[SI-LIST] Re: DC/DC converter output dip

  • From: "Pratt, Gary" <gary_pratt@xxxxxxxxxxx>
  • To: si-list@xxxxxxxxxxxxx
  • Date: Wed, 6 Aug 2003 14:15:36 -0400

Jean-Pierre, 
How is the common 1.2 reference voltage generated?  And, what supply is
used for the op-amps?  From your first description it sounded like a
control loop problem (Veneble being an excellent suggesting for solving
that type of problem).  But, now, it sounds like you may have a
relaxation loop related to the switching of the MOS transistor and its
affect on the 3.3 supply.  

Can you provide a JPEG of the schematic containing the supplies,
op-amps, MOS transistors and associated circuitry? 

Gary 



-----Original Message----- 
From: Jean_Pierre.Bouthemy@xxxxxxxxxx 
[ mailto:Jean_Pierre.Bouthemy@xxxxxxxxxx
<mailto:Jean_Pierre.Bouthemy@xxxxxxxxxx> ] 
Sent: Wednesday, August 06, 2003 12:21 PM 
To: si-list@xxxxxxxxxxxxx 
Cc: Tom.Currelly@xxxxxxxxxxx 
Subject: [SI-LIST] Re: DC/DC converter output dip 



Hi Chris, 

The design is a little bit complex. Let me try to clarify. 
A +5V isolated converter is used as a "housekeeping" voltage that comes 
first in order to supply the monitoring circuits. 
I have a 3.3V isolated DC converter whose output feeds 3 non-isolated DC

converters: 2.5V, 1.8V and 1.5V. 
The 4 voltages 3.3V, 2.5V, 1.8V and 1.5V are all monitored with the same

way. They each feed an op-amp with a resistor divider on the input+. The

input- is a common reference voltage of 1.2V. The feedback is made with
a 
1Mohm resistor between input+ and the output. 
The output of the op-amp are pulled-up together to the "housekeeping"
+5V 
and feed a circuit to drive the MOS transistors at the output of the DC 
converters. 
When all four voltage have correct levels, the common outputs are at
high 
level, and a signal is activated to open the MOS so that the devices on
the 
board can be supplied. 
Hope you better understand my design. 

The noise I have on the +2.5V output is not like ripple. The signal is
high 
at 2.5V during 90ms and then a 2ms pulse down to 1.5V. 
I have investigated and it seems like it is due to the 90ms ripple noise
on 
the +5V output. But the problem is still nearly the same: when I remove
the 
resistor of the the +5V monitoring circuit, I have no noise anymore. 
Strange isn't it? 

Could a simple RC filter at the output of the +5V DC converter do the
job 
of phase margin compensation? 
Any help will be welcome. 

Jean-Pierre 


-------------------------------------------------------------------- 
Jean-Pierre BOUTHEMY 
Hardware Design Engineer 
ALCATEL - Mobile Networking Division (MCG) 
Mobile Core R&D 
+33 (0)2 99 26 08 18 
mailto:jean_pierre.bouthemy@xxxxxxxxxx
<mailto:jean_pierre.bouthemy@xxxxxxxxxx>  
-------------------------------------------------------------------- 


|---------+------------------------------> 
|         |           Christopher        | 
|         |           Jakubiec           | 
|         |           <Christopher.Jakubi| 
|         |           ec@xxxxxxx>        | 
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|         |           lists.org          | 
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|         |                              | 
|         |           06/08/2003 17:52   | 
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Jean-Pierre, 

As I understand from your description you have a 2.5V output from the
dc-dc 

converter that feeds into an op-amp comparator by using a resistor
divider 
on 
the + and - inputs to the op-amp.  Is this correct?  It sounds like a
very 
basic 
comparator setup.  Are you using negative feedback with nothing
connected 
inbetween the output and input of the op-amp? 

Chris Jakubiec 
Sun Microsystems 

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