Can I use a diode to lower the voltage of a battery charger output?

[DarrenB]

I am a bit worried that my BMS might Y2K on me with all the error flags it will need to log. I think I might still stick to the diode. Just need to find one that can drop exactly 0.4V. Please could someone recommend one that I can find in the UK. It needs to be ok for 4A continuous and 7A peak and drop exactly 0.4V.

 

[alec_t]

The diode drop is current-dependent, so asking for "exactly 0.4V" is optimistic. You won't find any diode which drops 0.4V at both 4A and 7A.

 

[DarrenB]

The diode drop is current-dependent, so asking for "exactly 0.4V" is optimistic.

I have literally just discovered this as I was looking at the data sheet for this schottky diode. It looks like it will drop between 0.45-0.6V.

 

[alec_t]

If you really need to drop a fixed voltage, independently of current, you could use an active circuit with a MOSFET instead of the diode. Here's an example where an N-channel MOSFET is in series with the negative side of the supply Vreg :-

The graph shows a pretty constant reduced output voltage over a 4A to 8A load current range.

 

[crutschow]

The diode drop is current-dependent

True.
But the OP has talked about charging up a capacitor, so the current at the end of the charge will be determined by the added load resistor, and that current will determine the forward diode drop when the cap is charged..
The resistor value can be adjusted to tweak this drop.

 

[DarrenB]

If you really need to drop a fixed voltage, independently of current, you could use an active circuit with a MOSFET instead of the diode. Here's an example where an N-channel MOSFET is in series with the negative side of the supply Vreg :-
View attachment 139840
The graph shows a pretty constant reduced output voltage over a 4A to 8A load current range.

It is starting to get really complicated though if I go that route. Also I already have too many unanswered questions and this circuit will be far beyond my knowledge. Can I ask what will happen if I use a diode? Will the current, voltage drop, CC/CV all just balance out like it would if I used 2 wires to charge a battery rather than 4 wires to read the voltage and also charge?