How to measure the wattage and VA of Powersupplies and UPSes??

[Qaisar Azeemi]

HI;

i want to measure the wattage of my own designed power supply That i've designed usign 4 LM338 ICs in parallel. moreover how can i find the wattage/VA of different powersupplies/Invertors/UPSes that use Transistors or MOSFETs in parallel. is it enough that i just add the given wattage in the Datasheets of the components to get the total wattage of the device? or there other factors are also considerable? and how to convert that wattage in to VA of the equipment.

Thank you in anticipation.

 

[ALERTLINKS]

Maximum wattage, indicated in datasheet is the dissipation in device not output wattage. Wattage is VxA, as you know.
Now for example consider 75n75 fet, It can withstand 80 Ampere at max 75 volts. In a switching application it is passing 50 Amperes to a transformer load connected to 48V DC. Output Wattage is 48x50= 2400 watts. So you are operating 2400W load with a device which has a power handling of only 75 watts.
There is a small resistance in device when fully turned on called Rdson in datasheet. It is a 9.5 mohms.
http://www.unisonic.com.tw/datasheet/75N75.pdf
A voltage is developed across device when passing 50A through this resistance. 9.5 x 50 /1000 =.475V
So power dissipated in device = 0.475V x 50A = 23.75 watts. It reside in safe limit of 75 watts.
Actually voltage on load will have a drop. Voltage on load will be 48 - 0.475 volts.
Devices operating in analog region have more power dissipated due to high voltage on device. It is just like a resistor in series, less resistance less power loss.

 

[Qaisar Azeemi]

Maximum wattage, indicated in datasheet is the dissipation in device not output wattage. Wattage is VxA, as you know.
Now for example consider 75n75 fet, It can withstand 80 Ampere at max 75 volts. In a switching application it is passing 50 Amperes to a transformer load connected to 48V DC. Output Wattage is 48x50= 2400 watts. So you are operating 2400W load with a device which has a power handling of only 75 watts.
There is a small resistance in device when fully turned on called Rdson in datasheet. It is a 9.5 mohms.

A voltage is developed across device when passing 50A through this resistance. 9.5 x 50 /1000 =.475V
So power dissipated in device = 0.475V x 50A = 23.75 watts. It reside in safe limit of 75 watts.
Actually voltage on load will have a drop. Voltage on load will be 48 - 0.475 volts.
Devices operating in analog region have more power dissipated due to high voltage on device. It is just like a resistor in series, less resistance less power loss.

Thank you very much Alertlinks. as you calculated the wattage associated with a Mosfet are instantaneous or per hour wattage ratings? i am confused in it.

suppose a 1000W, 250V UPS output that can provide 4A current to drive the load. is this calculation for an hour usage.

second question is that for a 12V input 1000W Inverter/UPS. can i calculate the input current as 1000W/12V = 83.33A
and the output current is 1000w/250v = 4A. and suppose we consider a ceiling fan 120W, 220V needs 0.54A current. so we can say that we can drive 4A/0.5A = 8 Fans with a 1KW inverter/UPS.

And Suppose that commonly available UPSes that contain D1047 Transistors in parallel to drive the loads and each D1047 is suppose 100W and can pass 12A current (I am supposing because i cant find the data sheet of D1047) thus for 1000W and 83.33A input current 7 transistors are enough to switch the power?

Am i right in these calculations?

and how to find the VA of the Device?

- - - Updated - - -

 

[ALERTLINKS]

May be you are typing wrong for 2SD1047 as commonly called D1047. See datasheet here;
https://www.st.com/internet/com/TECHNICAL_RESOURCES/TECHNICAL_LITERATURE/DATASHEET/DM00026462.pdf
These all calculations are for instantaneous values expressed unit in watts ( W ). Watt_hour ( Wh ) means, wattage consumed for full one hour. This unit shows on watt_hour meter to show electricity consumed for billing.

for a 12V input 1000W Inverter/UPS. can i calculate the input current as 1000W/12V = 83.33A and the output current is 1000w/250v = 4A. and suppose we consider a ceiling fan 120W, 220V needs 0.54A current. so w le can say that we can drive 4A/0.5A = 8 Fans with a 1KW inverter/UPS.

