Dummy load

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You can remove the extra resistors R21, 22, 23, 24, 27 and 18. 1% are cheaper here than 5%.
 
Safe Operating Area. "S O A"
Even if you run at max safe amps AND max safe volts... you can still operate in an area that will break the FET.

For simplicity sake, I say that SOA = max watts. And for MOSFET, this is basically accurate.
 
You can also remove R24. Yes that FET is good. Please tell us what you are doing so we can be sure it will work.
 
What I doing? This is joke? DL1 offcourse
Actually I draw board(pcb) and this is not easy - many copied elements

And I have little problem with my free eagle, look - M10 and R10
 
I have no experience with Eagle (I use express pcb) , but many here do. Maybe they can help. I'm not sure I wouldn't build it on perf board or strip board and hand wire it as many of the lines are high current and will require wide traces or heavy wire.
 
I will add some copper rods on load traces(like in many server ps's), I need only little help with eagle, I find one person with full eagle what move these two components.
How many watts drops on 7w resistor? with 1200W load? these need good cooling?
 
Yes the .13 ohm resistors will be very hot. Each one will have 5 amps so 5^2 X .13 = 3.75 watts. So maybe 150C. Ten watt ones would be cooler, but bigger.
 
()blivion, Here is a pretty good write up on linear operation.

**broken link removed**
 
I add some aluminium profiles to 7W resistors connected with small line to main cooling copper tube.
 
Be careful. The copper pipe that the FETs are soldered to has 24 volts on it.
 
Huh, almost complete, only few elements to move on full version of eagle(on free version size of pcb is limited) and some plugs for 12v, temp and amp.

Any proposals/objections for PCB?

ronv said:
Be careful. The copper pipe that the FETs are soldered to has 24 volts on it.
Click to expand...
Thanks.
 
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I want use this display for amp, volts and temp. **broken link removed**
Amp. output is 5v-50A, so when I use this display I get 05.0, so how to get 50.0 on display?
Thanks.
 
You can't. That is one of the problems with DLI. It only measures one branch of the circuit so you must multiply the reading by 10 yourself. In DLII he used an actual current meter to measure the whole thing.
 

So I cut wire and display will not show dot...

Edit: Next question, how many power needs this circuit from 12V?

Edit2: here is actual board:
 
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If you can find the right wire that would work.

It does not need much more than the buzzer, but it should be a regulated supply.

**broken link removed**
 
Exactly this I want use
About dot, I want unsolder plug from voltmeter, so I can by the way cut the trace from display what is responsible for display dot.
Or meybe I add somewhere a resitor what result displaying correct value.
I could ask for it the Chinese but they probably will not answer, as always...

Thanks for help guys!

EDIT: I find one little and very cheap, but 100mA, It wold be ok?
 
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Should be ok, but a little bigger might be best. How much does the buzzer draw? The meter needs 25 ma, the circuits maybe 10 or 15 ma so all we are missing is the alarm.

Can you post a link?
 
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