help with pic delay and crystal + led connection

[FoxyRick]

The ULN2803 has open-collector Darlington transistors as the outputs. The emitters are tied to ground. So, it can only sink current to ground, not source it.

The circuit you show is functionally correct, but you do not need the Zener diode in yours. It is there to protect against back-EMF if you switch relay or motor coils. LED's don't cause back-EMF. Just leave it out and leave pin 10 unconnected.

Also, you will need to adjust the resistor value. You can connect the LED anodes to your 5V supply, you don't need 9V. The output Darlington transistors in the ULN2803 drop about 1.2V themselves.

So, if you use 5V, there will be 3.8V remaining across the LED/resistor. If the LED drops, say, 2V at 20mA, then we need a resistor to drop the remaining 3.8V-2V = 1.8V at 20mA.

Resistor = 1.8/0.02 = 90 ohms.

Check your data sheet for your LED's. If the characteristics are different from 2.0V at 20mA then change the above accordingly.

 

[kobre98]

The ULN2803 has open-collector Darlington transistors as the outputs. The emitters are tied to ground. So, it can only sink current to ground, not source it.

The circuit you show is functionally correct, but you do not need the Zener diode in yours. It is there to protect against back-EMF if you switch relay or motor coils. LED's don't cause back-EMF. Just leave it out and leave pin 10 unconnected.

Also, you will need to adjust the resistor value. You can connect the LED anodes to your 5V supply, you don't need 9V. The output Darlington transistors in the ULN2803 drop about 1.2V themselves.

So, if you use 5V, there will be 3.8V remaining across the LED/resistor. If the LED drops, say, 2V at 20mA, then we need a resistor to drop the remaining 3.8V-2V = 1.8V at 20mA.

Resistor = 1.8/0.02 = 90 ohms.

Check your data sheet for your LED's. If the characteristics are different from 2.0V at 20mA then change the above accordingly.

pin 10 is common the datasheet said that it must be connected to +5v
and from the datasheet specifications "Continuous Input Current, IIN ....... 25 mA" is that for each pin ?!
so it has 8 pins 8*25ma= 200ma ampere as a whole for each uln2803 ic
is that correct ?
what will happen if i use 220 ohm then the led's will have low light ?!
if i do used 9 volt source separated from pic and let the uln take 1.8 there will remain 7.2 volt and let the led take 3 volt at 25 ma then 4.2 volt at 25 ma >>
4.2/.025=168ohm let it be 220 ohm with slight change in the light did i calculate it correctly ?!
so if i increased the voltage i can have higher brightness since the uln can hold Input Voltage, VIN ..............................30 V and up to 50 volt as output ?

 

[FoxyRick]

Pin 10 is the COMmon connection for all the reverse protection diodes. It does not have to be connected. The reason it is externally accessible is because the ULN2803 allows the use of a separate supply for the driven loads (LED's, relays or whatever). If you are driving inductive loads (not LED's) these cause a back-EMF when switched off. That back-EMF would damage the ULN2803. So, a reverse protection diode is used connected across the load to its +V. This allows the back-EMF to flow through the diode, instead of forcing its way through the ULN2803.

In the ULN2803, each channel has one end of a protection diode connected to its output, and the other ends are all connected together to the COMmon pin. So, you can connect this pin to the load's +V if you need, and get protection for all the channels. The zener shown in your circuit is to allow some voltage development because just connecting the protection diode to +V slows down the turn-off of relays. The zener speeds it up again.

You don't need to connect pin 10 if you are just driving LED's, but it will do no harm if you connect it to your voltage. Don't bother with the zener diode though; it's a waste of money.

Personally, I would stick to powering the LED's from your 5V, unless you want to run blue or white LED's because these need up to 4V and the ULN2803 will give just below that from 5V. There is no advantage to more volts with normal LED's and there is more wasted energy as more volts will need dropping across the resistor.

That 25mA sounds more like the maximum base current - you don't have to worry about that. It's certainly not the maximum sink current which is generally 500mA per channel (but maximum of 2.5A through the ground pin for the total). So you can run your LED's at enough current to make them explode if you want. The ULN2803 will be fine.

Here is a part of TI's datasheet for their ULN2803:



You can see the Continuous Collector Current = 500mA and Total Substrate-Terminal Current = -2.5A.

 

[kobre98]

Thanks a lot, i'll try nd report back to u :d