4 ms·
> Next, there's a massive volume difference between the hot and warm modes. If I read the text and schematics correctly, that's because one uses 0.7V silicon d
by ableal 11y ago
> Next, there's a massive volume difference between the hot and warm modes.
If I read the text and schematics correctly, that's because one uses 0.7V silicon diodes and the other 1.7V LEDs in the opamp feedback loop. (At least red LEDs used to be 1.7V when I was messing with them.)
So that should clip the voltage in that branch to +/- 0.7 or +/- 1.7 volts, hence the power difference if nothing else is moderating it.
- cushychicken 11y agoThe turn on levels I measured were much smaller in discrepancy than that; the silicon diodes were clipping at 0.55 V, while the LEDs were only slightly higher, at about 0.8 V. Made me wonder if I was measuring them incorrectly.
- ableal 11y agoThat sounds OK, if a bit low. Perhaps the resistors are large, or you might be diverting some current through a low(ish) impedance voltmeter, but those should be rare nowadays. The trick with diodes is that the curve is exponential - so for a "0.7V" diode, and pulling numbers out of the air, you can get 0.3V for nanoamp currents, or 1.0V for power-station currents (for the brief moment before it melts ;-) The real equation is very simple, if anyone wants to play. For instance, it's well explained here, midway down the page: http://www.allaboutcircuits.com/textbook/semiconductors/chpt-3/introduction-to-diodes-and-rectifiers/ http://www.allaboutcircuits.com/textbook/semiconductors/chpt...
- cushychicken 11y agoThat's a good point; the DMM I used could have just been extrapolating "turnon" to a certain current. I'll have to check the DMM's specs.