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Considering that the voltage of the transformer will drop when it is loaded, and it can be 6.5V when it is unloaded, the effective value of the secondary voltage of the transformer is:

(6.5+0.7 * 2)/1.414 = 5.6v, where 0.7 is the voltage drop of the rectifier, and the bridge rectifier needs two diodes to work at the same time.

Turn off the diode 1N400 1.