PIC16F877P Adjustable digital power supply with LCD display

Adjustable microcontroller-controlled digital power supply project with all its details, volt and ampere values can be read on the LCD, PCB box design, 3D views, source software prepared in ASM language.

Feature

• Controlled ventilation according to current consumption.
• Current consumption reading with visual and audible overcurrent indication.
• Preset memories (2.5, 3, 3.3, 5, 9 and 12 volts at 0.5A, 1A and 2.5A).
• Ventilation control with temperature reading.
• Control circuit power supply lines separated from the power section.
• Information of all values via LCD 16×2 with contrast control and backlight.
• Control and communication with PC or another control card via RS232.
• Code written in C for the SDCC compiler.
   
   

 A power supply is one of the most used tools in an electronics laboratory, on the market there are various voltages (12v, -12v, 5V, -5V) with high amperages. The idea of this power supply is that it is an analog supply controlled/regulated by a microcontroller, easy to assemble, cheap and has multiple features with expansion possibilities. 

Forced ventilation must be controlled by temperature and consumption, as design constraints for 25V at 3 or 4A may compromise some components. 

External power lines should be isolated from the control section, this reduces possible voltage drops due to ventilation, LCD backlight or use of other devices.

In general the weight of two transformers plus the electronics can lead to the source having a significant weight (3kg in my case), so a 3mm aluminum cabinet will be used.

The circuit consists of two parts, one of which is stepper transistor analog, LM324N in operation, transformers, 7805 regulator, control inputs and outputs. The control part consists of the microcontroller, LCD, buzzer, ICSP connector, and connectors to the input buttons and encoder if selected for use.

The regulator circuit is dominated by the LM324N component; this is a quadruple process that gives us a maximum voltage of 32V and the possibility of using low tolerance resistors (1% is recommended). I divided the circuit into sections to understand and analyze it better

• Input voltage: 220VAC
• Output voltage: 0 25VDC to 0 2.5A
• Display Accuracy 1V / 100mA 100mV / 3mA
• 2×16 LCD volt ampere indicator
• RS232 Communication
• Heat fan control
• Short circuit current control

Power Stage

Power stage In this source the control stage (digital part) is the most elaborate, while the analog part is the classic part with the addition of control against short circuits provided by T1,T2. 0.51ohm resistors grounding the base of power transistors T3 (BC547) and Q1..3. Voltage regulation The operation will be as follows, the constant current source formed by D4, D5, R5, Q4 and R1 provides 2mA to the base of the darlingtons. The reference of said transistors is controlled by the operational ones of the control part.

This is done with the voltage divider formed by R13, R14 and R17 but if there is 25v it will be 25.25v as the reading is affected by the current supplied and drops across the resistor Rs (shunt resistor, 10 1ohm resistors in parallel) when at the output so IC1D which is the inverting amplifier is used, For example, if there is a consumption of 2.5A, a drop of 0.25V appears at the ends of Rs. IC1D is then used to perform said correction, so we will see that it has the formula:

Vo = -(R18/(R19+R20)) * Vin

then if Vin is 0.25V as we said

Vo = -( 10k/40k )*0.25

and since the ground reference of the LM324 amplifier is -0.0625 thanks to D9. Voltage remains at 5V ready to input into comparator IC1A (takes)

Vo = Vs+, when V1 > V2 Since Vo = Vs-, V1 < V2, compares the output of the divider voltage with the selected voltage. In this way it regulates the output voltage (or cuts off if the voltage exceeds the desired value). The same goes for current, if we follow the suggested example, if there is a consumption of 2.5A, there will be 0.25vy at the non-inverting operational input since Vo = (1+ 19k/1k) * 0.25 = 5V. And the output ends up at the inverter input of IC1C and the same thing happens as in IC1A, that is, the reference of the transistors is regulated or cut off if the pre-selected maximum is exceeded.  Analysis of the encoder Encoder for continuous rotation with three output pins. Two gray code output channels (http://es.wikipedia.org/wiki/Gray_Code). It creates three binary sequences for each spin: 11,10, 00, 01. Two-bit Gray code 00 01 11 10 The encoder has three terminals; one is the common terminal and the others are digital outputs generated by the internal contacts of the device. Creates the following sequence Channel EU Status1 0 0

The same is with the current, following the proposed example, if there is a consumption of 2.5A, we would have 0.25v at the input of the non-inverting operational and since
Vo = (1+ 19k/1k) * 0.25 = 5V
And the output goes to the inverting input of IC1C and the same thing happens as in IC1A, that is, the reference of the transistors is regulated or it is cut off if the preselected maximum is exceeded.

Encoder analysis

Encoder for continuous rotation with three output pins. Two gray code output channels (http://es.wikipedia.org/wiki/Gray_Code). Generates three binary sequences 11,10, 00, 01 for each spin.

Two-bit gray code

• 00
• 01
• 11
• 10

The encoder has three terminals, one is the common terminal, and the others are the digital outputs generated by the internal contacts of the device.

