I've gotta admit that I've not really thought about using diodes to drop voltage. You always instinctively reach for a couple of resistors. Thanks Bill for your informative videos. I'm 53 and have come to the subject a little late in life but I've always had an interest in electronics.
Yes, very interesting. Thank you. I will need to think on it, to sort it out. Seems like the diodes are "demanding" their typical voltage drop, no matter the arrangement. In the second arrangement - resistor then diodes... Your last reading was 0.68V. This is the typical v drop for one diode... So, I am wondering if you made the circuit as a single resistor to a single diode, would not the voltage still be 0.68 before the diode, regardless of the voltage supplied? I think what would change is the v drop on the resistor. It would have to drop the additional 1.4V that the two removed diodes were dropping.
I've gotta admit that I've not really thought about using diodes to drop voltage. You always instinctively reach for a couple of resistors. Thanks Bill for your informative videos. I'm 53 and have come to the subject a little late in life but I've always had an interest in electronics.
Diode voltage drop dissipates excatly as much heat as would a resistor of same voltage drop. There is no mystery in that, just ohms law.
Thank you for another great informative and useful video.
How can it use less power than resistors? Power =VI and that's not gonna change whether resistors of diodes are used.
Seems like I may have seen this method used to create a negative voltage rail for op amps in a very basic variable linear power supply circuit?
Yes, very interesting. Thank you. I will need to think on it, to sort it out. Seems like the diodes are "demanding" their typical voltage drop, no matter the arrangement. In the second arrangement - resistor then diodes... Your last reading was 0.68V. This is the typical v drop for one diode... So, I am wondering if you made the circuit as a single resistor to a single diode, would not the voltage still be 0.68 before the diode, regardless of the voltage supplied? I think what would change is the v drop on the resistor. It would have to drop the additional 1.4V that the two removed diodes were dropping.
Hmm... What about the current?
The ”division” ratio is not constant, as it is in a proper divider.
Voltage ”clamp” is a better name.
See my comment, above. Yes, "clamping" does seem to be a good descriptive term. Clamping is what my thinking would lead to, I suspect.