Minimum/Breakout circuit for R5F212L4SNFP

I have R5F212L4SNFP MCU & E8 debugger (not E8a).
I have made a breakout circuit for this MCU.

1. Is this the minimum circuit ok?
2. I have added 4.7K/0.1uF for reset circuit, it is ok? Hardware manual says 4.7k or more. It dont specify 0.1uF cap
3. Mode pull resistor is 4.7uF
4. Vcc & Vss decoupled by 0.1u+10u

Below image is for E8 used with R5F212L4SNFP. Do it give full step by step debug capability using E8 or any other pin required?

  • The E8 isn't supported with drivers on Windows 7 64 bit or newer. You'd have to use XP 32 bit for development.

  • In reply to Calvin Grier:


    I am using Window 7 32 bit. I have ordered some pieces of MCU, will solder & then try to use E8 debugger on it. However E8 get detected on window 7 32 bit.

    Also as I have asked is selection of 4.7k/0.1uF for reset & 4.7k for mode is ok?

    Is this is the minimum circuit with which device will work as I am making a minium breakout baord.