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How much maximum output current will be delivered by 12V DC DC Converter ??
How much maximum output current will be delivered by 12V DC DC Converter ??
- Thread starter datta2kale
- Start date
Half than the maximum current of a 6V supply using the same size (VA) transformer.
Last edited:
It depends on the input power and the efficiency of the DC-DC converter. For example:
Input power = 100 watts, Efficiency =92%
Output power = 100 watts * 92% = 92 watts
Maximum Output Current = 92 watts / 12 volts = 7.67 Amperes
Easy peasy calculation.
Output Power required is 5KW.
If I have to achieve 5KW output power then with 12V DC DC, 418A output current.
So that's why i want to know with 12V DC DC Converter 418A output current is possible?? If yes then how this system works, Is any parts will overheated/damage during operation
Output Power required is 5KW.
If I have to achieve 5KW output power then with 12V DC DC, 418A output current.
So that's why i want to know with 12V DC DC Converter 418A output current is possible?? If yes then how this system works, Is any parts will overheated/damage during operation
Output Power required is 5KW.
If I have to achieve 5KW output power then with 12V DC DC, 418A output current.
So that's why i want to know with 12V DC DC Converter 418A output current is possible?? If yes then how this system works, Is any parts will overheated/damage during operation
ronsimpson
- Joined Oct 7, 2019
- 4,646
You will probably input 5300 watts to get 5000 watts output.
Yes, anything is possible. It will be hard.
Yes, parts will be hot. Transistors, transformer, diodes, will be hot.
panic mode
- Joined Oct 10, 2011
- 4,864
just do the math...
assuming efficiency is 94% to get 5000W output power, input power will need to be 5000W/0.94 = 5319W
the difference between input and output power is lost as heat.
5319-5000 = 319W
that is a lot of heat... using soldering iron as a unit of heat, that is some 10 soldering irons.
assuming efficiency is 94% to get 5000W output power, input power will need to be 5000W/0.94 = 5319W
the difference between input and output power is lost as heat.
5319-5000 = 319W
that is a lot of heat... using soldering iron as a unit of heat, that is some 10 soldering irons.
There's no need to reply with the same information three times.
Yes, getting a DC-DC converter to deliver that kind of power is quite doable, but it will not be cheap. Contact vendors of commercial and industrial power supplies -- a starting point might be electroplating supplies. When I was at NIST we used an industrial supply that delivered 3000 A at up to 2 V, so 6 kW. The unit was something like five feet on a side. However, it was not a DC-DC converter because (1) those weren't very sophisticated in those days, and (2) we couldn't tolerate the noise associated with any kind of a switching supply since we were looking for nanovolt-scale voltage drops across the samples.
Yes, getting a DC-DC converter to deliver that kind of power is quite doable, but it will not be cheap. Contact vendors of commercial and industrial power supplies -- a starting point might be electroplating supplies. When I was at NIST we used an industrial supply that delivered 3000 A at up to 2 V, so 6 kW. The unit was something like five feet on a side. However, it was not a DC-DC converter because (1) those weren't very sophisticated in those days, and (2) we couldn't tolerate the noise associated with any kind of a switching supply since we were looking for nanovolt-scale voltage drops across the samples.
I see that the TS enjoys repeating himself which is not the best way to solicit help. Nevertheless, your project is doable, but it will require substantial monetary, engineering, and purchasing resources to design and build. OTOH, there may be a COTS (Commercial Off the Shelf) solution.
MisterBill2
- Joined Jan 23, 2018
- 27,164
Certainly such a power supply can be built. BUT without a lot more information there is no way to do more than guess.
For starters, the input voltage and input voltage range must be specified.
Next the output current and duty cycle, as well as the required output voltage stability must be specified.
Then there should be a specification as to efficiency
Then there is the issue of output voltage ripple and noise, which the acceptable limits must be stated.
I can easily provide a DC to DC supply that can deliver 12.0 volts, +- 0.1 volt, and only costs $300. But there is a limitation that the input must be 12 volts +-0.1 volt with at least 500 amps capability.
For starters, the input voltage and input voltage range must be specified.
Next the output current and duty cycle, as well as the required output voltage stability must be specified.
Then there should be a specification as to efficiency
Then there is the issue of output voltage ripple and noise, which the acceptable limits must be stated.
I can easily provide a DC to DC supply that can deliver 12.0 volts, +- 0.1 volt, and only costs $300. But there is a limitation that the input must be 12 volts +-0.1 volt with at least 500 amps capability.
