Min voltage drop in silicon rectifier?

[Ken]

I want to isolate flooded lead acid batteries from each other using
power rectifiers like these:
http://www.ruttonsha.com/New_Folder/Ruttonsha/diodepdf/60hm.pdf

I would charge them through the rectifiers.

From the graps, it appears that there is a 0.5V drop across the
rectifier even at low current.

Inasmuch as the batteries could be ruined (boiled off) over time by a
significant trickle charge, I need to know the voltage drop at, say,
10mA -- so I know where to set the float voltage.

How can this be ascertained, short of experimentation?


Ken
(to reply via email
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[John Popelish]

I want to isolate flooded lead acid batteries from each other using
power rectifiers like these:
http://www.ruttonsha.com/New_Folder/Ruttonsha/diodepdf/60hm.pdf

I would charge them through the rectifiers.

From the graps, it appears that there is a 0.5V drop across the
rectifier even at low current.

Inasmuch as the batteries could be ruined (boiled off) over time by a
significant trickle charge, I need to know the voltage drop at, say,
10mA -- so I know where to set the float voltage.

How can this be ascertained, short of experimentation?

The experiment is probably worth doing, because at 10 mA, the
rectifier is operating way below the bottom of the graph. It shows
that a room temperature diode drops about .65 volts with 10 amperes
going through it. As the current falls, the voltage keeps falling at
about 60 millivolts for each time you divide the current by 10 (for an
ideal diode). So to get from 10 amps (the bottom of the forward
voltage graph) to 10 mA, you have to divide by 10, 3 times, so the
forward voltage will be somewhere around .65-.06-.06-.06=0.47 volts.

 

[John Smith]

The experiment is probably worth doing, because at 10 mA, the
rectifier is operating way below the bottom of the graph. It shows
that a room temperature diode drops about .65 volts with 10 amperes
going through it. As the current falls, the voltage keeps falling at
about 60 millivolts for each time you divide the current by 10 (for an
ideal diode). So to get from 10 amps (the bottom of the forward
voltage graph) to 10 mA, you have to divide by 10, 3 times, so the
forward voltage will be somewhere around .65-.06-.06-.06=0.47 volts.

Which sounds garbage to me. The drop will remain fairly constant.

That said: I think what the OP needs to do, is have some feedback directly
from the batteries being charged, and use that feedback to control the
charging voltage. The drop across the diode becomes irrelevant.

 

[Jonathan Kirwan]

Which sounds garbage to me. The drop will remain fairly constant.

Why? I thought this effect is expressed by Ebers-Moll, or so. At
20C, 58.26mV/decade from (kT/q)*ln(10/1). But I'm a hobbist, so I'll
listen to why it's garbage with rapt attention.

Jon

 

[Bob Masta]

Which sounds garbage to me. The drop will remain fairly constant.

Ahhh, I'd say that going from 0.65 to 0.47 over a 3 decade range
of current is "fairly constant"!

Best regards,


Bob Masta
dqatechATdaqartaDOTcom

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