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Kai Klaas
02/14/06 07:01
Read: 1062 times
Germany


 
#109945 - Answers II
Responding to: Suresh R's previous message


Suresh said:
Let the o/p of regulator be for instance 10 V. since this is been fedback to the Adj Terminal through the 120E resistor , will this feedback, don't experience any effect due to R7?

No, LM338 produces a constant voltage of 1.24V between output and adjust terminal. It's time to have a look into datasheet...

Suresh said:
So how does this lead to a soft start when the device is turned 'ON'.
Is it that the 10mA current is being avoided when the device is turned 'ON'. why is it that C9 should be fully charged with the help of R8 when Vbe of Q4 becomes smaller than .7V.

Directly after connecting the input voltage to input pin, LM338 produces a constant voltage of 1.24V dropping across R7 and forcing a current of about 10mA to flow through it.
Because C9 is discharged, base of Q4 is on ground potential, directly after power on. This makes Q4 turning on, allowing most of the 10mA current flowing from emitter to collector (ground). In this moment voltage drop across Pot R6 and R8 is limited to about 0.7V. So, only a small fraction of the 10mA current is flowing through them.

The more now C9 becomes charged by the base current of Q4, the more the base potenial rises and with this the emitter potential, which is about 0.7V higher than base potential. By this, voltage dropping across Pot R6 increases and more and more current of the 10mA current is flowing through it. But across R8 voltage drop remains constant, about 0.7V, so far.

After a while Q4 can no longer conduct current because Ube becomes smaller and smaller. Without R8 Ube would not fall under 0.4V or so, because all the time a certain leakage current could flow. To fully turn-off Q4 R8 is needed, which charges C9 furtherly, resulting in a Ube of finally 0V.

Suresh said:
during plating operation i measured the following.
current set using R17 = 1A
voltage drop accross load = 1.5V
for this condition, i would like to know what would be the voltage drop accross the resistor R12 which would be appearing in the inverting i/p of U2A. (or) in short, how could i know this is the voltage dropping in R12. how should i calculate it.
Also kindly tell me how to calculate the power dissipated in MJE 3055( Q3 ).

If you want a current to flow of 1A, then you need a setting of Pot R17 so that 1V is reaching the non-inverting input of U2B. As this voltage is copied to R12, you will observe a voltage of also 1V across R12. As R12 = 1Ohm, this causes a current to flow through R12 of 1V / 1Ohm = 1A.
U2A controls Q2 and Q3 so that the voltage drop across R12 remains contant. This is the way how this constant current source works.

Heat dissipation of Q3:
If output voltage of LM338 is 10V, a current of 1A is flowing and 1.5V drops across the load, then a voltage of about 10V - 1.5V - 1V (voltage drop across R12!) = 7.5V drops across Q3. So, heat dissipation of Q3 is 7.5V x 1A = 7.5W.
Heat dissipation of LM338 is even higher: If input voltage of LM338 is 20V and output voltage is 10V, then heat disspation is (20V - 10V) x 1A = 10W! Good cooling is needed for the LM338.
It's a good idea to reduce input voltage of LM338 a bit, if ouptut voltage of LM338 need not to be higher than 10V.
Take care, also Q2 might need cooling!

Suresh said:
Due to its requirement for high i/p to produce a stable o/p.

No more than 3V between output and input is needed!

Kai

List of 21 messages in thread
TopicAuthorDate
3A constant current source      Suresh R      02/11/06 04:41      
   Where did you find it?      Andy Neil      02/11/06 05:56      
   Ripple      Andy Neil      02/11/06 06:00      
      plating purpose      Suresh R      02/11/06 09:45      
         I got it from a person known to me      Andy Neil      02/11/06 15:04      
            replies      Suresh R      02/12/06 09:24      
               plating circuit      Terry Lingle      02/12/06 11:40      
                  Compliance of source      Steve M. Taylor      02/12/06 13:51      
                     yes      Terry Lingle      02/12/06 16:21      
         Think about that      Neil Kurzman      02/11/06 16:08      
   Component Purposes      Bruce Bailey      02/11/06 12:58      
   This circuit will never work!      Kai Klaas      02/11/06 19:46      
      Could it be?      Neil Kurzman      02/11/06 23:55      
         Function      Kai Klaas      02/12/06 20:08      
            issues      Suresh R      02/13/06 02:42      
                plating quality control.      Terry Lingle      02/13/06 07:29      
               Answers      Kai Klaas      02/13/06 11:36      
                  followups      Suresh R      02/14/06 02:20      
                     Answers II      Kai Klaas      02/14/06 07:01      
                        followups & Back EMF's?      Suresh R      02/15/06 03:59      
                           Floating regulator      Kai Klaas      02/15/06 09:27      

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