Generate the following sequence
Channel A B

• Status1 = 0 0
• Status2 = 0 1
• Status3 =1 1
• Status4 =1 0

Clockwise rotation ->

00 01 11 10 00

<- counterclockwise="" in="" n="" rotation="" direction="" span="" style="font-weight: bold;">Connection to the PIC
The middle terminal goes to ground and the other two are each connected to a pullup resistor of at least 1k at VCC.

The source code below shows the sequence

; * * * * * *
; * BANK 1 Operations
; * * * * * *
BSF STATUS,RP0 ; Set Bank 1
MOVLW B'0000010' ; Set TMR0 prescaler to 8
MOVWF OPTION_REG ; Store it in the OPTION register
CLRF TRISB ; B all outputs
BSF TRISB,QUAD_A ; Except for Quadrature inputs
BSF TRISB,QUAD_B
; * * * * * * * * * * *
; * BANK 0 Operations *
; * * * * * * * * * * *
CLRF STATUS ; Back to BANK 0
BSF INTCON, T0IE ; Enable Timer 0 to interrupt
BCF INTCON, T0IF ; Reset interrupt flag
BSF INTCON, GIE ; Enable interrupts

Then the interrupt service should be:

; Interrupt Service Routine Pre-amble, save state,
; reset status to BANK 0
INTR_PRE:
MOVWF TMP_W ; Copy W to temp register
SWAPF STATUS,W ; Swap Status and move to W
MOVWF TMP_STATUS ; Copy STATUS to a temp
CLRF STATUS ; Force Bank 0
;
; State is saved, and we've expended 3 Tcy plus the
; 3 Tcy (4 worst case) of interrupt latency for a total
; of 6(7) Tcy.
;
; Now loop through until we've satisfied all the
;pending interrupts.
;
ISR_0:
; ... test bit to see if it is set
BTFSS INTCON,T0IF ; Timeer0 Overflow?
GOTO ISR_1 ; No, check next thing.
;
; Else process Timer 0 Overflow Interrupt
;
BCF INTCON, T0IF ; Clear interrupt
MOVLW D'133' ; Reset 1khz counter
MOVWF TMR0 ; Store it.
CALL QUAD_STATE ; Check Quadrature Encoders.
GOTO ISR_1 ; Nope, keep counting
ISR_1:
;
; Exit the interrupt service routine.
; This involves recovering W and STATUS and then
; returning. Note that putting STATUS back
; automatically pops the bank back as well.
; This takes 6 Tcy for a total overhead of 12 Tcy for sync
; interrupts and 13 Tcy for async interrupts.
;
INTR_POST:
SWAPF TMP_STATUS,W ; Pull Status back into W
MOVWF STATUS ; Store it in status
SWAPF TMP_W,F ; Prepare W to be restored
SWAPF TMP_W,W ; Restore it
RETFIE

As you can see, the TMR0 interrupt is reloaded first to ensure a tick rate (and this is also the first verified interrupt!) Then the state of the encoder is checked in the call to QUAD_STATE:

;
; QUAD State
;
; A quadrature encoder traverse a couple of states
; when it is rotating these are:
; 00 | Counter
; 10 | Clockwise
; 11 | ^
; 01 V |
; 00 Clockwise |
;
;
QUAD_STATE:
BCF STATUS,C ; Force Carry to be zero
MOVF PORTB,W ; Read the encoder
ANDLW H'6' ; And it with 0110
MOVWF Q_1 ; Store it
RRF Q_1,F ; And rotate it right.

RLF Q_NOW,F ; Rotate Q_NOW Left
RLF Q_NOW,W ; by two
IORWF Q_1,W ; Or in the current value
MOVWF QUAD_ACT ; Store at as next action
MOVF Q_1,W ; Get last time
MOVWF Q_NOW ; And store it.
;
; Computed jump based on Quadrature pin state.
;
MOVLW high QUAD_STATE
MOVWF PCLATH
MOVF QUAD_ACT,W ; Get button state
ADDWF PCL,F ; Indirect jump
RETURN ; 00 -> 00
GOTO DEC_COUNT ; 00 -> 01 -1
GOTO INC_COUNT ; 00 -> 10 +1
RETURN ; 00 -> 11
GOTO INC_COUNT ; 01 -> 00 +1
RETURN ; 01 -> 01
RETURN ; 01 -> 10
GOTO DEC_COUNT ; 01 -> 11 -1
GOTO DEC_COUNT ; 10 -> 00 -1
RETURN ; 10 -> 01
RETURN ; 10 -> 10
GOTO INC_COUNT ; 10 -> 11 +1
RETURN ; 11 -> 00
GOTO INC_COUNT ; 11 -> 01 +1
GOTO DEC_COUNT ; 11 -> 10 -1
RETURN ; 11 -> 11
INC_COUNT:
INCF COUNT,F
MOVLW D'201'
SUBWF COUNT,W
BTFSS STATUS,Z
RETURN
DECF COUNT,F
RETURN
DEC_COUNT
DECF COUNT,F
MOVLW H'FF'
SUBWF COUNT,W
BTFSS STATUS,Z
RETURN
INCF COUNT,F
RETURN

Figure below shows the circuit how should be

For more details regarding components, circuit Diagram, and PCB plate, do not hesitate to contact us. Also available the simulator and new modification with PIC24. The program for the PIC16F877 is all written in Assembly.