Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

[SOLVED] How to connect or wire the load (for load test power supply)

Status
Not open for further replies.

ShuKosugi

Junior Member level 1
Joined
Nov 22, 2013
Messages
19
Helped
1
Reputation
2
Reaction score
1
Trophy points
3
Activity points
148
Hello Guys...here I am again...am still getting unclear idea on how to load test my power supply PCF250-4001. Really needed help who can explain in "layman's terms". You can see the image below the power supply drawing...so how I am going to connect the load (using load resistor) to satisfy the condition written in user manual of this power supply...thanks.

PCF250_wiring.JPG

Conditions:
Total current available from V1 + V2 is 40 amperes, total current available from V3 + V4 is 6.6 amperes.
Minimum Load, V1 Min. load required to maintain regulation on V2 is 4 Amps
Minimum Load, V3 Min. load required to maintain regulation on V4 is 0.6 Amp
 

HI,

it´s just Ohm´s law.

R = U / I

Minimum Load, V1 Min. load required to maintain regulation on V2 is 4 Amps
I´m a bit confused why V1 and V2...

I expect the 4A are on V1: So U = 5V, I= 4A. R = U / I = 5V / 4A = 1.25Ohms.

This is the minimum load. To test with higher currrent, then you may add other resistors in parallel to this resistor.
The resistors will become hot. Very hot. Even this miminmum load of 4A @ 5V give a power dissipation of 4A x 5V = 20W.
This about the same power as my sodering iron.

More current means more heat.

Buy resistors that are rated for each power consumtion: P = U x I

Klaus
 

I saw a project for load testing a large battery, using a load made from steel bands used to wrap large shipments on skids. Several feet of steel band are wrapped from nail to nail pounded into a wood board. Air cools the steel. Ohm value is very low.

The aim is to achieve near-perfect conductivity between the steel and your power supply connections. It will be a challenge. Steel cannot easily be soldered the usual way. Possibly silver solder, or tack weld, etc. A nut and bolt could work if surfaces are very clean and tightly joined.
 

Thanks guys...i connect 1ohm resistor (25W) across V1 and across V2 connected 2.2ohm (25W).
THe result was not good. The V1 (5V) dropped to 4V and V2(12V) dropped down to 5V.
Any idea guys....do we have problem with the power supply? Or load set-up...remember the condition below....also V1 and V2 total current capacity is 40A, like thier load sharing...still really not clear to me...thanks a lot.

Conditions:
Total current available from V1 + V2 is 40 amperes, total current available from V3 + V4 is 6.6 amperes.
Minimum Load, V1 Min. load required to maintain regulation on V2 is 4 Amps
Minimum Load, V3 Min. load required to maintain regulation on V4 is 0.6 Amp
 

Guys thanks for the responses...I just directly installed the power supply into our system and it works. But the load testing i still have to study back if have time. This is one of the power supplies acting weird on load test. Need to close this thread...thanks again EDABOARD.
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top