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Lazybutt  
#1 Posted : Saturday, December 11, 2010 9:44:07 PM(UTC)
Lazybutt

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Hello,
This is possibly a stupid question, so forgive me if I didn't get something right (Brick wall ), but from the manual, we set VR1 and VR2 so that there is 250mV accross R1 and R2, so there is 250mA flowing through all components: R1, R2, VR1, and VR2.

VR1 and VR2 are set at around 50 ohms... doing the calculation, that means they dissipate 50*0.25*0.25 = 3.125W !!!! (and that's on only 25% of the pot surface !!!)
But these components are rated at 0.5W... what am I not getting right..?? shouldn't these pots be higher rated ??

Edited by user Monday, December 13, 2010 11:16:23 PM(UTC)  | Reason: Not specified

Lazybutt  
#2 Posted : Monday, December 13, 2010 11:15:26 PM(UTC)
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Hello again,
Incidentally, the parts list says these pots should be 20 Ohms, but the Manual and schematics state 200 Ohms.

In addition, I'd be interested to understand why these pots were moved from modulating the current at the base of QP1, QN3 in V2.1.1 to the emitter in V2.1.2 ?
Similarly, is there any reason for abandonning the constant current source provided by Q2 and Q6 in V2.1.2? Do LEDs provide better regulation by themselves ?
Thanks a lot!

Edited by user Monday, December 13, 2010 11:17:05 PM(UTC)  | Reason: Not specified

Russ White  
#3 Posted : Tuesday, December 14, 2010 4:41:43 AM(UTC)
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You are not doing the calculations right. :)
Lazybutt  
#4 Posted : Tuesday, December 14, 2010 6:02:13 AM(UTC)
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Russ White wrote:
You are not doing the calculations right. :)


Hahaha, I'm probably not doing the reasonning right, but the calculations..... Brick wall

P=R.I^2 (That one I'm taking for granted right ???)
R= 50 Ohms (that one from tha manual: "adjust the CCS pots so that the resistance across the CCS R positions (VR1 and VR2) is about 50ohm"))
I= 0.25 A (again, the manual "until the measured voltage across R1 and R2 are ~ 0.25V" R1=1 so I=0.25, and what flows through R1 flows through VR1)

Now I understand where my reasonning is wrong... I'm assuming VR1 stays at around 50 ohm.. but once adjusted, VR1/2 are probably nvery far from 50ohm, but more at around 1-2 ohm...

Am I correct ?

Edited by user Tuesday, December 14, 2010 6:02:50 AM(UTC)  | Reason: Not specified

Brian Donegan  
#5 Posted : Tuesday, December 14, 2010 6:15:56 AM(UTC)
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I=V/R and P=IV
P=V^2/R
P=.25^2/50
P=.00125W
Lazybutt  
#6 Posted : Tuesday, December 14, 2010 6:31:05 AM(UTC)
Lazybutt

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Brian Donegan wrote:
I=V/R and P=IV
P=V^2/R
P=.25^2/50
P=.00125W


Sorry, but aren't you mixing apples and carrots there ??? (i.e the voltage accross R1 with the resistance of VR1...)

if the voltage across R1 is 0.25V (as we aim as per the manual), then it just cannot be the same across VR1 (which is in series with R1) if VR1=50ohm ....

.... Or I didnt understand the voltage divider theory... Whistle

Edited by user Tuesday, December 14, 2010 6:35:29 AM(UTC)  | Reason: Not specified

Russ White  
#7 Posted : Tuesday, December 14, 2010 7:20:57 AM(UTC)
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The solution is really really simple. :)

While the placid is running measure the voltage drop across the pot (only the pot - it is in rheostat mode).

Then measure the pot resistance with the placid off. Now you can work out the actual power dissipated.

Edited by user Tuesday, December 14, 2010 7:26:00 AM(UTC)  | Reason: Not specified

Russ White  
#8 Posted : Tuesday, December 14, 2010 7:23:58 AM(UTC)
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Also we include a 20R and 200R pot for the CCS, but the 20R should always be used unless you are providing very low current of say less than 100ma.
Russ White  
#9 Posted : Tuesday, December 14, 2010 7:27:39 AM(UTC)
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Another point, one you have worked out how much current you need, you can easily replace the pot with a fixed resistor if you like, but there is no need to do so.
Lazybutt  
#10 Posted : Tuesday, December 14, 2010 7:58:42 AM(UTC)
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Got it !
Of course would have done that but I don't have my Placid (yet!) and was into thinking that VR1 should be at 50 ohm to deliver 250mA.. Now I understand thats not the case ! Thx !
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