# Change the current from a voltage source?

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So what I understand so far, if I want to have 16 A out from a voltage source (230 V RMS) which is 325.269 V in amplitude (peak), can I simply add an resistor in parallel to achieve that?

R = 325.269 V / 16 A = 20.329 ohm

Do you think guys if I am on the right track? So if I am want simply to have another current from the voltage source, I just change the resistor to a different value?

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325V is the peak momentary voltage. The 20.3 ohm resistor will heat with 2713W continuously and must be huge. A heater?
EDIT: Wait a minute. You calculated a resistor with the peak voltage so with the RMS voltage the current will be only 234.66V/20.329 ohms= 11.54A average.

I just want to simulate the AC grid. I wanted first to try these values above,
and than I will try for lower currents (4,6,8 A) but for 230 V (RMS). So I don't regarding the heat, just to get the inputs correct for the my converter to receive.

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"16A" from a "230V" mains source would normally be interpreted as 16A RMS from a 230V RMS source.
Why are you bothered with 3 places of decimals? Mains voltage isn't that accurate, and is constantly varying as different loads are switched in and out.
If you connect your converter across the mains (which would be very dangerous in the absence of isolation), the 20Ω resistor in parallel will have no noticeable effect on the converter. If, on the other hand, the resistor is in series with the converter then the resistor will make a good electric fire element and the converter will get little voltage.

It seems like the question the original poster is asking is so unclear that there is no question to answer and the posted replies are meaningless.

It sounds like the OP wants to draw 16A from a 230v circuit using a 20-some ohm resistor. 16 amps at 230v = 3680 watts. If he's planning on connecting a standard resistor, it's going to disappear in a flash of smoke.

If he really does want to draw 16 amps, a bunch of light bulbs in parallel, a space heater, an electric kettle.... is a practical and safe way to draw the much current.

I think the OP is asking for a confirmation of Ohms Law?

I = E/R, so rearrange it to be

R=E/I = 230Vrms/16Arms = 14.375 Ohms.

You can also get the same answer by using peak values:

(230*1.414)/(16*1.414) = 230/16 = 14.375 Ohms.

So what I understand so far, if I want to have 16 A out from a voltage source (230 V RMS) which is 325.269 V in amplitude (peak), can I simply add an resistor in parallel to achieve that?

R = 325.269 V / 16 A = 20.329 ohm

Do you think guys if I am on the right track? So if I am want simply to have another current from the voltage source, I just change the resistor to a different value?

You can get any amount of current from a voltaqe source if it will supply the load current, and a correct resistor is chosen. In your case, the dissipation of the resistor will be 230 times 16 = 3.680 kW. That is about the heat equivalent of 4 clothes irons. Double that wattage for good design practice. You probably need a carbon pile resistor. https://www.electro-tech-online.com/threads/questions-from-newbie-carbon-pile-load-tester.15676/

Ratch

I just want to simulate the AC grid.

that's a bit of a tall order... since "the grid" includes a lot more than a single 240V line. there are usually 3 phases, step up/down transformers, load balancers, substations, transfer switches, etc...

I mean, the AC grid doesnt need to be accurate. But rememeber, I simulating an OBC (On-board charger) for a electric vehicle so it should be high power in and out from the converter. I did now tried for lower currents (, but do you think it looks reasonable? See the attached image (table).

I think the output currents looks so low for the AC/DC converter..

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i'm still not really sure what the question is... you have about 1/2A output from the AC/DC supply, then 8A out from a DC/DC converter at the same voltage? are you charging a battery at 1/2A and then discharging it at 8A? without any idea what you are trying to accomplish, and with what hardware i might as well tell you the answer is "42"... or maybe that these go to 11. most of us here are "tech geeks", and when questions are asked where somebody is looking for a solution to an ambiguous problem where little or no data is available, we tend to get a little upset about it, especially when the input consists of more and more ambiguous data... to put it in perspective...
Caller: "i need help figuring out the length of one side of a triangle...
Me: "describe the triangle"
Caller: "one corner is a 90 degree angle"
Me: "ok, and?....."
Caller: "one side is 30ft"
Me: "ok, and?....."
Caller: "isn't that enough?"
Me: "i need another angle or another side.... what are you trying to figure out?"
Caller: "im trying to figure out how long it will take to cross the street here....."

Okay, I am sorry, that I was not so clear on about what I want to accomplish.

This is my case, I am trying to simulate an on-board charger to charge a HV battery inside an electric vehicle. This on-board charger is consist of two parts:

-AC/DC converter with PFC boost interleaved
-DC/DC converter

The on-board charger is supplied by the AC-grid, and all I know, the charger should be supplied with 230 V, 50 Hz, 4-8 A.
The battery is a high voltage battery, where the maximum current is about 12 A, with an charging voltage range of 320-380 V.

I also know that maximum voltage and current from the AC/DC converter is 420 V and 12 A.
DC/DC converter have also a maximum voltage and current of 420 V and 12 A.

At the moment, I just want to see if what output power can I achieve from the AC/DC and DC/DC converter when using lower input currents (4-8 A).

Hope it is clearer now, otherwise let me know.

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230 volts (RMS) at 8 amps is 1840 Watts. 420 volts at 12 amps is 5040 Watts. You can't get more power out of a converter than you are putting in.
I agree with the comments from others that your original question makes no sense.

Les.

230 volts (RMS) at 8 amps is 1840 Watts. 420 volts at 12 amps is 5040 Watts. You can't get more power out of a converter than you are putting in.
I agree with the comments from others that your original question makes no sense.

Les.
I know that. I just want to see what output power can I achieve if I have lower currents in the input. I assuming it will not be 420 V and 12 A, it will be much lower than that.

So does your question boil down to "What is the best power conversion efficiency I can hope to achieve ?" ?

Les.

230*8*efficiency/420 is your available output current.

Mike.

if your charger is fed by 230V@8A. you won't get 380V@12A. you will get about 4.8A out of the 380V charger (actually, more like 4.2A if the efficiency is 90%). the current is going to be inversely proportional to the voltage. if you get 420V out of the charger, then the current will be 3.9A, again assuming 90% efficiency.

Is this just an academic exercise? I hope so, given the lethally high voltages and currents involved and the difficulties you are having just with the simulation.

Is this just an academic exercise? I hope so, given the lethally high voltages and currents involved and the difficulties you are having just with the simulation.
It is just a simulation for an academic project, nothing that will be built in real-life.

I added a current source in order change the input current? Do you think its correct to do that?
However, the output looks really bad. It should be more constant. The values for the components are calculated
and should be correct.

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See my comment in your other thread re the current source. The mains will provide all the current the circuit demands.

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