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New design

Posted by: mike_mccrady on
I have attached the schematic for a new power supply design and would like for some feedback. It is designed for 6.5VDC @ .5A w/peak output at 3.5-4 Amps for no more than 400ms and 12VDC @ .01A w/peak output at 2 Amps for 250ms. I also attached the PIXI Designer file with the added extension of ".doc". Thanks, Mike

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DisconnectPS.pdf 45.22 KB
DisconnectPS .pxlpk_.doc 37.43 KB

评论

Submitted by PI-Spock on 09/15/2008
Issues with the design 1) On the design spreadsheet the total power at the top of spreadsheet (~ 22W) does not match the total power at the bottom (~ 46 W) Thats a problem and you may not be able to deliver the power! 2) Your output snubber caps C1 and C6 are too big Need to be about 1 nF 3) No need to use glass passivated diode on input bridge. regular diodes are ok. 4) You are using a 160 V TVS in porimary snubber... You can use a 180 V or 200 V zener instead 5) You end up with way too much leakage with secondary winding foil and 17 turns. I'd recommend you stick with quad filar AWG 25 wire for both outputs instead 6) No need for R2 resistor. In any case 200 ohms is too much. I would not use more than 22 ohms. 7) BP pin capacitor is 0.1u. This needs to be at least 0.22u
Sorry I haven't gotten back to you sooner, other projects. I appreciate the comments and I will make the changes. On 1)The peak power on the 6.5VDC is 22W for only about 25-100 ms & on the 12V output it is 24W for 250 ms and the loads will never be at the same time.
Submitted by husee on 11/20/2009

Aimed at system designers whose interest focusses on other fields, this note reviews the basic power supply design knowhow assumed in the rest of the book.
In mains-supplied electronic systems the AC input
voltage must be converted into a DC voltage with the
right value and degree of stabilization.
Figures 1 and 2 show the simplest rectifier circuits houston homes for sale.
In these basic configurations the peak voltage
across the load is equal to the peak value of the AC
voltage supplied by the transformer’s secondary
winding. For most applications the output ripple produced
by these circuits is too high. However, for
some applications - driving small motors or lamps,
for example - they are satisfactory.
If a filter capacitor is added after the rectifier diodes
the output voltage waveform is improved considerably car finance.
Figures 3 and 4 show two classic circuits commonly
used to obtain continuous voltages starting
from an alternating voltage. The Figure 3 circuit uses
a center-tapped transformer with two rectifier diodes
while the Figure 4 circuit uses a simple transformer
and four rectifier diodes.

Figure 1 : Basic Half Wave Rectifier Circuit.

Figure 2 : Full Wave Rectifier Wich uses a Center-
tapped Transformer bad credit loans.

Figure 3 : Full Wave Rectified Output From the
Transformer/rectifier Combination is filtered
by C1.

Figure 4 : This Circuit Performs Identically to that
Shown in Figure 3.

Figure 5 shows the continuous voltage curve obtained
by adding a filter capacitor to the Figure 1 circuit.
The section b-c is a straight line. During this time it
is the filter capacitor that supplies the load current.
The slope of this line increases as the current increases,
bringing point c lower. Consequently the diode
conduction time (c-d) increases, increasing ripple.
With zero load current the DC output voltage is equal
to the peak value of the rectified AC voltage los angeles mortgage loans.
Figure 6 shows how to obtain positive and negative
outputs referred to a common ground. Useful design
data for this circuit is given in figures 7, 8 and 9. In
particular, the curves shown in Figure 7 are helpful
in determining the voltage ripple for a given load current
and filter capacitor value. The value of the voltage
ripple obtained is directly proportional to the
load current and inversely proportional to the filter
capacitor value.