Voltage divider resistor values question.

[Buzz]

Hi guys.

Vout = Vin . R2 / R1+R2 . I'm making a divider to drop my nice free 20v wall wart to 18 volts. Giving R1 a nominal value of 1 gives me a neat R2 value of 9.

So I have a ratio, R1:R2 = 1:9.

Is there any difference in how I apply that ratio? To take my point to an extreme, would a 10R-90R combo work the same as a 1M-9M ?

Apart from picking handier values, is there any other rule of thumb to be applied here?

Edit: For no reason apart from it feeling right, I'm thinking in the ballpark of 33K:300K. What else is at play here, heat dissipation?

This relates to my finding of an old AD 80 analog delay, which IIRC hungrily chewed through two nine volt batteries at a pace to be admired. I don't know what the power consumption is, but my 20v adaptor puts out 1.5 amps, which is "plenty big".

[bluebunny]

Assuming the 20V is regulated, would it hurt the AD80?  (Or rig up a 7818?  Datasheet shows a typical 2V dropout, so it might be a little close!)

[Kipper4]

If It's using 2x 9v battery. isnt it a bipolar supply. wouldnt it be better with a charge pump?

[samhay]

> I'm making a divider to drop my nice free 20v wall wart to 18 volts

Why?

Assuming this is a good idea, a voltage regular is probably your best choice, if you can get it to work with a 2V drop-out (3V is better).

Another good option is to place a series resitor between your power suply and circuit V+. Rember V = I*R, so if you circuit draws 20 mA you will drop 2V across a 100R resistor. This is a good option if your circuit draws a fairly constant ammount of current and should give you a good idea why a voltage divider almost certainly won't work.

The other option to consider involves an 18V Zener diode.

[Buzz]

Cheers guys.

Why? I just want to see if the board works. I dont have an 18v wall wart, but I have a 20v one. I assume it's regulated, it's a big box I've kept from an old computer peripheral. I guess I should look at the spec sheet or open it up to tell. I don't want to spend money on it until I know it works. If the MN 3005 or any other hard to source / expensive part is cactus you'll probably see it in the trade thread. Although it would be a nice pedal to have, I wouldn't really have a use for it

The original battery wire connectors are gone, and the way this thing chews batteries I don't care to replace them.

I have plenty of resistors on hand and this seemed at first glance to be the easiest/ cheapest way to test the board. I do have a charge pump on hand, but not a 7818. I might have an 18v Zener, I'll have a look in my diode stash.

Why would a voltage divider be a bad idea?

[antonis]

Apart from picking handier values, is there any other rule of thumb to be applied here?

A rule of thumb is also that the current flowing through R2 should be - at least - 10 times greater than the current flowing through the load...
(so you may "ignore" the load loss and calculate divider resistor values with same current flowing through R1 and R2, resulting in Vcc*R2/R1+R2 formula..)

If the load resistor is less than 10 times greater than R2 you have to take it in mind and calculate the voltage using KCL & KVL (R1 in series with R2//Rload)..

[anotherjim]

Crazy as it may sound, I'd put x3 (or 4 to be cautious) 1Amp diodes in series with the positive feed to drop almost 2 volts.

This found 20v supply could be lower under load. Or, it could go higher, if it's poorly regulated. If there's a lot of ripple, your meter reads an average voltage, capacitors on the AD80 power rails will turn that average to peak which will be higher. Your meter won't tell you how well regulated it is on it's DC range. If you measure on AC, it will show you the ripple voltage.

[Buzz]

Schematic:

http://www.studio250.fr/docs/ibanez/DELAY_IBANEZ-D80.jpg

That's it or a very similar version. I don't know if it requires a bipolar supply, the power input plug is a simple mini headphone jack, mono, 18v.

The supply I intended to use is a logitech scp60

[Buzz]

I looked inside the 20v adaptor. It has a hefty AC/AC transformer followed by a bridge rectifier.

