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Part Number: TPS65150
I'm thinking about using the TPS65150 device.I am checking the loss of the circuit and I want to ask you one question.In the data sheet, the efficiency of the Boost DCDC circuit is shown as follows.Can you tell me the loss of the TPS65150 device itself?Since the device of TPS65150 does not have WEBENCH, it can not calculate the loss of the device itself.
It is hard to distinguish the real losses inside the device and the losses in the inductor or diode. A rough estimation of the losses can be done by having a look on the DC losses in the system. But for this you need to know the duty cycle of the converter and you need to keep in mind that this is a rough estimation, not an accurate calculation.
You can use the power stage designer for such a rough estimation. Please download it out here: stage designer tool&tisearch=Search-EN-Everything
Here you get the calculation for the duty cycle and based on the duty cycle and the average current and rDS(on) of the transistor you can estimate the loss in the TPS65150.
For more information on buck-boost devices have a look at .
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In reply to Brigitte:
Thank you for your reply.Regarding the loss of the BOOST circuit, I understood about the loss of Q1_FET.Loss of the device itself is almost the loss of Q1_FET?Or is it necessary to consider the loss of charge pump circuit, logic circuit and gate driver?
In reply to Kaji@PAN:
In most LCD applications, the boost converter is the main source of power. The charge pumps often just deliver 1mA or less in average, so the biggest loss is in the MOSFET and driver circuitry of the boost converter.
There are losses in the charge pumps as well and often the biggest loss is due to the topology of the charge pump. Charge pumps can only deliver a multiple of the input voltage and if you regulate it, it is regulated by creating losses as in a linear regulator. So the biggest loss is often due to the regulation. The loss inside the IC is the maximum possible output voltage times the output current minus the regulated output voltage times the output current. Please have a look into this app note:
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