CY8C5888AXQ-LP096100-TQFP Supply

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Anonymous91
Level 1
Level 1
First reply posted First question asked Welcome!

Hi all,

i am a little confused regarding the supply concept of a Psoc5 CY8C5888AXQ-LP096100-TQFP and  have a question regarding the supply.

In the data sheet of the psoc i have found a figure for an example schematic for a 100 pin TQFP part with power connections. In our application we have a DC/DC converter for a supply voltage of 5V. Regarding the supply i have come across the regulated and unregulated mode of the psoc. Due to our 5V supply, as far as i understood i have to apply the 5V to all the VDDx (VDDio,VDDA and VDDD) of the PSoc in order to use the unregulated mode. In this case, i dont have to supply the VCCx pins, but i have to connect both of those without any supply. Can someone please confirm that this consideration is correct? How does the Psoc "know" that im using the internally unregulated mode? Do i have to do any further settings in the Psoc Creator itself, or is this mode recognized automatically by the Psoc MCU?

I also saw pins for VBAT/VBOOST. As far as i understood those are for powering the Psoc with an integrated boost converter in order to supply the psoc with a single cell battery or something like that. Since we have a DC/DC converter, we dont need those pins. My problem ist, that in the example schematic the VBOOST, VBAT and VSSB pin are simply connected to ground, but in Figure 6-7 of the Datasheet (Application of Boost Converter not powering PSoC device) there is an external circuit with the note, that all components and values are required. Since we are not using the internal boost converter and a supply voltage connected to VBAT, is it enough to simply connect VSSB,VBAT and VBOOST to ground as shown ine the example schematic?

Many thanks for your help.

 

 

 

 

 

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1 Solution
BiBi_1928986
Level 7
Level 7
First comment on blog 500 replies posted 250 replies posted

Hello.

For your 5V PSoC application, use the schematic you attached, "Psoc5_Example_schematic_Power_Connections.PNG".

Since 5LP is powered by 5V, it will use unregulated power mode.  So, you don't need to do anything with the internal 1.8V regulator. There's nothing to program/initialize the internal 1.8V regulator when operating in unregulated power mode.  What's important here is, only connect capacitors to the pins Vcca and Vccd.  Do Not Connect Vcca nor Vccd to any other voltage rail.  MANY-MANY people make this mistake!  All other capacitors will connect to their respective pins as shown in "Psoc5_Example_schematic_Power_Connections.PNG".

As for VBAT/VBOOST, make the ground connections as shown in schematic "Psoc5_Example_schematic_Power_Connections.PNG".
The text/figure of the other schematic shows what must be connected IF the VBAT/VBOOST feature is being used.   Since you are not using this feature, the associated pins are tied to ground.

For a more complete description of PSoC 5LP power modes, see the 5LP Arch TRM
https://www.cypress.com/file/123561/download

FYI
Vdda has to be greater than or equal to the highest voltage applied to the PSoC.  There's nothing preventing PSoC from using Vddd, Vddio, at a lower voltage, like 3.3V, 1.8V.  And, different ports can use different voltages.  Just keep in mind all 8 associated GPIO's will operate at the connected Vddio voltage.  This is handy if you have for example, an LCD that operates at 3.3V (or other peripherals).

BTW, KIT-059 is a good reference schematic for 5LP (68-pin package though).
https://www.cypress.com/documentation/development-kitsboards/cy8ckit-059-psoc-5lp-prototyping-kit-on...

Along with this checklist when making new designs
https://www.cypress.com/documentation/application-notes/an81623-psoc-3-psoc-4-and-psoc-5lp-digital-d...

Good luck with your project.

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1 Reply
BiBi_1928986
Level 7
Level 7
First comment on blog 500 replies posted 250 replies posted

Hello.

For your 5V PSoC application, use the schematic you attached, "Psoc5_Example_schematic_Power_Connections.PNG".

Since 5LP is powered by 5V, it will use unregulated power mode.  So, you don't need to do anything with the internal 1.8V regulator. There's nothing to program/initialize the internal 1.8V regulator when operating in unregulated power mode.  What's important here is, only connect capacitors to the pins Vcca and Vccd.  Do Not Connect Vcca nor Vccd to any other voltage rail.  MANY-MANY people make this mistake!  All other capacitors will connect to their respective pins as shown in "Psoc5_Example_schematic_Power_Connections.PNG".

As for VBAT/VBOOST, make the ground connections as shown in schematic "Psoc5_Example_schematic_Power_Connections.PNG".
The text/figure of the other schematic shows what must be connected IF the VBAT/VBOOST feature is being used.   Since you are not using this feature, the associated pins are tied to ground.

For a more complete description of PSoC 5LP power modes, see the 5LP Arch TRM
https://www.cypress.com/file/123561/download

FYI
Vdda has to be greater than or equal to the highest voltage applied to the PSoC.  There's nothing preventing PSoC from using Vddd, Vddio, at a lower voltage, like 3.3V, 1.8V.  And, different ports can use different voltages.  Just keep in mind all 8 associated GPIO's will operate at the connected Vddio voltage.  This is handy if you have for example, an LCD that operates at 3.3V (or other peripherals).

BTW, KIT-059 is a good reference schematic for 5LP (68-pin package though).
https://www.cypress.com/documentation/development-kitsboards/cy8ckit-059-psoc-5lp-prototyping-kit-on...

Along with this checklist when making new designs
https://www.cypress.com/documentation/application-notes/an81623-psoc-3-psoc-4-and-psoc-5lp-digital-d...

Good luck with your project.

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