Maybe I'm barking up the wrong tree.

I have two 9v transformers here I could double up...

Or maybe I should just scrounge two 9v batteries out of other devices?

I only really need this for the testing stage. If it works I'll find a more suitable power source.

[Buzz]

Crazy as it may sound, I'd put x3 (or 4 to be cautious) 1Amp diodes in series with the positive feed to drop almost 2 volts.

Would 1n400X Diodes fit the bill for this method?

The data sheet says: Average Rectified Output Current  1.0A for these diodes. Is that the qualification for a 1Amp diode?

[Buzz]

OK, I really have been over-thinking this.

Looking over the schematic the two batteries are run in series. The single red + lead for the battery connection is still there.

Two battery snaps and some scrounged 9vs is all I need for board testing purposes. I have those ( and the required jacks and momentary switch )

Thanks for the help guys, If it runs I'll buy a suitable adaptor.

[ashcat_lt]

The answer to the original question is that there are never actually only two resistors in a divider that you're hacking into a circuit.  There is always (at least) some impedance from the source in series with R1, and there is always an impedance from the load in parallel with R2.  You need your divider to be big enough that the source doesn't matter and small enough that the load doesn't matter or else you have to take them into account in your calculations. 

In power supply stuff, you usually do need to include them.  And then when you end up with very small resistors, you have to do your power calculations and specify them hefty enough to handle it. 

Also, I'm pretty sure that 1n400x is pretty much the go to for this sort of thing.

[samhay]

The AD-80 has a 12V voltage regulator - look bottom left of the schematic. I would imagine it will run fine from your 20V supply as long as you get the polarity right. You might want to add a series protection diode while you have it on your bench.

[anotherjim]

Yes, 1N400x series. Any of them would do - but....

Just a transformer and rectifier means it isn't regulated at all. So the voltage you read with a meter is the average. It will be about 1.4x more than that when it's smoothed by the circuit capacitors - over 25volt! You probably also will have to add a large capacitor to smooth the ripple as the designed capacitance in the circuit will be intended for a smoother supply and won't be large enough. It may well hum if you don't.

Anyway, probably too much voltage to loose than a diode string can sensibly do.

I wouldn't use a divider for this, I'd go with a proper regulator and 18volt has a standard 3 terminal type...

You need the caps and chip, you already have the bridge rectifier.

Don't forget what I said before about testing a DC supply on the meter AC range. A large (volts) AC reading on a DC supply tells you that there's no regulation. A regulated (or at least capacitor smoothed) supply should only read a few millivolts on AC.
Just for laughs, fit a cap across the output (at least 50v rating) and see how the DC voltage goes up compared to without.

Anyway, I mention the above for completeness, since you may just buy an adapter instead.

[Buzz]

All great information.

This is obviously an area where my knowledge is severely lacking.

I'll have a play with the adaptor on the DMM for the sake of experimentation, and to help absorb the concept.

If I were to walk down the path of voltage reduction and regulation via a setup using a 7818, would I be correct to assume a 24v transformer of the same ilk ( which I also have ) would be a better choice? That would give me a drop of ~6v rather than ~2v. More headroom to give me a clean 18v from the 7818?

I have previously made a homebrewed 9v power supply. I used a power adapter that read 9v on the packaging and tested at 12v on the DMM.

After rigging it up with 7809 based circuit ( I think I took it from the beavis page ) it gave me a steady supply voltage... of 8.99v. It was useless for the purpose I built it for, which was RG's Ge transistor tester. Since I had space in the box I had also fitted a potentiometer in a voltage divider with a separate output to use as a "dying battery simulator" ( which I now believe is more correctly referred to as a voltage sagger? ) Good learning experiment, but I have yet to come across a pedal which sounds better being under-powered. Useless for my fuzz faces, I always leave trimpots installed and have pretty much tuned them that way to my liking before being boxed.

[PRR]

Use a voltage-divider for VERY SMALL loads.

