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### Network # Average Current

J

#### Jack

Jan 1, 1970
0
hi all

For a 1n540x diode average forward current is given 3amps, read in the data sheet.

Is this average current the same what i measure in my dmm.

J

#### John Popelish

Jan 1, 1970
0
Jack said:
hi all

For a 1n540x diode average forward current is given 3amps, read in the data sheet.

Is this average current the same what i measure in my dmm.

Almost certainly. But I never use a component at its absolute maximum
rating, voltage, current, or power.

J

#### Jack// ani

Jan 1, 1970
0
John Popelish said:
Almost certainly. But I never use a component at its absolute maximum
rating, voltage, current, or power.

Thanks John

I probably knew that, just wanted to get assured of it.

Ok now it will be wise to ask, suppose i want to get 3amp dc out of it
using bridge rectifier, what should be the transformer amps rating.

By calculation it should be 3.33amps,

Iav=2.82*Irms/pi

so, Irms=3.33amps

It that correct? And will it depend upon full/half wave rectifier?

J

#### John Popelish

Jan 1, 1970
0
Jack// ani said:
Thanks John

I probably knew that, just wanted to get assured of it.

Ok now it will be wise to ask, suppose i want to get 3amp dc out of it
using bridge rectifier, what should be the transformer amps rating.

By calculation it should be 3.33amps,

Iav=2.82*Irms/pi

The diodes have an average current rating because they drop a nearly
constant voltage over a wide range of current. So the heat produced
is fairly closely proportional to the average current, as long as you
don't get into extreme cases (of very large pulses widely spaced).
Transformers are heated by the square of current, like resistors,
because their windings are resistive. So they have an RMS current
rating. RMS stands for square root of the mean (average over a
representative time period) of the square of the instantaneous
current. The proportion of the RMS current from the transformer to
the average DC rectifier output current depends a lot on what follows
the rectifier. If the load is a resistor, then the RMS load current
is also the RMS transformer current. In this case, the RMS load
current is 1.11 times the rectifier average current.

If the load is primarily capacitive, all the rectifier current is
confined to the brief periods when the transformer voltage is higher
than what is stored in the capacitor, so the RMS current can get quite
a bit higher than the average current. This affects the current
rating required in the transformer.
so, Irms=3.33amps

It that correct? And will it depend upon full/half wave rectifier?

It is correct for a resistive load and a full wave rectifier (load
current proportional to absolute value of instantaneous transformer
voltage). Half wave rectifiers pose additional problems for
transformers that involve core saturation that don't occur with full
wave rectification. And the ratio of RMS to average current is also
different (higher).

What load will your rectifier be driving?

J

#### Jack// ani

Jan 1, 1970
0
John Popelish said:
If the load is primarily capacitive, all the rectifier current is
confined to the brief periods when the transformer voltage is higher
than what is stored in the capacitor, so the RMS current can get quite
a bit higher than the average current. This affects the current
rating required in the transformer.

Just two more questions, John

1. Reactance offered by a capacitive load is always higher that a
resistive load, then why should it extract more current?

2. Why do capacitor charges to peak ac voltage if placed across a
bridge rectifier? Without a cap i measured 11v, and it increased to
16v, after hooking a cap?

Thanks for your help

J

#### John Popelish

Jan 1, 1970
0
Jack// ani said:
Just two more questions, John

1. Reactance offered by a capacitive load is always higher that a
resistive load, then why should it extract more current?

Reactance is a linear concept that applies to sine waves. To get your
mind around a capacitive input rectifier filter, you have to think in
the more general differential description of capacitance. I=C*(dv/dt)
current equals the capacitance times the time rate of change of
voltage across the capacitor.

As long as the capacitor voltage is equal to or greater than the
transformer voltage, the rectifier isolates the two. But the moment
the transformer wave rises above the capacitor voltage, the rectifier
is essentially a short circuit, and the voltage on the capacitor must
rise as fast as the transformer wave is rising, regardless of how much
current that takes. So the current into a capacitive filter is narrow
sort of half sine wave pulses that occur on the part of the
transformer voltage wave just before the peak voltage. Since the
transformer windings are heated by the RMS current, a pulse waveform
like this has a much higher RMS value than the average of the current
in those pulses. This is what the 'squared' part of the RMS does. It
is not unusual to have to double (or more) the transformer RMS current
rating relative to the DC output average current when using a
capacitor input rectifier filter to take care of this higher winding
RMS current. The exact ratio depends on how much leakage reactance
there is between primary and secondary windings that tends to spread
out the charging pulses, by sagging the waveform a bit while the cap
is charging, lowering the slope a bit. If you look at the transformer
waveform with a scope, you can see the flattened spot on the wave
where the cap is charging just before peak voltage.
2. Why do capacitor charges to peak ac voltage if placed across a
bridge rectifier? Without a cap i measured 11v, and it increased to
16v, after hooking a cap?

The diodes act as check valves, pumping the cap up all the way to the
transformer waveform peak voltage and then turning off, leaving that
voltage trapped in the capacitor. A resistor load on the rectifier
keeps it on the whole waveform so the resistor voltage is the same as
the transformer waveform (except for the inversion of one half). your
meter reads the average of the rectified transformer waveform instead
of the peak value.

J

Jan 1, 1970
0
Thanks a lot.

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