Please help me calculate this drop

[ayayay!]

I can't believe I don't know this, but I'm suffering from a huge brain fart here.

I want to use a typical buff-n-blend circuit for a pedal.  Something like Beavis's.  That's not the issue.  What I need is the circuit to-be-blended from a toy that requires only 3.2 volts.  How do I drop my 9.6v supply to proper voltage & current?  Please show me the math so I can slap my forehead.  

My thought was a simple voltage divider, with a current limiting resistor in series.  Do I even need that current limiter?

And while I'm at it, I need a simple stomp switch control:  SPST, latching, momentary, normally closed.  I've found one on Mouser, but not Smallbear, Mammoth, etc.  

My

[ashcat_lt]

3.2 = 9.6 / 3 which is pretty convenient.  The ratio for the divider will be 2:1 in favor of the "top" series resistor.  That resistor will work to limit current all on its own if valued correctly.

[ayayay!]

Actually I don't think I'll need a current limiting resistor.  A 10k/4.7divider should do.  Sorry for thinking out loud.  Sometime you (I) just gotta put it out there to get past the mental block.  I was hung up on that current limiter.

Anyone got any suggestions for the switch?  

Edit:  late post.  You're right ashcat, 2:1 would be better.  Thanks! 10k/3.3k it is!

[Mike Burgundy]

The rough version is 3.2V is 1/3 of 9.6 (convenient) so the voltage devider  looks like:
9.6V
|
R1
|
------Vref = 3.2V
|
R2
|
ground

and R1=2R2.
If the circuit draws no current that'll be spot on. You can choose R1,2 for low consumption (remember theyre always shunting some power to ground) and theoretically use 2M and 1M.
However, real circuits draw current, so that will pull the Vref down. Decreasing both resistors lowers the circuit's influence on the average level, but increases overall current draw on your power supply. Increasing R2 a bit could compensate for the circuits impedance, so you get exactly the right voltage under load - this is where (R2||circuit load impedance)=0.5R1. Still here?
Now most (audio) circuits don't draw a constant current - it fluctuates and changes Vref. Constantly. Higher values for R1 and 2 decrease this influence.
R1 already acts as a current limiter.
The exact values of R1 and R2 depend on how much current you want to draw from your power supply, how much current the extra circuit needs and how much variation in Vref is acceptable.
You'll need to know current draw.
Or just experiment  - see if 100k and 200k works. I'd start there. Then tweak.


edit...late post too. Thats what I get for being long-winded ;P

[ashcat_lt]

Edit:  late post.  You're right ashcat, 2:1 would be better.  Thanks! 10k/3.3k it is!

Where do you live that 10 = 2 * 3.3 ?  I think I may want to do some business there.

[ayayay!]

R1 already acts as a current limiter

That's what I figured out after I posted.  Derp!

Edit:  late post.  You're right ashcat, 2:1 would be better.  Thanks! 10k/3.3k it is!

Where do you live that 10 = 2 * 3.3 ?  I think I may want to do some business there.

See my tag line.   

[Johan]

9.6v-----collector..........emitter---~3volt
         l              base
         l                    l
   4k7 resistor        l
         l__________l
                             l
                    3.5volt "white" LED
                             l
                         ground                 

[ayayay!]

9.6v-----collector..........emitter---~3volt
         l              base
         l                    l
   4k7 resistor        l
         l__________l
                             l
                    3.5volt "white" LED
                             l
                         ground                 

Neat little trick.  Could use that LED for an always on perimeter light, or other...

Anyone got a switch recommendation? 

[blueduck577]

You can use an op-amp follower after the Vref and it'll supply all the current you need without drawing current out the divider :

[R.G.]

The real question you need to be asking is how much current the 3.2V toy needs.

This is yet another instance where the thinking in my article on picking resistors for bias networks applies. A voltage divider string made from resistors needs to pass about 10 times as much current through the divider resistors as is drawn from the junction to have any hope of keeping the voltage within bounds. If that 3.2V toy is intended for battery operation and pulls half an amp, a resistive divider is out of the question. If it's a CMOS noisemaker, you may be able to use 1M and 330K.

