Am I just being an idiot? Maths...!

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Swedish Chef

Well-known member
Joined
Jun 7, 2004
Messages
351
Location
London
I was doing some stuff about rectifiers this morning and afterwards I tried to do this exercise:

With this figure:


Now I know i am mentally challenged, but this really isn't that hard is it?

10 years since A levels and my brain has turned to mush! :sad:

chef

p.s. Help :oops:
 
although the text states that it's 1 amp averaged over time, the graph does not state the duty cycle of the waveform nor the times for high or low, so you cannot truly judge anything by the graph. 1 amp average, but the highs are clearly higher than one amp on the graph. you would need to get more specs on your rectifier to make sure that it can handle a peak that high for that long. the next question would be what type of circuit is it in? if it's in some kind of inductive circuit(motor drives etc..) then you will get current surges when the field on the inductor is turned on/off. these can be up to and more than 50 times the nominal operating current, yet another place where you need to verify true operating conditions before you spec!

:thumb:

ps: I've seen a setup where a highly inductive load (motor and electromag clutch, 5 amp total draw at full load according to the datasheets) was put on a circuit being fed by 150VDC and switched by SSRs(solid state relays, nothing but triacs really). this setup was first attempted without an output snubber on the reversing power contactor... causing HUGE arcing and 30A rectifiers getting obliterated! once the design got to me to verify what was causing it, i found the arc was caused by not only current spikes of 40-50A but also a voltage spike of near 300V.. (similar in theory to flyback transformers in TVs) some of the arcs actually jumped over 3/4 inch to ground on a screw! turns out, the engineers who came up with this idea forgot the need for high powered snubbers for the contactors to keep the spiking down to a minimum, especially with relay contact bouncing(more sub-second arcs). snubbers now installed, very little arc, current spikes within maximum of rectifiers and all is well.

:guinness:
 
Assuming standard textbook practise of posing basic problems the diagram shows the actual current waveform in question: regular symmetric squarewave with 2A peak value, also a resistive linear load is assumed.

The same excersise can be modified to a 10/90% pulse ratio and 20A peak current to approximate current pulses in a rectifier. What rating should the fuse be now ?

Otherwise you're right Svart that reallife situations are far from basic and linear.
 
I also assumed the basic "Easy as possible!!!" conditions.
Is that 10% for low or high?
And is it really as simple as
max current x (Time current flows in 1 cycle/Period of 1 cycle)?
:? :oops:
What about the P=I^2R

chef
 
Yes! I was wondering about that.... if the fuses behave with I<2>R and not just with current, and the duty cycle is 50%, then the answer should be sqr rt (0.5 x 2 sqrd)... = squre route of 2 ?????

:?:

I realise that I know nothing about this.
(and I need some sleep)
 
Hmm,

I'm not doing your homework for you- you've only been back at school 3 days...

:wink:

Looks familiar...AOE, p.47

'aving a cuppa and a think...

Mark
 
[quote author="SSLtech"]I were thinkin' ?2 (root two) 1.414...

But I weren't doin' the math, just gut reacting...

Keef[/quote]

keef,
how did you get the root-sign in?
stewart
 
depends on the fuse you use. slow blows can take short high current pulses, fast blows can shorter pulses and of course resetable fuses can work in PICOseconds and may trip constantly even though their average current rating may be sufficient.. but of course this is more real world stuff, not ideal condition stuff.

cheers all! :guinness:
 
Well...

If you apply the rules for a "complex" waveform- i.e. non-sine but equal mark:space ratios-

1. Square waveform over 1 cycle- waveform doesn't go below 0V so cycle is always positive:

Sum Of Squares= 2^2 + 0^2

= 4

2. Mean of this value= 4/2 =2

...because this value of "4" came from 2 readings. It doesn't matter that one of these happened to be "0".

3. Final operation- take the Square Root of this value:

Irms= (sqrt)2

Irms= 1.41A

(I can't do Keiths sqrt symbol either....perhaps he got it for his birthday?)

:grin:

Mark
 
Go back to school. It is a different world than Real Life. (Hey, that's a reason to NOT go to school....)

> the highs are clearly higher than one amp on the graph.

The graph shows 2A peak current.

> you would need to get more specs on your rectifier to make sure that it can handle a peak that high for that long.

It is clearly "textbook", not "real life". If they don't ask about rectifier melt-down, it isn't part of the problem or the "right answer".

> what type of circuit is it in? if it's in some kind of inductive circuit....

Not mentioned, probably has not been covered in the course yet, not part of the problem. And I don't think an inductance radically changes the fuse problem (though it sure blows-up voltage sensitive parts).

> depends on the fuse you use. slow blows can take short high current pulses, fast blows can shorter pulses

The problem actually states that the fuse time constant is so long it may be ignored. (Schools are not real-world.)

1A average, but 1.414A RMS.

The fuse rating, REAL-WORLD, has to be a lot higher, yet lower than the smoke-point of the transformer or load. Fuses are poor protection against slight overload, their only real use is is to turn a dead-short into an open.
 
Ah. I've been here before, and the Venerable Mr PRR pointed out that not all computer/forum systems will cooperate, but none the less, here's how:

It's one of the (press + hold [alt]) [x], [x], [x] (release [alt]) sequence. (I use a windows 2000 at work, XP at home, PC computers...)

Ω = (press + hold [alt]) [2], [3], [4] (release [alt])
£ = (press + hold [alt]) [1], [5], [6] (release [alt])
± = (press + hold [alt]) [2], [4], [1] (release [alt])
√ = (press + hold [alt]) [2], [5], [1] (release [alt])
ⁿ = (press + hold [alt]) [2], [5], [2] (release [alt])
² = (press + hold [alt]) [2], [5], [3] (release [alt])

... Often times, as PRR observed (this was on the 'old' forum, before the big tree got blown down by the winds of change) this might display correctly in the "post text" window, but get mutilated in the final page assembly, but I find enough symbols in the list to be useful.

As a suggestion if it works for you and you like this method, spend a few minutes in "word" or something similar, and type a list of all the results that you get with various number combinations next to the 2 or 3-diit number that produces that character; then print it and tape it to the inside of a cabinet door near your monitor, or somewhere similar... it's what I did while I learned many of the most useful number combinations.

Keith
 
sorry PRR, it took me forever to UN-learn my schooling..this textbook stuff just seems WRONG now.. guess i'm just thinking TOO much!

yes i agree about fuses.. i use breakers on power inputs and TVS/varistors on sensitive parts, etc since they work faster...

we get a lot of recent tech school grads through here.. they strut through the doors like they own the place, usually with diploma in hand. once they settle down a bit and their egos get bashed a bit by what they didn't learn and they start to open their eyes instead of their mouths things are ok.. :green:
 
Note: You must use your numeric keypad, not the regular numbers to get the ascii stuff.
This is on my Windows box at least.

_Çüéâäà1åçêëèïîìÄÅÉæÆôöòûùÿÖÜ¢£¥P?íáóíóúñѪº¿_¬½¼¡«»___¦¦

°?n__CPaÔr&B *ÿ&˼

That math problem was poorly stated and they did not show the circuit.
 
You can also just use the character map, and just copy and paste while in symbol font, OR (the one I prefer), switch from standard 101 keyboard to English international. ä×÷åÉ®þüÚíÍ.çæ©¢ ¿¿¿¿¿ I noticed while typing those, some are unavailable, but you get the idea! :wink:
 
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