Power rectifier calculation for telecom equipment with battery backup?
Armughan asked:
I have two devices which need to be powered up by -48V DC power. One device has two components IDU and ODU (microwave equipment) the combined power consumption is 40.5 Watt(18.5+22), the second equipment has power consumption of 60Watt (multiplexer). The rectifier also has to charge the batteries for backup.
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I have two devices which need to be powered up by -48V DC power. One device has two components IDU and ODU (microwave equipment) the combined power consumption is 40.5 Watt(18.5+22), the second equipment has power consumption of 60Watt (multiplexer). The rectifier also has to charge the batteries for backup.
What will be the rating of the rectifier i should use? also if i need backup of 4 hours what will be the rating of the battery?
Also it will be very helpful if someone can recommend some book or pdf for better understand the power concepts.. like what if i have to make the same calculations for AC power..
Can you mention the required rectifier without battery backup.
If we take battery backup into calculations the backup time required will be 4 hours and recharge time will be 10 hours.
March 11th, 2010 at 5:28 am
You can’t answer this question without some further information.
You need to start with how long the back-up power might be needed.
This allows the battery size to be calculated.
Then find out how long you can allow for a re-charge.
This determines the battery charge rate and therefore the current to the battery. Now, you beed to know what current limiting there is in the battery charger circuit, since it needs to cover a flat battery case.
In the end, I estimate that a rectifier in the range 10 to 30 amps would be used in most practical applications. Simon R
March 14th, 2010 at 8:17 am
It’s just addition. Calculate the current for each part of the load, as power/voltage, and add them up to get the max current. Then get a diode rated for that current, or preferably about 50% higher.
You are missing the battery charge current and battery size. Battery charging is a complicated process, with current changing with state of charge, and monitoring the charge to prevent overcharging and damage. so it takes more than a simple rectifier, it takes a charge controller circuit.
Your load is 100 watts, which equates to 2.1 amps. If you double it to allow for battery charge current, then it is about 4 amps. A 5 or 10 amp diode will do fine.
. billrussell42
March 15th, 2010 at 4:22 am
The load is 100.5W
The backup time is 4 hours, so the energy requirement from the battery is 402Wh.
Battery capacity..
The 48V battery is actually 52.08V on float charge, but use 48V to calculate the Ah rating of the battery. This assumes the equipment uses constant power with varying battery voltage, using the nominal voltage. It is 402Wh/48V so only 8.375Ah. However the current rate is 100.5W/48V = 2.1A and smaller batteries should not discharge faster than the 20h rate. Thus for 20 hours it is a 40Ah capacity. This gives a longer backup than required, but meets the needs of the battery, which is presumably a telecom type for long life and high reliability. It is likely that is actually a 50Ah battery when you go to buy it, so the battery is defined as 50Ah capacity.
The rectifier (which means float battery charger in Telecomms speak) will need to charge the battery at the 10h rate as well as run the load. That means 5A for the charge, and 2.1A for the load, so the required current is 7.1A. If you prefer to charge at the 20 hour rate that is 2.5A and so the total is 4.6A. This choice depends on how long you are willing to wait for recharge. It means you can use a 5A or a 10A rectifier unit, as these are the likely sizes when you come to buy. The actual charge time is longer than expected 10 or 20 hours for a full charge, because the battery needs 1.4 times the capacity used for recharge, and the charge tapers off as it approaches float voltage. However the battery should not be run that way. Alternative poower should be provided before the battery gets anywhere near empty.
Charging…
The rectifier runs at the float voltage (52.08V) but during a charge it will be initially current limited to the rectifier rating, so 5 or 10A as above. The voltage is reduced accordingly. When the battery approaches full, the voltage rises towards 52.08V, and the current will taper off until it is just the load current and a slight battery current to account for “local action”.
Backup time
This battery (50Ah) would not be expected to operate less than 50% charge for a long life, so it would be used to 25Ah, and so hold the load for 10-12 hours.
Obviously you can also use a battery of only 10Ah capacity and get 4 hours, and a 5A rectifier probably, but the life will be shorter than the expected 20 years and the capacity very marginal. It is always good in the long run to go bigger. It also allows for a bit of expansion or the inevitable deterioration. Ecko