A voltage divider looks to its load like a voltage source equal to the unloaded voltage at the junction of the resistors, but in series with a resistance equal to the parallel combination of the two resistors.

Case in point: the proposed 10K/3.3K divider driven from 9V gives 9V * (3.3K/(10K +3.3K)) = 2.23V unloaded, but in series with R = (10K * 3.3K)/(10K +3.3K) = 2.48K.

This is the Thevenin equivalent circuit for the voltage divider. A load of 22K attached to the junction pulls I = 2.23V/ (2.48K +22K) = 91uA and the voltage across the 22K is 2.003V. A 2.2K resistor attached to the junction pulls 2.23/(2.48K +2.2K) = 476uA and the voltage across the 2.2K is 476uA * 2200 = 1.048V.

Active devices like the emitter follower can do better, because the active device needs so little current from the divider that runs its base. The opamp will do even better - until it runs into its own survival-current-limit at about 20ma.

[ayayay!]

The real question you need to be asking is how much current the 3.2V toy needs.

Didn't I?  

How do I drop my 9.6v supply to proper voltage & current?

No seriously though R.G., thanks for the detailed answer.  That's what I was asking for in my first post:  The math.  

Active devices like the emitter follower can do better, because the active device needs so little current from the divider that runs its base. The opamp will do even better - until it runs into its own survival-current-limit at about 20ma.

That is exactly what I was looking for.  I figured it needed very, very little current, but wasn't sure how to proceed.  Thank you!  I'm recording all weekend, but hope to have this cobbled together next week and report back.  

[PRR]

The real question you need to be asking is how much current the 3.2V toy needs.

No. You mention current but did not measure it, and did not ask how to measure it, only: 
a toy that requires only 3.2 volts.  How do I drop my 9.6v supply to proper voltage & current?

And that is THE KEY to this problem. (Also how much the current varies.) (and how much voltage tolerance it has....) (And which polarity will interface best.)

You got the toy. You got it on a battery, right? Break a battery lead and insert a resistor. You want the resistor drop to be "small" compared to battery voltage. So the problem is circular: you need an estimate of current to measure the current.

So think. Does the toy eat 100 Amperes? No, that would be 320 Watts of heat, and would burn the kid except a toy battery won't supply 100 Amperes.

Could it be 1 microAmp? Watches run on that, but circuits which "do work" (lights, speakers, even line outputs) usually eat more.

I'd assume 1mA. A "small" drop on a 3.2V supply might be 0.3V. 0.3V/0.001A= 300 ohms. Stick in a 330 ohm resistor and measure the voltage drop. If way-large, use a smaller resistor like 100 or 47 ohms. If way small so you can hardly read the drop, go larger.

Work the toy through all modes and volumes, standby to max-annoy. See if the current changes.

If it pulls say 2mA all the time, and you want to drop 6V (9.6V-3.2V= 6V), then 6V/0.002A= 3K resistor.

However if current ever goes to zero (standby), this won't work, the standed-by toy would get the full 9V. Since several common chip processes have <7V breakdowns, and breakdown is not always benign even if limited, it might be ruined.

If your 9V is ample (has lots of excess capacity), a resistor and Zener is simple. Figure 0-2mA for toy. Allow decent current in Zener, maybe >2mA (depends what you use for "zener"). Your resistor is then 6V/0.004A or 1.5K. The zener gets 2mA normal 4mA when toy stands-by, you should check Zener dissipation (this is one place Zener supplies give trouble).

3.3V Zeners are available. Five Si diodes in series will be very close to 3.2V and may be laying around your bench. The voltage can't be too exact because the battery toy must allow for battery fade.

Zener worst-case dissipation might be 9.9V supply, 1350 ohms ("1.5K 10%), 3.5V in Zener. 9.9V-3.5V= 6.4V dropped, 6.4V/1350= nearly 5mA, 5mA*6.4V= 30 milliWatts. You can't hardly buy less than 250mA per Zener, OK. With 5 Si diodes it is 6mA each, diodes often don't have dissipation ratings but are clearly good for 50mA-100mA forward current, 5mA is OK.