About Light Load Disadvantage of Resonance Converters

Thread Starter

Trunke10

Joined Apr 1, 2022
3
Recently I've bee studying Robert W. Erickson's Fundamentals of Power Electronics. At the chapter of resonant conversion, the book says that "The circulating tank currents are independent, or weakly dependent on, the load current, and lead to poor efficiency at light load, which is confusing for me. According to my understanding, the effective resistance Re of the dc-dc converter increases at light load. Therefore, the peak magnitude of the tank current would decrease, which conflicts to what the author states. Could someone tell me where I'm wrong?
 

Papabravo

Joined Feb 24, 2006
18,968
The efficiency of a converter is determined by the power delivered to the load divided by the power consumed from the source. For a fixed amount of input power, the efficiency will decrease as the power to the load is reduced.
 

Thread Starter

Trunke10

Joined Apr 1, 2022
3
The efficiency of a converter is determined by the power delivered to the load divided by the power consumed from the source. For a fixed amount of input power, the efficiency will decrease as the power to the load is reduced.
First of all, thank you for your reply. I agree of what you said, efficiency is load power/source power. But, in light load conditions wouldn't input current (input power) decrease in resonant converters?
 

Ian0

Joined Aug 7, 2020
5,819
First of all, thank you for your reply. I agree of what you said, efficiency is load power/source power. But, in light load conditions wouldn't input current (input power) decrease in resonant converters?
Yes, but not as much as it should. The circulating current in the resonant circuit remains, and it must pass through some resistance, creating loss.
 

Thread Starter

Trunke10

Joined Apr 1, 2022
3
Yes, but not as much as it should. The circulating current in the resonant circuit remains, and it must pass through some resistance, creating loss.
Thanks a lot. I think I understand what you're saying. Since we operate at frequencies slightly above/below resonant frequency, we get an inductive/capacitive load. Therefore, input impedance of the tank doesn't act "purely resistive", which implies that input current wouldn't decrease "as much as it should". Am I correct?
 

Ian0

Joined Aug 7, 2020
5,819
The resonant circuit is constantly transferring energy from the capacitors to the inductors and back again. Think of it like a flywheel. You put energy in and you take energy out. But if you stop taking energy out of the system you still have to put some energy in to keep the flywheel spinning.
You don’t have to put that energy in to a non-resonant system. So why not simply use a non-resonant converter? Because there are other losses associated with the non-resonant converter, and a high load those losses are higher than the losses from the resonant converter.
 
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