Almost never the Main Power for "a box".

Specially a box accused of "chewing batteries".

> would a 10R-90R combo work the same as a 1M-9M ?

As Ashcat says, there's always another resistor (the load).

If this is a 1Meg resistor to an input, you might pick voltage-divider values under 100K and ignore the load. Millions of "Vref" networks do.

However I would think a "chews batteries" pedal eats 50mA or 0.050 Amps. At 18V this is 18V/0.050A= 360 Ohms equivalent load.

Take your 1M-9M net and hang 360 Ohms across the 9Meg. Is it still a 0.9 ratio? No, more like 0.00036, or way-way under a Volt out (0.007V).

Take your 10R-90R combo. Hang 360r across the 90r. That's not very different: 90r||360r= 72r, 10+72 is a 0.88 ratio or 17.56V out. BUT to get your 0.050A pedal powered you are pulling 20V/82r or 0.244 Amps out of the power source. Nearly 4/5th of your power is wasted in the divider resistors. The 90r resistor must dissipate *3.43 Watts* of heat.

If you have the quarter Amp, if you have 5 Watt 100r resistors, then it "will work". Running hot, wasting power, and for long-life better make it a 10W resistor.

And as you now know, this pedal really runs on all-12V internally, the regulator will probably stand 20V input (but did you check that "20V" wart UN-loaded?), or shut-down safely; and batteries are still the safest way to check the pedal.
____________________________________

> a steady supply voltage... of 8.99v. It was useless for the purpose I built it for, which was RG's Ge transistor tester.

? ? ? How do you know it was 8.99V? Affordable DMMs aren't that precise. And surely it will give the same results with 8.99V as it would with 9.00V? Ge technology isn't like a rocket to Mars. Anyway IIRC R.G. was probably thinking "9V batt", and a fresh 9V batt makes 9.36V.

And about 2 or 3 Ohms in the "gnd" lead of the 7809 wudda put you as close to 9.00V (indicated on your DMM) as you could ever want.

[R.G.]

And about 2 or 3 Ohms in the "gnd" lead of the 7809 wudda put you as close to 9.00V (indicated on your DMM) as you could ever want.

And then we'd have to drift off into the sins and sadnesses of meters. Ugh.

[Buzz]

Use a voltage-divider for VERY SMALL loads.



> a steady supply voltage... of 8.99v. It was useless for the purpose I built it for, which was RG's Ge transistor tester.

? ? ? How do you know it was 8.99V? Affordable DMMs aren't that precise. And surely it will give the same results with 8.99V as it would with 9.00V? Ge technology isn't like a rocket to Mars. Anyway IIRC R.G. was probably thinking "9V batt", and a fresh 9V batt makes 9.36V.

Well I should admit it wasn't useless. It ball parked me enough to get suitable Q1s and Q2s to make some nice fuzz faces. And it was the only method I had at hand to take leakage into account. Overthinking and OCD... I went through quite a lot of resistors finding ones with the exact value. Which brings us to...

And about 2 or 3 Ohms in the "gnd" lead of the 7809 wudda put you as close to 9.00V (indicated on your DMM) as you could ever want.

And then we'd have to drift off into the sins and sadnesses of meters. Ugh.

...the multimeter debate. Sorry about opening that old can of worms

[Buzz]

Use a voltage-divider for VERY SMALL loads.

Well now I know this and move forward.

By the way, I just rigged up the AD-80 board and it works!

Sweet. Really wish I hadn't junked the enclosure back in the day. I checked out how much these units go for on ebay 

[Rob Strand]

Fresh batteries are 9.5V.
So 2x9.5 = 19V.

Honestly just put a 1 or 2 diodes in series and be done with it.

You could also use a single series resistor but you have to tune the value.
Vdrop = = I * R => R = Vdrop / I = 1V / I;  if I=50mA then use R=22R
There's nothing wrong with this Boss did it for years.