For the first look, your calculations are right but practically there are other factors which determines the maximum load for UPS. For example fans take some double the current when gaining speed from zero. As it is a inductive load, you have to consider "power factor". It is about 0.6 for UPS unless corrected. So from 1000VA UPS, you can draw only 1000 x 0.6 = 600 Watts. There is difference between UPSes made for continuous use and for a few minutes usage of equal wattage.
It to better to give margin upto 50% -70% not to operate devices at there full stated capacity. Otherwise they will be more
prone to occasional damages. You have to consider short term excessive loads. For example degassing coil in a tv can draw 10 times more current for a few seconds.

 

[Qaisar Azeemi]

Thank you very much Alertlinks. How can i improve the powerfactor of the Devices (Electrical devices and UPSes)?

kindly clear the following confusions related to DataSheet :

Rds(on) is not given in the data sheet of 2SD1047 and according to datasheet Vceo=140V and Icm=20A ---> so can we say Rds(on)=Vceo/Icm=7ohms....... ??

and the the maximum power decipated in the device is Ptot=100W but Ptot comes out to be Ptot=Vceo X Icm = 140 x 20 = 2800W...... ???(i think i am calculating Wrong values)

 

[Tahmid]

2SD1047 is an NPN transistor. RDS(on) is only applicable to MOSFETs. For 2SD1047, you have to consider the VCE(sat). Provided you drive the transistor into saturation mode (which you should aim to do), the power dissipated by the transistor is VCE(sat) X Current flowing through.

P = VCE(sat) X I

Hope this helps.
Tahmid.

 

[Qaisar Azeemi]

oh sorry for asking the silly question i didn't think that Rds is for drain to source resistance. ...... anyhow according to datasheet Vceo=140V and Icm=20A --->so the the maximum power decipated in the device is Ptot=100W but Ptot comes out to be Ptot=Vceo X Icm = 140 x 20 = 2800W...... ???(i think i am calculating Wrong values)....... kindly guide me about that

- - - Updated - - -

oh sorry for asking the silly question i didn't think that Rds is for drain to source resistance. ...... anyhow according to datasheet Vceo=140V and Icm=20A --->so the the maximum power decipated in the device is Ptot=100W but Ptot comes out to be Ptot=Vceo X Icm = 140 x 20 = 2800W...... ???(i think i am calculating Wrong values)....... kindly guide me about that

 

[ALERTLINKS]

Reduce voltage or current, not to exceed power to maximum allowed in datasheet. Note it instant value. When device is on in saturated condition the voltage drop on device drops low and current increase and when device is off its vice versa. Consider it as a switch. When switch is off, voltage is at full on its contacts and when it is turned on the voltage drop on its contacts is minimal, in fact near zero. Small voltage drop on switch is due to internal resistance of contact and interconnecting wire.
Transistors and IGBTs have no fix internal resistance as compared to FETs but they try to maintain a small voltage drop on there current carrying terminals called as saturation voltage (Vsat). So you see when device is off, there is no power lost in device. There is a graphical representation of safe operating area for selecting voltage and current for safe operation in analog mode, as in many audio power amplifiers.

 

[Qaisar Azeemi]

sorry i cant get it. kindly explain it further.

 

[ALERTLINKS]

A bulb of of 60 watts 240V will draw 60/240 = 0.25A
A bulb of of 60 watts 12V will draw 60/12 = 5A
A switch is connected in series with bulb having a resistance of 0.1 ohm.
With 240V, the power lost in switch due to resistance is 0.1 x 0.25 = 0.025 watts when switched on.
With 12V, the power lost in switch due to resistance is 0.1 x 5 = 0.5 watts when switched on.
Consider replacing switch with a FET having Rdson of 0.1 ohm. Although load is 60 watts but power on FET is different.
Now if switch is replaced with a transistor having Vsat of 1.5V.
W=V x A
In first case W = 1.5 x 0.25 = .375
In second case W = 1.5 x 5 = 7.5


http://techhouse.brown.edu/~dmorris/projects/tutorials/transistor.switches.pdf
**broken link removed**
http://www.microchip.com/stellent/groups/designcenter_sg/documents/market_communication/en028089.pdf
http://www.microsemi.com/en/sites/default/files/micnotes/APT0403.pdf
**broken link removed**