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Re: MOSFet Bridge Rectifier problem...
The schematic is just for a linear regulator with a bypass transistor to assist the LM317.
Each regulator is set to 24V which is fine but to get 48V you have to stack one above the other. In other words, join the ground of one to the output of the other and take your 48V from the other two connections. In this configuration you must use two isolated transformer secondaries. They can be on the same transformer or on two separate transformers but they must not connect directly to each other.
I see a problem though, assuming you are using lead-acid batteries, the charging voltage needs to be about 55V. If you really need a steady 30A, you will need transformers rated at no less than 1.7KW which will be big and heavy.
Remember that you cannot defy the laws of physics, it makes no difference whatsoever if you use BJT or FET devices in a linear regulator. Your intention is to convert unused energy into heat and the amount will be the same regardless of transistor type.
I'm not sure what you think you will gain by using an LM338 instead of an LM317. The transistor is to carry the excess current that the regulator can't handle so it does most of the work. You would share the current slightly better but the overall power would be the same.
Brian.
The schematic is just for a linear regulator with a bypass transistor to assist the LM317.
Each regulator is set to 24V which is fine but to get 48V you have to stack one above the other. In other words, join the ground of one to the output of the other and take your 48V from the other two connections. In this configuration you must use two isolated transformer secondaries. They can be on the same transformer or on two separate transformers but they must not connect directly to each other.
I see a problem though, assuming you are using lead-acid batteries, the charging voltage needs to be about 55V. If you really need a steady 30A, you will need transformers rated at no less than 1.7KW which will be big and heavy.
Remember that you cannot defy the laws of physics, it makes no difference whatsoever if you use BJT or FET devices in a linear regulator. Your intention is to convert unused energy into heat and the amount will be the same regardless of transistor type.
I'm not sure what you think you will gain by using an LM338 instead of an LM317. The transistor is to carry the excess current that the regulator can't handle so it does most of the work. You would share the current slightly better but the overall power would be the same.
Brian.