powerconversion efficiency of cc-rectifier

[vijith133]

hi ,
i am designing a cc-rectifier with the help of cadence spectre, but am confused how to calculate the power conversion efficiency(PCE), can anyone help me with this issue, the output power is (Vdc)^2/Rout but how would i calculate the input power?
note : its a floating input

thank you

 

[erikl]

Your (caps + rectifier circuit + load) must have a (time-dependent) impedance, so there will be a time-dependent potential difference over your cc-source. Just integrate and multiply both the time dependent voltage × current.

 

[vijith133]

Your (caps + rectifier circuit + load) must have a (time-dependent) impedance, so there will be a time-dependent potential difference over your cc-source. Just integrate and multiply both the time dependent voltage × current.

Hi,
thanks for the reply
but the input power is it the difference voltage and current in the input to be integrated like

integral((vplus(t)-vminus(t))*(iplus(t)* iminus(t)))

and output power

integral(vout(t)*iout(t))

within the same interval of time
but the power conversion efficiency computed is very low compared to the previous works

 

[erikl]

... the input power is it the difference voltage and current in the input to be integrated like
integral((vplus(t)-vminus(t))*(iplus(t)* iminus(t)))

If you replace the latter "*" sign by a "-" sign, and take the absolute values of the *minus(t) values, this should fit:
integral((vplus(t)-|vminus(t)|)*(iplus(t)-|iminus(t)|))

... the power conversion efficiency computed is very low compared to the previous works

Yes, unfortunately. It's a pity, isn't it? :-(

 

[vijith133]

thanks for the help ,
sorry for the "*" sign , what i meant was "-"
i can fix on that equation its giving me better results..... when i sweep the resistance values, the efficiency at a particular vale of resistance was about 76% , but i am not able to judge if it is right. besides i tried another equation to calculate the input power.... but i dont know wheather it sounds reliable

integral((vplus(t)*(iplus(t)) +(vminus(t)* iminus(t)))

if you dont mind could please tell me if it is worth , because i dont get the same results as the equation which you have suggested

thanks

 

[erikl]

... the efficiency at a particular vale of resistance was about 76% , but i am not able to judge if it is right.

Not too bad! You've probably found the max. efficiency value for an optimum load resistance. But of course the efficiency also depends strongly on the value and form of the input power.

You know you have to integrate over a full period of time, and then divide by this time period:

t=t_period
∫((vplus(t)-|vminus(t)|)*(iplus(t)-|iminus(t)|)*dt) / t_period
t=0

integral((vplus(t)*(iplus(t)) +(vminus(t)* iminus(t)))

No, I don't think this is correct.