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suggestion:
add three pull-up resistors (10K) from A0, A1 and A2 on the PCF8575 and also add three 'jumper' links to ground from them. That will ensure the address is fixed and allow it to be changed if more I2C devices are connected to RC3 and RC4. It makes it more versatile.
Add a resistor (suggest 47K) between RB0 and 5V. This is because the PCF8575 INT pin can only pull the line low. If RB0 is left floating you will get random levels on it and false interrupts will occur.
Change R1 and R2 to 4.7K as per manufacturers recommendations.
Add a 100nF capacitor between the 'In' pins of both regulators to their GND pins as close to the regulators as you can manage.
Add 100nF capacitors across the VCC/VDD pins of the other ICs and their GND/VSS pns as well, again as close to the IC as you can manage.
I edited the lines and disconnected the feed from 5V regulator. Now both regulators are fed from Vin.At the moment any current you draw through the 3.3V regulator also has to pass through the 5V regulator. If you feed it's input from the point you call VCC instead, through a diode it will increase the load you can place on it.
Brian.
Almost right!
1. A0, A1 and A2 on the PCF8575 don't need R12, R13 and R14 but they should be wired to pins 4, 5 and 6 on the address switch. The R9, R10 and R11 will make the pins high unless the switch is closed to ground one or more of them. It is worthwhile doing to avoid address clashes if other I2C devices are connected to RC3 and RC4.
2. You have wired a resistor in series with VEE on the LCD module, the way it is wired will give you zero contrast setting so nothing will be visible. instead, connect VEE to the wiper of a potentiometer with one end grounded and the other at 5V, the control will let you set the contrast to a suitable level. If you intended to limit the current to the backlight as I suggested earlier, the resistor goes in series with the 'A' pin (Anode of the backlight LEDs).
oh ! what a mistake !3. You have added U5, but not connected VDD to it and the capacitor between GND and A2 should be between GND and VDD.
Yep, I know that. Most I2C modules comes with smd pullup resistors.R1 and R2 are the pull up resistors on the SDA and SCL lines, they are needed for the I2C bus to work.
You can connect V_USB directly to 5V, they should both have 5 volts on them but be careful because if you provide external power (through PWR_IN1) the regulated 5V will also appear on the USB socket and therefore could feed backwards to anything plugged in to the USB socket.
You could avoid this by using a three pin header and jumper so only one or the other can be used at a time. Like A - B or B-C but avoiding A -C.
Brian.
Close but not quite there yet!
Connect the address pins A0, A1 and A2 of the PCF8575 like you wored them on U5. At the moment they are permanently pulled high and the switches do nothing.
VEE needs to go to the wiper of RP1 and doesn't need a series resistor. RP1 must however be connected between GND and 5V. The idea is that adjusting RP1 will allow VEE to go between 0V and 5V.
Brian.
omg what I'm thinking ! I got back to the design, it's a mess. But I corrected it.You still have the address lines on the PCF8575 wrong. Each line should have a resistor to 5V and a switch to GND. They are logic level inputs but draw almost no current so if the switch is open the address line is pulled to 5V (logic high) and if the switch is closed if connects it to GND (logic low).
The power switching isn't quite right. You are aiming to get one voltage out from two possible sources so think of the switches being the other way around. Remember a change-over switch gives you two alternative connection routes through it, it isn't just two on/off switches linked together.
Seriously - do not even consider making anything like that PICKIT3 clone. It only has half the functionality of a real one and would only work with some PIC devices, it could damage others.
Yep got this part of voltage backup to the IC as they should be near as possible.The idea behind capacitors close to the power inputs of devices is that they are a nearby source of power when the device draws a pulse of current. Do not underestimate the effect of even a few cm of wiring or copper track. It isn't simply that a voltage drop is caused by the resistance of the wiring (Ohms Law) because the resistance might actually be very small,
inductance plays a large part as well. Inductance limits the speed at which current changes can flow though a conductor so when a device draws pulses of current, as most do, the voltage at one end of the track might be stable but at the other end it might jump up and down. A capacitor placed physically close to the device act like a reservoir of power, it fills up at its own pace but can release energy exactly where it is needed and very rapidly.
Brian.
A schematic is just a diagram of how the components connect to each other, it doesn't show their physical location with respect to each other. The capacitors would electrically be in parallel but on the board they would be spaced close to the ICs as explained earlier.Still my questions about the schematic of the project, I want to know why there are 2 caps on the input side of the
3.3V regulator and 3 on the output side ?
ok yeah, so if in the diagram it's not specified by the symbol of whether the capacitor is electrolytic or ceramic then I would assume by their values. Good to know that.A schematic is just a diagram of how the components connect to each other, it doesn't show their physical location with respect to each other. The capacitors would electrically be in parallel but on the board they would be spaced close to the ICs as explained earlier.
There are no fixed rules about the type of capacitors, ceramic are usually the best but they can be big and expensive beyond a few uF in value. In general, values of 1uF or lower will be ceramic or plastic based and ones above 1uF will be electrolytic.
Brian.
Hint:
View attachment 166484
I didn't show the capacitors but they are needed. With the same switches wired this way, either both regulators are fed from the external power through D1 (top input line) or the 5V comes directly from the USB socket and the 3.3V is regulated from the USB voltage (bottom input line). The USB socket input (USBI) should always be 5V if fed from a proper USB host but beware that some phone chargers are either unregulated or produce much higher voltage so please avoid using them.
Brian.
You're right, I'm not planning to do all that for a simple course, it uses I2C chips and SPI .. etc. A bit complicated for starting the course.the measure of good design is good performance with low cost
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