Paralleling Power Supply Units Made Easy ???
I am considering paralleling a couple of 20a variable adjustable psu's for 40a charging. According to what I have read on the world wide inter-web, it can be done quite easily, provided the following ...
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No, that doesn't work well. The one that's in cv will taper current and quickly be a non-factor.
Here's the story behind tapering.
You got 2 20A supplies, one set at 14.0V the other at 14.7V. Both supplies are connected to the battery, each having a cable resistance of 15 mohm. The battery is discharged to 12.2V.
At the onset, both supplies are in cc, outputting 20A each. At this time the cable resistance isn't a factor. As the battery voltage increases, so does the supply voltage.
V_ps = V_bat + I_cc * R_cable
For example, when the battery is at 12.7V, the supply will be at:
V_ps = 12.7V + 20A * 15 mohm = 13.0V
When the battery voltage gets to a specific point, the lower voltage supply will taper. In this example the battery will be at:
V_taper = V_ps - I_cc * R_cable
V_taper = 14.0V - 20A * 15 mohm = 13.7V
When the battery reaches 13.7V, the 14.0V supply will begin to taper. This supply will output zero amps once the battery reaches 14.0V. The other supply will begin to taper when battery is at 14.4V.
Your best bet for efficient charging is to set both supplies at the same voltage. The lower the cable resistance, the longer the supplies will be in cc, and that's the objective.
Here's the story behind tapering.
You got 2 20A supplies, one set at 14.0V the other at 14.7V. Both supplies are connected to the battery, each having a cable resistance of 15 mohm. The battery is discharged to 12.2V.
At the onset, both supplies are in cc, outputting 20A each. At this time the cable resistance isn't a factor. As the battery voltage increases, so does the supply voltage.
V_ps = V_bat + I_cc * R_cable
For example, when the battery is at 12.7V, the supply will be at:
V_ps = 12.7V + 20A * 15 mohm = 13.0V
When the battery voltage gets to a specific point, the lower voltage supply will taper. In this example the battery will be at:
V_taper = V_ps - I_cc * R_cable
V_taper = 14.0V - 20A * 15 mohm = 13.7V
When the battery reaches 13.7V, the 14.0V supply will begin to taper. This supply will output zero amps once the battery reaches 14.0V. The other supply will begin to taper when battery is at 14.4V.
Your best bet for efficient charging is to set both supplies at the same voltage. The lower the cable resistance, the longer the supplies will be in cc, and that's the objective.
jrnymn7 wrote:
"When in voltage mode, power supplies do not share current well. The higher voltage supply will back bias the other supply. Depending on the cable resistance, you may get something from the lower voltage supply."
Yes, this is the inherent issue with trying to parallel psu's, and this is why it was recommended to use one in c.v. and the other in c.c.
