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Catecory: TheoryHit Count: 869
Long calble application issue of dc-dc converter
In many applications, the dc-dc converter will be fed by a long input cable from the dc power source. In such case, your maximum load will be limited physically by the cable resistance. You have to keep below equation being true:
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Vs>=2sqrt((Po*Rcable)/eff)) or Po<=Vs^2*eff/(4Rcalbe)
Where, Po: output power Rcable: Cable resistance; Vs: input supply voltage; h: Converter efficiency. For example, for 100m of AWG16. The resistance is about 1.32 Ohm. Therefore, to power 80W, you need to keep the supply dc voltage higher that 21V (assuming h=0.95); if your input could go as low as 16V, then your maximum load would be 46W (3.8A |
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Please note that the above equation is for steady state operation. During startup, the input current would be higher and it may try multiple times to succeed startup even if the input voltage is higher than mentioned above. Please also note that this issue is not related to the dc-dc converter's performance. |
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Comments: 3
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Leo QinTitle should be: long cable application...
2Reply[dcy] Reply:[florawu] Reply: -
mrpowerthis fit better for "Knowledge", not for discussion.
0Reply -
phoenix.liu@netpowercorp.cn
Can the problem of long cable impedance be solved by introducing shutdown delay or increasing delay?
0Reply
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