Diodes in series

Thread Starter

Blackbull

Joined Jul 26, 2008
70
Looking, in a magazine, at the circuit diagram for a lead-acid battery maitenance charger and quoting from the text: '' The reference voltage is derived from the 5V line, which is regulated by IC1, via potential divider R2-R3-P1-D3-D4. Diodes D3 and D4 provide the requisite temperature compensation.'' Is this enough information for anyone to tell me how these diodes ( in series ) work and would this be circuit or ambient temperature?
Many thanks.
 

beenthere

Joined Apr 20, 2004
15,808
Not enough information. The identity of "P1" is open to doubt (probably a variable resistor, but the standard designation of any resistance is 'R'). The location of the diodes is what tells if they are compensating for ambient or for the circuit temp.

Diodes change conduction with temperature, by the way.
 

SgtWookie

Joined Jul 17, 2007
22,201
Can you post a link to the schematic, or post the schematic?

The Vf of the diodes will change as the ambient temperature changes. It's a NTC (negative temperature coefficient), so as the temperature changes, the Vf will decrease.

The current flowing through the diodes will tend to heat them, as they'll be dissipating power. This will tend to make an offset from the actual ambient temperature.

However, regulating voltage to a lead-acid battery using ambient temperature is a poor method. The internal temperature of the battery is what needs to be monitored; it will vary considerably less than the ambient temperature. The best way to do this without penetrating the battery case is to monitor the temperature of the positive terminal.

Consider using an LM35 (precision temp sensor) instead of diodes; it'll be much more accurate. You could use a permanent epoxy such as JB-Weld to secure the LM35 to a battery clamp.

Read this tech note about temperature compensation for battery charging:
http://rep.mgeups.com/edg/edg/technote/battch.pdf
 

Thread Starter

Blackbull

Joined Jul 26, 2008
70
Posting a schematic is a bit of a problem at the moment, but you have answered my question anyway. Many thanks.
 
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