2021-09-25 05:30 AM
2021-09-27 02:51 AM
You should correct the feedback voltage divider because with the current sizing of R73 and R74 you will only see 1.055V at the output. However, if you swap the two values of R73 and R74, it will fit and you will get 3.3V.
The capacitor C52 can also perhaps be reduced to 10nF or omitted. If the signal MEM_3V3_EN is controlled by another device, the discharge current from C52 may become too high.
Good luck!
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/Peter
2021-09-27 04:18 AM
Could you please elaborate below comment.
The capacitor C52 can also perhaps be reduced to 10nF or omitted. If the signal MEM_3V3_EN is controlled by another device, the discharge current from C52 may become too high.
2021-09-27 04:28 AM
The capacitor C52 is charged to 3.3V via R86. The pin EN is also connected to the signal MEM_3V3_EN, so that one can assume that e.g. a microcontroller or another logic gate is connected there to pull EN to low in order to deactivate the LDO. Depending on the output impedance of this pin, a very short but very large current could flow from C52 into the driver pin, which without current limitation can be dangerous for it.
2021-09-27 04:34 AM
Actually MEM_3V3_EN is just a signal name. It is not connected to anywhere. Only NVCC_3V3 power rail has role on the EN pin. So value of C52 is OK..right?
2021-09-27 04:36 AM
Yes. The capacitor can also be omitted because it is not necessary - R86 is just a pull-up.
2021-09-27 04:37 AM
Actually Resistor and capacitor is added as RC circuit
2021-09-27 04:44 AM
If you have intentionally built in the time delay of about 1ms, C52 can of course remain.