What is the effect of ripple on the power dissipation of a resistive load in a dc-dc converter step up-down config?
The effect of ripple on the power dissipation of a load
- Thread starter leodavinci90
- Start date
In a resistive load the current and the voltage are in phase. As such a slight increase in voltage and current will produce a slight increase in power dissipation. A slight decrease will produce a slight decrease in power dissipation. If the average DC output is the same as the DC output there should be no measurable difference.
Is this the effect of "Voltage Ripple" ?
Attachments
circuit is attached
That is not strictly a resistive load because C1 is in parallel with the load. Is your generator connected to the gate of the MOSFET putting out a sine wave or a square wave? As such it is not really ripple but converting DC to AC for no apparent purpose. I'm not sure what your circuit is trying to do because the inductor has no place to dump the current when the switch turns off. The diode is blocking it.
Hi, Are u a power electronics specialist?
This last statement isn't true.
Consider the extreme case -- Vdc = 0 V. If there is ANY ripple, then power will be dissipated in the load while if there is no ripple then no power will be dissipated. This is also true for ripple riding on a much larger average and is due to power being nonlinear, so an excursion of ΔV above the DC level results in MORE additional power than an excursion of ΔV below the DC level results in less.
Consider Rload being 10 Ω and Vdc = 100 V and ΔV being 10 V. The power dissipated without ripple is 1000 W. On the up side voltage is 110 V and the power dissipated is 1210 W (an increase of 210 W) while on the downside it is 810 W (a decrease of 90 W). So the total power dissipation goes up. In fact, if you have a sinusoid riding on a DC current then the total power dissipation is the sum of the power that would have been dissipated individually.
According to circuit attached I am calculating power efficiency as attached in this new image.
Is this correct, comment please?
Is this correct, comment please?
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Hello,
Without looking at your circuit yet, the affect of a ripple voltage in a resistive load is that at the peaks the power is greater but at the valleys the power is less, so the load sees more power at the peaks. The average power is given by the integral for average power, and very often this relates to the RMS values of the voltage and current.
The average power is:
Pavg=(1/T)*Integral i(t)*v(t) dt [integrated from 0 to T]
where T is the period of the signal.
That integral works for any waveshape including triangle wave, curved triangle waves, etc., but you can usually approximate converter outputs as triangle waves.
Without looking at your circuit yet, the affect of a ripple voltage in a resistive load is that at the peaks the power is greater but at the valleys the power is less, so the load sees more power at the peaks. The average power is given by the integral for average power, and very often this relates to the RMS values of the voltage and current.
The average power is:
Pavg=(1/T)*Integral i(t)*v(t) dt [integrated from 0 to T]
where T is the period of the signal.
That integral works for any waveshape including triangle wave, curved triangle waves, etc., but you can usually approximate converter outputs as triangle waves.
