[SOLVED] Problems with residual voltage to PWM.

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Mondalot

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Hello,

I want to regulate a motor with PWM and regulate I'm using a handlebar equipped with accelerator Hall effect (View photo).
The problem is the residual voltage of 0.8V.
The handlebar is fed to 5V.
At full throttle at the output provides a voltage 4.3V only.
Using this formula this means to regulate the PWM with 0.8V
I'm 84 steps of 255, without doing anything.

As I can make as smooth regulation of 0-255.
Without having to start 84.

Code: 
Device = 18F26K22
    Xtal = 16   

Config_Start
 FOSC = INTIO67   
 PLLCFG = On 'On      ' Oscillator multiplied by 4
 PRICLKEN = On     ' Primary clock enabled
 FCMEN = OFF           ' Fail-Safe Clock Monitor disabled
 IESO = Off            ' Internal/External Oscillator Switchover mode disabled
 PWRTEN = On           ' Power up timer enabled
 BOREN = SBORDIS       ' Brown-out Reset enabled in hardware only (SBOREN is disabled)
 BORV = 190            ' Brown Out Reset Voltage set to 1.90 V nominal
 WDTEN = Off           ' Watch dog timer is always disabled. SWDTEN has no effect.
 WDTPS = 128           ' Watchdog Timer Postscale 1:128
 CCP2MX = PORTC1       ' CCP2 input/output is multiplexed with RC1
 PBADEN = Off          ' PORTB<5:0> pins are configured as digital I/O on Reset
 CCP3MX = PORTC6       ' P3A/CCP3 input/output is multiplexed with RC6 
 HFOFST = On           ' HFINTOSC output and ready status are not delayed by the oscillator stable status
 T3CMX = PORTC0        ' Timer3 Clock Input (T3CKI) is on RC0
 P2BMX = PORTB5        ' ECCP2 B (P2B) is on RB5                   (EXTMCLR = Enable MCLR/Disable RE Input)
 MCLRE = INTMCLR       ' MCLR pin enabled, RE3 input pin disabled  (INTMCLR = Disable MCLR/Enabled RE Input)
 STVREN = Off          ' Stack full/underflow will not cause Reset
 LVP = Off             ' Single-Supply ICSP disabled
 XINST = Off           ' Instruction set extension and Indexed Addressing mode disabled (Legacy mode)
 Debug = Off           ' Disabled
 Cp0 = Off             ' Block 0 (000800-001FFFh) not code-protected
 CP1 = Off             ' Block 1 (002000-003FFFh) not code-protected
 CP2 = Off             ' Block 2 (004000-005FFFh) not code-protected
 CP3 = Off             ' Block 3 (006000-007FFFh) not code-protected
 CPB = Off             ' Boot block (000000-0007FFh) not code-protected
 CPD = Off             ' Data EEPROM not code-protected
 WRT0 = Off            ' Block 0 (000800-001FFFh) not write-protected
 WRT1 = Off            ' Block 1 (002000-003FFFh) not write-protected
 WRT2 = Off            ' Block 2 (004000-005FFFh) not write-protected
 WRT3 = Off            ' Block 3 (006000-007FFFh) not write-protected
 WRTC = Off            ' Configuration registers (300000-3000FFh) not write-protected
 WRTB = Off            ' Boot Block (000000-0007FFh) not write-protected
 WRTD = Off            ' Data EEPROM not write-protected
 EBTR0 = Off           ' Block 0 (000800-001FFFh) not protected from table reads executed in other blocks
 EBTR1 = Off           ' Block 1 (002000-003FFFh) not protected from table reads executed in other blocks
 EBTR2 = Off           ' Block 2 (004000-005FFFh) not protected from table reads executed in other blocks
 EBTR3 = Off           ' Block 3 (006000-007FFFh) not protected from table reads executed in other blocks
 EBTRB = Off           ' Boot Block (000000-0007FFh) not protected from table reads executed in other blocks
Config_End


Declare Adin_Res    = 10   ' Resolution 10 BITS 
Declare Adin_Tad    = FRC  ' Oscilator RC AD
Declare Adin_Stime  = 100  ' AD samples 100uS
Declare CCP1_Pin PORTC.2   ' Select Hpwm port and bit for CCP1 module (ch 1)


ANSELA = 00001111  ' Port A4-A7 Digital=0 / A0-A3 Analog
TRISA  =   00101111  ' Port A  I/O 
ADCON1 = %1000000    ' Right justified
ADCON2 = %1000000    ' Right justified


Dim DUTY As Byte                       ' Duty PWM
Dim TENSION As ADRESL.Word      ' Voltage       
Dim VALOR_GAS As Float                    
Dim VOLT As Float                         




VOLT=0
DUTY=0
VALOR_GAS=0


CLS


Main:


GoSub ADC_Handlebar
    Print At 4,6,Dec2 VOLT ," - ", Dec DUTY     'Display 
     HPWM 1,DUTY,20000 
 
 DelayMs 100
Goto Main

ADC_Handlebar:
   TENSION = ADIn 1                         ' Conversor AD
   VALOR_GAS = 489 *(TENSION / 10)  ' 489 = (5V/1023) 10Bits / Vref 5V
   VOLT = (VALOR_GAS / 10000)          ' Value in Volts
                  
         DUTY=(255/4.3)*VOLT             ' Calculates PWM (4.3V  full throttle not 5V )
Return

END

Picture:


THX.
 

Hi,

I´m confused by the values.

You talk about 0.8V and 5.0V and 4.3V... how do they fit together?

