 values, with R1=240ohms & R2=1.5MOhms, output will be 9V.
Now, I have a 12V in, which still doesn't meet that requirement I should have about 13.5V in to the LM317.
Argh, it's getting very technical, and I still can't work out the values for A,B,C,D & the diode...

The LM317 family regulates voltage, but not the voltage you think. They regulate the voltage between the output pin and the adjust pin. They let through current until this voltage is nominally 1.25V. So by putting a resistor between these two pins, and then allowing that resistor to get current to ground somehow, the chip forces 1.25V across the resistor. If you put another resistor from the bottom of the "1.25V" resistor to ground, the current is drained away; the adjust pin does not eat current, just senses voltage, so you have to have some way for the current to get out of the "1.25V" resistor.

The voltage across that resistor is 1.25V, so the current through it is 1.25V/R. And since the adjust pin takes essentially none of this current, all that same current goes to ground through the bottom side "drain" resistor. Let's call the "1.25V" resistor R1 and the drain resistor R2. The current through both resistors has to be the same because the adjust pin only senses voltage, does not eat current. So the current through both resistors is 1.25V/R1, and the voltage across both of them is V = (1.25V/R1)*(R1+R2). This is algebraically the same as V = 1.25V*(1+R2/R1).

What you do to figure voltage is to pick an R1, generally 180 to 330 ohms; I tend to use 240 ohms. The output voltage is then V = 1.25V*(1+R2/240).

If R2 = 470, just to pick a number out of the air, the output voltage is V = 1.25*(1+470/240) = 1.25*(1+1.958) = 3.69V.  If R2 = 1K, the output voltage is V = (1.25*(1+1K/240))=6.458V. If R2 is 1M, the output voltage is V = 1.25*(1+1M/240) =5,209V.

ACK! 5.2kV? How is that possible? The good news is it's not. Your regulator stopped regulating when it tried to go above the input voltage minus a couple of volts. The maximum R2 value that does you any good is that value that makes the output be 9.6V (with the limits on the input voltage we first calculated, Vin minimum is 11.65V).

That value for R2 makes this equation true: Vout = 9.6V = 1.25*(1+R2/R1). If we rearrange this, 9.6V/1.25V = 1+R2/R1, and (9.6/1.25)-1 = R2/R1.

Then R2 = R1*((9.6/1.25)-1) and if R1=240, then R2 = 240*(6.68)=1603.2, about 1.6K. R2 can be varied from 0 ohms to 1603 ohms, and the output voltage will rise from 1.25V to 9.6V as R2 rises.

You can also make R2 be a fixed resistor plus a pot in series. The fixed resistor will be a minimum that the pot can't get below even if set to zero ohms, so the voltage won't go below the value set by the fixed resistor. That's what I did in the design.