Voltage dividers with resistors
Jul 1, 2001 at 5:05 PM Thread Starter Post #1 of 17

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Does it matter what range of resistor values you use for a voltage divider, whether it's in the tens, hundreds, or thousands of ohms range?

I've run into this problem twice, once a few months ago and once just last week. If I used a certain range of resistor values for a voltage divider, the voltage at the output would not be what is expected. However, if I changed this range, the voltage would be what is expected from calculations.

How does this work? My circuits course never covered low level things like this, only things as high level as Vout = Vin * (R-from-out-to-ground / (R-from-out-to-ground + R-from-in-to-out)).
 
Jul 1, 2001 at 5:18 PM Post #2 of 17
The ratio of resistor values determine voltage but the absolute values determine the current through them. In some cases you want to draw as little current as possible. So you put large values. However, large values of resistors generate more noise. So, say, you have a resistor divider and have the middle point going to the input of an opamp. Usually you would prefer not to use resistors in 100's of kiloohms to avoid noise. Also, if the current is too small, then the offset current of the opamp may screw up the balance, for example if the input of the opamp is BJT technology you are likely to have to pay attention.

In some other cases you may drive a Zener diode or a LED or something and need to have larger currents through resistors. So the purpose of the divider will determine the values of resistors...
 
Jul 1, 2001 at 5:25 PM Post #3 of 17
What if I wasn't getting expected values with nothing attached to the output except my volt meter?

The purpose of the voltage divider in my last application was to provide the 3-5V to the Vcc pin of my DS1802 digital pot. I don't think it needs any specific current, or that could just be my stupidity and ignorance together (I should check the datasheet again).

I remember, in my previous application, needing to adjust the resistors to drive a LED, as you mentioned.
 
Jul 3, 2001 at 6:51 AM Post #4 of 17
Hmm, I don't know what kind of a digital chip is that pot but unless it explicitly allows unregulated supply voltage, I would use a voltage regulator. It's a very specific chip though so it might not. But in any case, chip might need substantial current and the current would likely change as the volume and/or signal level goes up and down. Whatever the case, you'll need datasheet. That, or ask apheared (I assume you're using chip he suggested, no?).
 
Jul 3, 2001 at 7:02 AM Post #5 of 17
Yeah, this is the same chip Apheared mentioned using in a thread started in the summer of 2000.

So far I've tried several voltage divider combinations and have flamed up many resistors (oohhh! aahhh! light show!). Is your voltage regulator suggestion to use a voltage regulator to provide the 5V (from my 17V regulated power supply) instead of using a voltage divider? If so, just say the word and I'll have a LM317 in there for testing when I get home tomorrow.
 
Jul 3, 2001 at 10:51 AM Post #6 of 17
edit: Hey! Here ya go: Check out Bill Bowden's page... scroll down to the JavaScript Calculators section for a resistive divider calculator. He's got some other cool stuff too.
biggrin.gif


http://ourworld.compuserve.com/homepages/Bill_Bowden/

edit2: removed explanation, go play with the calculator instead. heh.
 
Jul 3, 2001 at 2:02 PM Post #7 of 17
Awesome! This is a page to bookmark
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Looks like I've been trying to dissipate way too much power in R1 through my 1/4 watt resistor, causing them to burn up. So much for the idea of paralleling them. Maybe I'll use my extra power resistor(s).

Apheared, now about that DS1802... what did you supply yours so you wouldn't get clipping in any situation?
 
Jul 3, 2001 at 5:05 PM Post #8 of 17
Hehe, have you been using single or double digit resistors for divider
smily_headphones1.gif
? Thank goodness for flame-retardant coating, eh?
 
Jul 3, 2001 at 5:26 PM Post #9 of 17
I actually tried both, in an attempt to cure the sound signal clipping. It seemed that the lower I went, the less the sound would clip. So I've tried resistor combinations (for a 17V power supply) of 3.3 and 10 ohms, 5 and 16.5, 10 and 33, 22 and 68, 100 and 330, 330 and 1k, 15k and 47k, and maybe a few more. All produced clipping (well the lowest few burned up before I could actually test them
tongue.gif
) Then I decided to try trimmer pots to just twist until I heard a sound difference (and just ignore trying to get the right divider for the IC's Vcc). High resistances were ok, but had a more obvious clipping. Lowering the resistance started a glow of the familiar burning, so a quick turn of the trimmer stopped that before the plastic parts of the trimmer melted.
 
Jul 3, 2001 at 5:35 PM Post #10 of 17
I think you should try to use the voltage reference then. A cheap 7805 would do just fine and doesn't need any extra parts except maybe a capacitor or two. LM317 requires more parts but it regulates better (I think) and voltage is of course adjustable.
 
Jul 3, 2001 at 10:13 PM Post #11 of 17
Argh, still have clipping, with the LM317T regulating to 4.97V.

(2 minutes later)

Hmm, I may now have just busted this chip by moving the voltage way above 5.5V now. Doh. I'll have to wire up another one. Anyways, what other suggestions for the clipping?
 
Jul 3, 2001 at 10:37 PM Post #12 of 17
That sucks
frown.gif
. Too bad it didn't help. Maybe you should contact apheared?

By the way, it is possible that the digital pot has limit on the input voltage it can process. What are you feeding the pot with, line out of a CD? What's the name of the chip?
 
Jul 3, 2001 at 10:45 PM Post #13 of 17
This is the Dallas Semiconductor DS1802 chip (www.dallassemiconductor.com). Feeding it anything from a computer soundcard's headphone out jack (Turtle Beach Santa Cruz), or a portable CDP's headphone out (clips at all volumes on the CDP, given the right bassy sound).
 
Jul 3, 2001 at 11:59 PM Post #14 of 17
Hm, according to app notes, the max ratings on the chip are 0.5V above Vcc and 0.5V below ground which would be unusable for audio signals without DC component. Application notes on the other hand have plots where the input voltage is specified up to 2Vrms and claim specifications up to 80kHz for 1Vrms, which makes sense.

If there is clipping then probably your input level is too high. Headphone outs are amplified. Line outs are usually specified 1-2Vrms and should work fine. Otherwise try to reduce volume on the headphone output (you are using this pot as input to your own headphone amp, right?).

By the way maximum supply current is only 2mA, and usually much less than that! So resistors should do the job (even though I prefer using regulators, since 7805 is so simple and you can get them in a small TO-92 case too).
 
Jul 4, 2001 at 12:42 AM Post #15 of 17
Yep. It does audio fine. I have only used 2 of them. I used both around 5V via a voltage divider off my Apheared-ish battery packs that were powering the amp.

Still, even if it is only .5 from vcc and ground, you're powering it at 5V! No headphone out is gonna put out 6V from a portable. (although I'll try to overdrive the inputs and measure where it gets funky)

It was really easy to use, actually.

If you're thinking it's no good for audio, heh you might enjoy the Audio Characterization Report on it, App Note 88 from Dallas (and maxim, whoohoo, didn't even rename it!)

http://pdfserv.maxim-ic.com/arpdf/AppNotes/app88.pdf
 

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