I did the performance measurement of the
switching regulator which used LM2575-5.0.
The way
of measuring is the way of increasing the output power, making the load
resistance value small gradually. The measurement item is the change of
the output voltage and the input electric current.
The characteristic when the load is rapidly changed is the
important characteristic of the regulator. However, because I don't have
the measurement receptacle, I am not measuring.
I calculate the output current, the input power, the output power
and the efficiency by the formula below each.
| Output current(A) |
= |
Output voltage(V) / Load resistance
value(ohm) |
| Input power(W) |
= |
Input voltage(V) x Input current(A)
[Input voltage = 12V] |
| Output power(W) |
= |
Output voltage(V) x Output current(A) |
| Efficiency(%) |
= |
(Output power(W) / Input power(W)) x
100 |
_dosyalar/blank_1_nv.gif)
_dosyalar/bol_pnk.gif) The change of the voltage and
the current by the load change
_dosyalar/ckt22_71e.gif)
When decreasing the load gradually, the output is
maximized at about 2 ohm. At the load resistance value below it, the
output voltage declines and the output power falls too. I think this
change according to the limitation circuit of the output current to
prevent from the damaging of the regulator by the over-current. At
the data sheet, the limitation value is 2.2 A.
The guarantee value of the output current of LM2575-5.0 is 1
A. Because it is, when pouring the 2-A output current into the
continuation, the normal operation of the regulator isn't
guaranteed.
Be
careful. |
The electric power change by
the load change
_dosyalar/ckt22_72e.gif)
The efficiency of LM2575-5.0 is about 80%. It is
the very good value. Because the input electric power in case of 1 A
of the output currents(5 W output) is 6 W, the loss of the regulator
is 1 W.
In the measurement, about 11 W(2.2 A
of output currents) are measured as the maximum output. Because the
regulator loss in this case is about 1.5 W, the efficiency is about
88%. It is the very high efficiency.
This
time, I attached the heatsink to the regulator but if being this
loss, the heatsink is unnecessary.
In the
result of the thermometry, the temperature rise
is the small value. |
The ripple
voltage
In case of the switching regulator, the
ripple voltage occurs to the output in the relation of the
operation.
The output wave form
in the no-load
condition |
The output wave form
in the 5-W
output |
With the photograph above, the left side
shows the ripple voltage which appears in the output in the
no-load condition. Because the ripple voltage is about 10
mVp-p, it becomes about 0.2% of the ripples at the 5-V output
voltage.
The photograph in the right
shows the ripple voltage when the load current is 1 A (5 W
output). Because the ripple voltage is about 100 mVp-p, it
becomes about 2% of the ripples at the 5-V output voltage.
However, by putting the secondary filter which combined the
several-µH coil and the hundreds-of-µF capacitor, the ripple
can be decreased. |
|
_dosyalar/ckt22_73e.gif) |
The secondary filter output wave form
in the 5-W output |
The photograph above is the one to have observed
the output voltage of the secondary filter which connected with the
output of the regulator.
The ripple voltage
is hardly observed. It has less than 5-mV voltage(It is 0.1% at 5
V). | I observed the
photographs above in the same range.
|
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_dosyalar/ckt22_71p.jpg)
_dosyalar/ckt22_72p.jpg)
_dosyalar/ckt22_73p.jpg)
_dosyalar/blank_1.gif){kind=spacer}