And you talk about 84 steps of 255. How does this fit together?

****
Let´s say it is 0.7V and 5.0V and 4.3V (5.0V - 0.7V = 4.3V)..
84 Steps of 255 gives about 33%. But 33% of 5V is 1.65V.

Please clarify.


Klaus
 

Hi,

The handlebar is comprised of 3 wires:
1 = VCC (5V)
2 = VSS (GND)
3 = Output

When the handle is connected to 5V, the output of Hall, automatically gives 0.8V (residual)
When the handlebar give Gas output Hall gives 4.3V = (Voltage Maximum all Gas) not 5V


The steps 255 is for:
Syntax
Hpwm Channel, Dutycycle, Frequency

Dutycycle:
is a variable, constant (0-255), or expression that specifies the on/off (high/low) ratio
of the signal. It ranges from 0 to 255, where 0 is off (low all the time) and 255 is on (high) all the
time. A value of 127 gives a 50% duty cycle (square wave).

The Dutycycle is calculate on la formula of code:

DUTY=(255/4.3)*VOLT ' Calculates PWM (4.3V full throttle not 5V )

as you regulate the handlebar gives you 0 to 255 but as I have the residual voltage
0.8V represents that automatically go to 84 = 33% DutyCicle no 0 that would be ideal.


I hope you clarified the doubt.

THX.
 
Last edited:

Hi,

Ok i think it is clarified..

So you want the input of 0.8V ... 4.3V should give 0...100% duty cycle.

If this is the case, then

DutyCycle = 100% * (hallVoltage - 0.8V) / (4.3V - 0.8V)

For 8 bit ADC values and 8 bit duty cycles:
Offset = 0.8V * 255 / 5V = 41
Gain = 5V / (4.3V - 0.8V) = 5V / 3.5V = 1.43

DutyCycle = (ADCvalue - 41) * 1.43

Klaus
 

Hi,

I changed the formula the AD converter is in 10 Bits.

Offset = (0.8V * 1023)/ 5V = 164
Gain = 5V / (4.3V - 0.8V) = 5V / 3.5V = 1.43


TENSION = ADIn 1
VALOR_GAS = 322 *(TENSION / 10) ' 322 = 3.3/1023 (10Bits)
ADCvalue = (VALOR_GAS / 10000)

DutyCycle = (ADCvalue - 164) * 1.43

But now 0 DutyCycle = 23% not 0%
255 DutyCycle = 27% not 255%

THX
 

Hi,

But now 0 DutyCycle = 23% not 0%
255 DutyCycle = 27% not 255%
Click to expand...

What does that mean? Where is the ADC?

Klaus
 

Hi,

ADC = ADIn 1 ' Read the value from the on-board Analogue to Digital Converter.
VALOR_GAS = 322 *(ADC/ 10) ' 322 = 3.3/1023 (10Bits)
ADCVolts= (VALOR_GAS / 10000) ' Passes Volts

DutyCycle = (ADCVolts- 164) * 1.43

But now 0 DutyCycle = 23% not 0%
255 DutyCycle = 27% not 255%

THX.
 

Hi,

me: --> DutyCycle = (ADCvalue - 41) * 1.43
you: --> DutyCycle = (ADCVolts- 164) * 1.43

why you now use "volts"?

You also wrote:
ADC = ADIn 1 ' Read the value from the on-board Analogue to Digital Converter.

******
please describe detailed what this means:
But now 0 DutyCycle = 23% not 0%
255 DutyCycle = 27% not 255%
Click to expand...


Klaus
 

Hi,

I used "ADCVolts" to be better understood on the display is displayed in Volts.
not what comes directly from the AD pin converter.

ADC = It would Pin pic which is assigned the AD with the ADIN command (gross value of the converter).

The 164 : Offset = (0.8V * 1023)/ 5V = 164 (1023=10Bits AD)

THX.
 

Hi,

I wrote:
For 8 bit ADC values and 8 bit duty cycles:
Offset = 0.8V * 255 / 5V = 41
Gain = 5V / (4.3V - 0.8V) = 5V / 3.5V = 1.43

DutyCycle = (ADCvalue - 41) * 1.43
Click to expand...
As said in the headline it is for 8 bit ADC values.
But you have 10 bit ADC.
So 0.8V input gives 164 LSB
4.3V gives 881 LSB.
From min to max this is 881LSB - 164LSB = 717LSB
At the duty cycle side this 717LSB give a range of 255 (0 to 255).
So the new gain is 255/717 = 0.356

DutyCycle = (ADC_value - 164) * 0.356.

Now from i put 0.8V ... 4.3V the ADCvalue is 164 ... 717 and after the calculation the duty cycle is 0 ... 255.

Klaus
 

Hi,

Sorry for the delay, I have been doing the final circuit board and functioning properly.

Thank you.
 

Hey again,

I have a little problem last minute

All I have as we can do it if you can do ...

The handlebars as I said is equipped with an inverter hall effect. when giving more gas or speed, the engine begins to start when I have a 1.8 or 2V almost 41% of duty.
and I have to go by turning the handlebars very very slowly so I would not be fired and is very upset because you're afraid.
If we subtract the 4.3 at full power - the residual offset 0.8, 3.8V are working for 100% duty.
As I can make this go much smoother without having to worry about going slowly giving gas.

or that other solutions may have.


The command works HPWM 8 bits = 0-255) would happen if 10 bits (0-1023) although not reach 1023 are about 881 +/-, this influence something. !!

THX, a lot
 

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