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### Topic: P=V*I applied on DC resistor circuits. (Read 3829 times)previous topic - next topic

#### afremont

#15
##### Mar 07, 2013, 04:40 pm

This might be redundant for many, but I want to point out that Power is also proportionate to the square of the Voltage.

ie. Volts is directly proportional to amps.

It's the law!

Only as long as Xc = Xl
Reminds me of 300,000km/S, it's not just a good idea.   It's the law!
Experience, it's what you get when you were expecting something else.

#16
##### Mar 07, 2013, 04:57 pm
Quote

Reminds me of 300,000km/S, it's not just a good idea.   It's the law!

Well, in a vacuum anyway ...
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

#17
##### Mar 07, 2013, 05:35 pm
Power dissipation often comes in handy in transistor selection also.
P=IV=I^2*R = V^2/R
For MOSFETs, current flow & Rds is usually known, so P=I^2 * R is good
(and if you current & Rds, V=IR so you can determine Vds also)

For BJTs, current flow & Vce-sat is usually known, so P=IV is good
Vce-sat of ~0.7V vs Rds of 0.035ohm shows why MOSFETs are preferred for high current loads:
example: 1A & 0.7V = 700mW, while 1A & 0.035ohm = 35mW
typically ignored is the power from the base current - say for an NPN 15mA was used, and Vbe is 0.7V -that's another 10.5mW that is dissipated.  1.5% of the total from the example above, so ignoring it is fairly safe.  At Arduino limits, say 35mA, max would be 0.035A * 0.7V = 24.5mW, still a fairly small amount.

For resistors, voltage & resistor value is known, so P=V^2/R is good.
For an LED for example: (Vs - Vf-led - Vtransistor)/current = resistor, or  (Vs - Vf-led - Vtransistor)/ resistor = current, depending on what is being solved for (what resistor do I need for 20mA? if I use this 220 ohm resistor, what current will I get?)
Vtransistor is Vce-sat for BJTs, and current * Rds for MOSFETs
How is this applicable? Say you're wondering how much can you do with a 1/8w resistor,as you notice a warm smell coming from your circuit? 0.125W = I^2 * R, so with 20mA (max continuous for most LEDs), smallest R you can get by with is 0.125/(0.02^2) = 312 ohm, so a standard 330 would be safe.
That same 312 ohm resistor with a 1/4W rating could handle more current: P=I^2R, or Sqrt(P/R) = I, so Sqrt(0.25/312) = 28mA.

All things to consider as you select components.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

#### albuino

#18
##### Mar 07, 2013, 09:12 pm
Thanks everyone. The help received has been excellent by far.
Actually it was a general electronic question. There is no project around this. So I have really learnt that I must believe electronic laws since I didn't.

#### MarkT

#19
##### Mar 08, 2013, 06:32 am

This might be redundant for many, but I want to point out that Power is also proportionate to the square of the Voltage.  P=V2/R  This is something many people learn the hard way after raising the input voltage to a linear regulator.  This results in an exponential power dissipation increase just the same as increasing the current flow.  What worked coolly with a 9V battery as input, now puts the regulator in shutdown after switching to a 12V supply.

Actually the input to a voltage regulator is a poor example here as linear regulators look like constant current loads (the regulator
keeps the output voltage constant, typically the load then draws a fixed current, thus the input to the regulator is a constant current.)

Increasing the voltage to the linear regulator does not increase the current (unlike the resistor case).  Dissipation in regulator is (Vin - Vout) * current.

Also a square law is not exponential - exponential increase is what happens as you increase the voltage across a diode, and makes
a square law look tame!
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### fungus

#20
##### Mar 08, 2013, 07:03 am

Also a square law is not exponential

Yes it is. The exponent is 2.
No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

#### JimboZA

#21
##### Mar 08, 2013, 07:15 amLast Edit: Mar 08, 2013, 10:58 am by JimboZA Reason: 1

Also a square law is not exponential

Yes it is. The exponent is 2.

My memory is getting old, but if I remember correctly, exponential means ax not xa, so x2 is not exponential. If exponential simply meant having an exponent, xa, then x1 or even x0 would be exponential.
Johannesburg hams call me: ZS6JMB on Highveld rep 145.7875 (-600 & 88.5 tone)

#### afremont

#22
##### Mar 08, 2013, 08:47 am
Yeah, I shouldn't have used a linear regulator for the example as the dissipation increase is linear as the input voltage increases.  I picked it without thinking it thru I suppose.  It's such a common occurrence for beginners to find that the regulator seems to suddenly get hot out of the blue.  More often it is from increasing current by adding things like LEDs or other loads to the circuit as they build it out.  Still, changing the input voltage often results in the same type of surprise even if it's not exponential in that case.

Somebody better tell these guys that a square law is not exponential:
http://www.platinumgmat.com/gmat_study_guide/exponential_equations

Experience, it's what you get when you were expecting something else.

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