Need help with a window comparator for a fuzz face.

[jimmybjj]

First let me preface with i do not have much electronics knowledge just i have learned from these forums. i am trying to implement a LM393 to light a LED when the trim has reach a predetermined voltage (like in this thread http://www.diystompboxes.com/smfforum/index.php?topic=86008.0 ) Seemed simple enough. I made a layout before i breadboarded it, mistake. Well i put it together and it doesn't work. The circuit is a PNP fuzz face and bias voltage is somewhere around -4.5v. On the LM393 datesheet it states the voltage range as 2.0-36.0v.This is one of those questions you already know the answer to but need confirmation Will the PNP circuit work with the LM393? I hope i am wrong but i am guessing in this case Vin is -4.5 so it will never be in the positive voltage window create by the comparator. Other informatioin you may or may not need. The fuzz if powered by a charge pump converter for -9 and +9 for the lm393.

If i am right and this will not work in the standard way is there a work around?

[jimmybjj]

just got to thinking the datasheet says:

Voltage range: 2.0V to 36V
Single or dual supplies: ±1.0V to ±18V

does that mean it works upto +18v and -18v?

if that is the case, can I just power the LM393 circuit with the fuzz supply and be good to go?

[earthtonesaudio]

When datasheets say the supply is + and - something, they mean above (+) and below (-) ground.  When they say "voltage range" is 2-36V, that means the magnitude of the voltage across the device's power pins.

So for a fuzz operating from supply voltage (GND, -9V), you can take this circuit and replace all the +9V points with GND, and all the GND points with -9V.  Well, except for C1.  It should probably be returned to actual ground rather than -9V.

Wired this way, all the comparator cares about is the voltage at its inputs RELATIVE to the voltage supply.  So when your fuzz is biased to -4.5V, it sees a voltage that is 4.5V higher than its most negative terminal and lights the LED.

[jimmybjj]

When datasheets say the supply is + and - something, they mean above (+) and below (-) ground.  When they say "voltage range" is 2-36V, that means the magnitude of the voltage across the device's power pins.

So for a fuzz operating from supply voltage (GND, -9V), you can take this circuit and replace all the +9V points with GND, and all the GND points with -9V.  Well, except for C1.  It should probably be returned to actual ground rather than -9V.

Wired this way, all the comparator cares about is the voltage at its inputs RELATIVE to the voltage supply.  So when your fuzz is biased to -4.5V, it sees a voltage that is 4.5V higher than its most negative terminal and lights the LED.

First, thanks for taking the time to write that explanation. So the supply can be anywhere between -18v to +18v? When you say magnitude across the power pins, does that mean operational voltage?

I will try and update my circuit when I get home and make a schematic to see if everything looks ok. One thing I am unclear on is when you have a positive ground circuit aren't all the electro's (c1) reversed.

[earthtonesaudio]

The maximum +/- voltage specification may be confusing in this instance.  Think of it in terms of magnitudes.  You can have the + pin voltage higher than the - pin by up to 36V.  This 36V maximum can be between 0V and +36V or between -18 and +18, or any arbitrary pair of voltages.  If the - pin is at +10,000V and the + pin is at +10,036V the difference between the two pins is still just 36V so you're fine.


Also check this out (you may have to resize Javascript window).
A. Typical representation of single 9V positive supply (sometimes called "negative ground")
B.  Another way of drawing A.
C.  Single supply, -9V (often called "positive ground")
D.  Dual or Bipolar supply.  Plus 4.5V on top, -4.5V on bottom.  0V in the middle.
E.  Dual supply, but with +6V on one side and -3V on the other.  1.5V is in the middle.

Note that each resistor has exactly 9V across it, or stated another way, the difference between the two ends is 9V.  With respect to the lower terminal, the voltage across each resistor is +9V.

[jimmybjj]

The maximum +/- voltage specification may be confusing in this instance.  Think of it in terms of magnitudes.  You can have the + pin voltage higher than the - pin by up to 36V.  This 36V maximum can be between 0V and +36V or between -18 and +18, or any arbitrary pair of voltages.  If the - pin is at +10,000V and the + pin is at +10,036V the difference between the two pins is still just 36V so you're fine.


Also check this out (you may have to resize Javascript window).
A. Typical representation of single 9V positive supply (sometimes called "negative ground")
B.  Another way of drawing A.
C.  Single supply, -9V (often called "positive ground")
D.  Dual or Bipolar supply.  Plus 4.5V on top, -4.5V on bottom.  0V in the middle.
E.  Dual supply, but with +6V on one side and -3V on the other.  1.5V is in the middle.

Note that each resistor has exactly 9V across it, or stated another way, the difference between the two ends is 9V.  With respect to the lower terminal, the voltage across each resistor is +9V.

Thanks so much that clears things up alot more. How does this schematic look? I think that C1(C7 in my schematic) should be turned around but other than that i think its correct.

[earthtonesaudio]

Yep, turn C7 around.  Otherwise looks good!