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PSoC™ 4 Forum Discussions

Not applicable
        PSoC4 Flash writes and reads are very much similar to that much similar to that in PSoC3/5.Here is a simple project.   
7 Replies
        Attaching the missing project. --NAMN   
Level 4
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Level 4

Flash writes are useful for storing application specific parameters for persistance across power cycles. Make sure the capacitors in the power supply have enough energy after a power fail to complete the flash write.

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A way of computing external cap size as follows (note done for PSOC 1, method can be used for any PSOC)-




Power Supply 
  If you are in a critical loop, whether it be controlling machinery, or writing internal EEPROM,
power supply and UP have to handle detection and orderly termination of processing threads upon loss
of power. Some points –
1)  Use LVD to alert UP there is an imminent loss of power about to occur. Process that interrupt to
create an orderly shutdown. Point of pickoff for LVD is also of consideration, closer to the
energy source generally better control and maximization over power loss. If external LVD is
desired then use A/D converter as LVD device. But also potentially presents circuit design
problems associated  with pin injection current limitations, eg. when power is absent on PSOC,
but coming up on LVD sense node. A simple resistor in series with input may resolve
that issue.
2)  LVD alone is not enough, some processes need time during a loss of power to complete. That’s
where short term energy storage is critical, eg. electrolytics, bulk caps, battery…..
3)  Circuit design, make sure energy cap does not supply energy back to line or
regulator by diode 
isolation. Schottky preferred (switched power MOSFET best)  where possible
due to lower Vf, 
less wasted Pdiss, and allows greater V compliance range in design.
4)  To compute cap size needed ( this is an approximation, assuming load looks like a constant current
source, in reality load typically resistive so one would use exponentials to calculate. UP looks like a
constant I load due to CV**2 x f Pdiss, whereas loads, like LEDs with I limiting resistors, look more resistive
) -
Q = C x V, I = C x dV / dT, C = ( I x dT ) / dV, dT = ( C x dV ) / I
I is max current needed
dT is time needed to complete processor thread completion, like EEPROM write as an
dV is allowed voltage drop to insure UP stays within operation specs, basically (
Vloss_of_power_detect – Vupmin )
All computations have to take worst case values to insure C min is computed, over V and T.
Stated another way as C and dV go up, I down, we maximize time UP allowed to continue
running on loss of power. So we can use LVD to shut down  processes using power, like LED
display driving as an example, shut off high speed clks to modules like PWM, counters, etc. to
lower I needed. Then finish orderly shutdown. Obviously you shutdown critical processes, like
ones controlling safety for humans, ticks, bugs,etc.. first……
5)  A typical calculation (29466, nominal values, you would need to worst case the example values)
a.  Assume 12 Mhz clock, this allows PSOC to operate over 3.0 V to 5.25 V, assume Vdd
Is 5V, +/- 5%. You could use LVD to switch clock down from 24 Mhz to 12 Mhz to
take advantage
of wider operating Vdd range while optimizing speed when normal power in effect.
Choose LVDlow_trip = 4.64 V, this is a warning power is about to fail, so dV = 4.64 – 3.00 =
Sum total current required, note PSOC specs show UP core current separate from modules
current, so you need to sum all the currents. Also PSOC currents speced as tytpicals, so you will
need to use a “fudge” factor scale them up. Or characterize design at temp extremes to get a
value with confidence.
So I = ( Iup + Imodules ) xFfudge, lets choose 50 mA as an example, fudge factor 2 x, so I = 100
mA,. Again you will have to worst case this as best as you can.
Lastly assume our main problem is writing EEPROM, we want to write one block with confidence
just as power is failing. 200 mS is max needed. Also we need to add some code execution time
for rest of orderly shutdown, say 50 mS. Then dT = .25 secs
Equation is Amps, Volts, Farads
So we have (nominals)  C = ( .1 x .25 ) / 1.64 = ~ .015 Farads, or 15,000 uF. Electrolytic tolerances
quite poor, very temp dependant, so factor this in to value needed.
Lastly this capacitor has to be charged on power recovery, so use defining equations and
estimates of available current to restore capacitance charge, in consideration of startup power
supply slew rates. Also protect 
regulator outputs from discharging this capacitance with a diode or switched MOSFET in series
with regulator 
output. Note its voltage drop factors into calculations.  Spice useful to get a handle on overall
supply design. 
The capacitor can be on the input or output side of your regulator, certain advantages and
considerations for either topology.
Additionally more energy may be available stored in  transformer fields. Spice good for
estimating this behavior.
Reminder you have to worst case all values over T and V for a successful/reliable design. And
don’t forget component aging effects.
Regards, Dana.

Not applicable

Thanks for sample code.


I can't see flash block declaration. Like const uint8 flash[] _at_ 0x00FF;


Also code areas are not blocked for writing.


Without this You can easily overwrite the code.


Terrible example

Level 10
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I think your initial post was made befor the Emulated EEProm usermodule was released. Now everything is a bit easier. But I have to admit that your code is really dangterous since you would have to watch yourself for not getting your code overwritten.





Not applicable





I'm trying to use the emulated EEPROM example.  It reads incompatible on my top design schematic how can I update the EEPROM for CY8C4247LQI-BL483?  

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In the "System Reference Guide" (from Creator help menu) are APIs listed to access flash. There are some side-effects regarding clock frequency, lookout for those.