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Sinking and sourcing with a darlington driver

I asked a question about sinking and sourcing LEDs while using a darlington driver. The images were slightly unclear so i have redrawn it. I had too many LEDs in my circuit so i needed to create a relay. I am going to use a darlington driver and when i was testing it, i was fine with all other 7 outputs. However, I am unsure how to be able to sink and source LEDs from 1 output pin while using a Darlington driver

My project is a traffic light. I have 8 available outputs. I have 9 groups of LEDs. Currently there is 5v availible for the LEDs. The maximum number of leds per group is 8. I also have a loudspeaker and a motor in there. Please see comments for LEDs data sheet This is where i am at the moment

ww.picaxe.com/Circuit-Creator/General-Outputs/Darlington-Driver-ULN2803A/

That is the link for the Darlington Driver i am using which is represented by the 18 - DIL

This is the schematic with the ULN2803A in place

Is it possible to sink and source at the same time with the driver?

I only have 8 outputs and i realised that either the green man or the red man can be on at the same time. So to maximise outputs i sunk the red LEDs and sourced the green LEDs. However, I realised i had too many LEDs to work with a 5v PIC, so i needed to create a relay

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  • \$\begingroup\$ Can you describe what you are trying to achieve? Are you trying to create an LED matrix of some sort? \$\endgroup\$ – Transistor Jan 3 '16 at 15:35
  • \$\begingroup\$ Its a light display \$\endgroup\$ – Hasan Imtiaz Jan 3 '16 at 17:24
  • \$\begingroup\$ You need to give a much better response than that if you want help. Of course it's a light display. The giveaway is in 'LEDs'. \$\endgroup\$ – Transistor Jan 3 '16 at 17:28
  • \$\begingroup\$ Its a traffic control system. Basically a traffic light, with a countdown, and motor to aid the visually impaired. The waiting times are controlled by light levels. \$\endgroup\$ – Hasan Imtiaz Jan 3 '16 at 17:42
  • \$\begingroup\$ @transistor I only have 8 outputs and i realised that either the green man or the red man can be on at the same time. So to maximise outputs i sunk the red LEDs and sourced the green LEDs. However, I realised i had too many LEDs to work with a 5v PIC, so i needed to create a relay \$\endgroup\$ – Hasan Imtiaz Jan 3 '16 at 17:44
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No. That chip is just a bunch of transistors with their emitters all connected to comm (pin 9). You can't do high-side switching with them.

enter image description here

Figure 1. Image from SE.

  • The ULN2803 consists of eight Darlington transistors connected to a common pin.
  • Each output can switch up to 500 mA to common.
  • The maximum current the common can handle is 2.5 A (but I recommend operating at 1 A max to improve reliability).

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 2.

Arrange your groups of LEDs as shown above. Since you have a 5 V supply that's enough to power two LEDs in each string. The 47 Ω resistor will limit current to about 20 mA on each pair of LEDs so the total for a group of eight will be 80 mA. You need to check how many will be on at one time. Then make sure your power supply can supply enough mA.

If you had a few more outputs your job would be much easier. The simplest way would be to use a relay on one channel.

schematic

simulate this circuit

Figure 3.

The sketch above shows two groups of LEDs sharing the low-side Darlington driver. The active group is selected by the high-side relay (which can feed other pairs of groups as well). The problem with this is that you need to group any LEDs that are on at the same time on the same relay circuit because it is only a low-speed multiplexer.

It is possible to do this with high-side transistor switching and achieve high-speed multiplexing. I'll leave that to you to research.

Note that the Darlington common has to be connected to the micro ground.


Current per LED

Current ratings

Figure 4. Absolute maximum ratings for TruOpto 55-1814.

The datasheet, page 1, clearly shows the LED maximum continuous current rating as 30 mA. If the LEDs are pulsed, as in a multiplexing application such as yours, they can take 100 mA for 10 ms max. every 100 ms. If running a faster multiplexer use 100 mA for 10% of the time.

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  • \$\begingroup\$ So should i connect the sinking LEDs to the 5v power suplly \$\endgroup\$ – Hasan Imtiaz Jan 3 '16 at 17:25
  • \$\begingroup\$ Edit your original question to explain what you are trying to achieve, why you designed the circuit the way you did and what result you expect. \$\endgroup\$ – Transistor Jan 3 '16 at 17:29
  • \$\begingroup\$ Answer updated. \$\endgroup\$ – Transistor Jan 3 '16 at 19:53
  • \$\begingroup\$ Is it possible to use a 9v supply if i keep the darlington driver on a different board as the 9v won't be passin through the pic thus not blowing it \$\endgroup\$ – Hasan Imtiaz Jan 3 '16 at 20:06
  • \$\begingroup\$ Yes, and if you have 9 V you can put four LEDs in each string. Remember that you need to connect the 9 V common to the 5 V common. \$\endgroup\$ – Transistor Jan 3 '16 at 20:09

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