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28th April 2017, 01:35 AM  #1 
diyAudio Member

Diode voltage drop question
OK, loosing my mind, I think over a bridge rectifier.
I'm using a bridge rectifier to convert AC to DC with a 200uF cap post rectifier and I have a 100K load resistor. As I do my math calculation, at low voltage (under 10 Volts AC), my diode drop is about .56 volts for each leg of the bridge (that is about what the diodes measure when I check them with my meters). However, when I increase my AC voltage, the math and measured values do not match any longer. So at 50 Volts AC, it looks like the voltage drop from the diodes is closer to 1 volt for each leg. As I get closer to 70 Volts, the drop appears to be 1.2 volts for each leg. Is this correct? I'm not sure why this is happening. Ultimately, I'm building a measurement circuit to measure AC volts via DC voltage. I'm not sure why this is not a linear drop across the diodes. Thanks 
28th April 2017, 01:43 AM  #2 
diyAudio Member
Join Date: Apr 2017

As the voltage increases, the current at the 100k load resistor also increases. The voltage drop of the diode also depends on the current flowing through it. Just check the diodes datasheet and look for the Id vs VF graph to have better understanding. cheers!

28th April 2017, 02:33 AM  #3 
diyAudio Member
Join Date: Apr 2011
Location: Upper midwest

For high diode currents it does more linearize, and there's an ohmic component.
At low currents, the relationship is more log. https://en.wikipedia.org/wiki/Shockley_diode_equation You might want to look into a "precision rectifier" circuit that will eliminate this problem. http://sound.whsites.net/appnotes/an001.htm Last edited by rayma; 28th April 2017 at 02:45 AM. 
28th April 2017, 05:04 AM  #4 
diyAudio Member
Join Date: Jun 2003
Location: Maine USA

> about .56 volts ..(that is about what the diodes measure when I check them with my meters).
At What Current?? Vf will rise around 60mV for every 10X of current. (This is what rayma is saying.) > 10VAC 0.56V > 50VAC 1.0V > 70VAC 1.2V If "voltage drop" were truly constant, it would be 0.56V for any current zero to infinity. If drop were resistive, then we might expect 2.8V at 50V and 3.92V at 70V. The truth lies *between* constant voltage and steady resistance. Shockley would predict, if 0.56V at 10V drive, then 0.61V at 50V and 0.66V at 70V drive. Your numbers do not agree, which is odd. FWIW, rectifier diodes diverge from strict Shockley's Law, being doped for low drop (or perhaps same drop on smaller die). There is a kink at low currents, and you may be crossing this zone. Still your numbers seem odd. Also recall that drop will reduce around 2mV per degree C. But anyway: *assume* there is severaltenths uncertainty in a naked diode's drop. For 1% accuracy, your signal must be 100X this uncertainty. If as bad as 0.6V (unsorted diodes in wide temperatures) then you want 60V signal. You are in this range. If lowlow level precision is needed, fudge it. You can mark the scale of a needlemeter, or add a lookup table to a digital system, to make the answer right despite the nonlinearity. Or as rayma also says, any greybeard analog designer can sketch 13 precision rectifier schemes without thinking. (There is one in your DMM.) Now the 0.5V1.2V drop of the diode is divided by amplifier gain. For 60Hz it is trivial to have 0.1% accuracy on 1V signal. For 20KHz you want a fast opamp for 1% accuracy around 1V. (Signals known to be well under 1V should be preamped.) 
28th April 2017, 12:10 PM  #5 
diyAudio Member

Thanks everyone for your feedback. I have homework to do.

28th April 2017, 04:07 PM  #6 
diyAudio Member
Join Date: Apr 2011
Location: Upper midwest


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