B
Ban
I wanted to share some insights I gained by spicing up a normal supply, in
this case a 12V/4Adc, consisting of a transformer, bridge rectifier and
smoothing capacitor.
In another thread some of us gave their opinions about the choice of the
required components, and I tried to verify.
1. The transformer can be modeled with an ideal ac-source plus interiour
resistance, as we know from any DC-supply. Almost all losses are due to
resistance of the windings. The manufacturer specifies a certain obtainable
output power, consisting of the product Vrms * Irms = P(VA). The input power
is by some percentage higher and the difference is due to losses. It is the
difference between the open circuit voltage Vo and the rated voltage Vrms .
Ri = (Vo-Vrms) / Irms.
2. With the rectifier and cap the load current is not sinusoidal. Only when
the momentary voltage is higher than the remaining voltage on C plus the 2
diodes Vf, the current starts charging the cap. it reaches the max. before
the voltage peak and goes symetrically down to zero. The max. current is a
multiple of the average and the rms value is much higher as with a sine.
Both are mainly dependent on the ratio of the interiour resistance to the
load resistance, not much on the capacitor size. A higher Ri reduces the
peak current a lot, since Ipeak is proportional to 1/sqrt(Rload*Ri)
3. The required size of the transformer is mainly dependent on the Irms of
the current thru its windings, which dissipates in the Ri and heats up the
transformer. So a transformer with a very low Ri has not much advantage,
since the Ipeak and Irms go up accordingly. The losses in the diode bridge
raise too and can be estimated in the datasheet where there are curves for
Ploss (Ipeak/Iav). Since this is for a single diode you have to double the
ratio and multiply with 4. The rectifier losses plus the losses in the cap
Irms^2*ESR plus the output power have to be lower than the transformer
rating.
4. The ripple voltage is not so much lower with a low Ri, mainly dependent
on the cap size.
to be continued
this case a 12V/4Adc, consisting of a transformer, bridge rectifier and
smoothing capacitor.
In another thread some of us gave their opinions about the choice of the
required components, and I tried to verify.
1. The transformer can be modeled with an ideal ac-source plus interiour
resistance, as we know from any DC-supply. Almost all losses are due to
resistance of the windings. The manufacturer specifies a certain obtainable
output power, consisting of the product Vrms * Irms = P(VA). The input power
is by some percentage higher and the difference is due to losses. It is the
difference between the open circuit voltage Vo and the rated voltage Vrms .
Ri = (Vo-Vrms) / Irms.
2. With the rectifier and cap the load current is not sinusoidal. Only when
the momentary voltage is higher than the remaining voltage on C plus the 2
diodes Vf, the current starts charging the cap. it reaches the max. before
the voltage peak and goes symetrically down to zero. The max. current is a
multiple of the average and the rms value is much higher as with a sine.
Both are mainly dependent on the ratio of the interiour resistance to the
load resistance, not much on the capacitor size. A higher Ri reduces the
peak current a lot, since Ipeak is proportional to 1/sqrt(Rload*Ri)
3. The required size of the transformer is mainly dependent on the Irms of
the current thru its windings, which dissipates in the Ri and heats up the
transformer. So a transformer with a very low Ri has not much advantage,
since the Ipeak and Irms go up accordingly. The losses in the diode bridge
raise too and can be estimated in the datasheet where there are curves for
Ploss (Ipeak/Iav). Since this is for a single diode you have to double the
ratio and multiply with 4. The rectifier losses plus the losses in the cap
Irms^2*ESR plus the output power have to be lower than the transformer
rating.
4. The ripple voltage is not so much lower with a low Ri, mainly dependent
on the cap size.
to be continued