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Old 28th April 2017, 01:35 AM   #1
skidave is offline skidave  United States
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Default 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.

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Old 28th April 2017, 01:43 AM   #2
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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!
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Old 28th April 2017, 02:33 AM   #3
rayma is online now rayma  United States
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Quote:
Originally Posted by skidave View Post
I'm not sure why this is not a linear drop across the diodes.
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.
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Old 28th April 2017, 05:04 AM   #4
PRR is offline PRR  United States
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> 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 several-tenths 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 low-low level precision is needed, fudge it. You can mark the scale of a needle-meter, or add a look-up table to a digital system, to make the answer right despite the nonlinearity.

Or as rayma also says, any grey-beard analog designer can sketch 13 precision rectifier schemes without thinking. (There is one in your DMM.) Now the 0.5V-1.2V 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 pre-amped.)
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Old 28th April 2017, 12:10 PM   #5
skidave is offline skidave  United States
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Thanks everyone for your feedback. I have homework to do.
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Old 28th April 2017, 04:07 PM   #6
rayma is online now rayma  United States
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Quote:
Originally Posted by PRR View Post
For 20KHz you want a fast opamp for 1% accuracy around 1V.
Fast diodes are needed too.
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