Multiplexing with 74HC595 and BJT's

Using Arduino LEDs and Multiplexing
1

Hi everyone!

I have a question about multiplexing with the 74HC595 (as the title says, of course...).
I read that the 74HC595 isn't suited for driving leds, because of the 6mA output limit.
Since I want to experiment with multiplexing without having to buy new chips (am just a poor student... :grin: ), I was thinking about driving BJT's with the SR.

The circuit would look something like in this picture:

Note that I've never ever drawn a circuit before, so if there are mistakes, please tell me :drooling_face:

Would this circuit work? The first 4 bits would then become negative logic.

Thanks in advance!

2

You need resistor's in the base of all the transistors. You also need resistors to limit the current through the LEDs, just on one side.

3

3.3V
How are you going to control the 74HC595?
Place a .1uF capacitor from the VCC pin to GND.

4

That may give you problems.
The path for each LED contains 2 transistors. Each transistor will have a Vce drop, maybe 0.7V.
3.3V - (2 x 0.7) = 1.9V for the LED. Most need 2.2V, 2.5V, 2.7V for RED LEDs, and more for other colors.

I'd suggest a 3xAA battery back for the LED power.

And don't forget current limit resistors, I'd suggest place one each in series between the common cathodes and the NPN collectors. 4 total.

Then your mutiplexing can be: drive pattern for first 4 anodes, drive one cathode, hold for a few mS, turn cathode off.
Repeat for pattern for next 4 anodes, drive next cathode, hold, turn off.
etc.

5

Thanks for the replies everyone!

Well, I thought resistors would be unnecessary since an LED can be driven by 3.3 volts...

@LarryD, I will control it with an arduino UNOR3. I tried to drive the SR without capacitor, and it works fine untill now...

@CrossRoads, thanks for the info! However, where can I find this voltage drop? (Im veryveryvery new to transistors) I searched for a datasheet of the 2N3904 and 2N3906, but could find anything that looks like something around 0.7...

6

Michiel_:
Note that I've never ever drawn a circuit before, so if there are mistakes, please tell me

Sure! Looks great!

First confusing aspect for those reviewing it is that you have drawn it upside-down - convention is to draw the ground side - almost always the negative - on the bottom and the positive power supply on the top.

Michiel_:
Well, I thought resistors would be unnecessary since an LED can be driven by 3.3 volts...

So you do not understand what a LED is - indeed, it is a diode, with the characteristics of a diode, specifically the "threshold" effect whereby up to a certain voltage it passes essentially no current, and at or around that threshold voltage, it will pass however much current you feed it with rather little increase in voltage and that voltage threshold is dependent on temperature. So a resistor is not used because you have "more voltage than you need", but to determine the current to be the correct value.

Similarly, the base-emitter junction of a transistor is a diode with an approximately 0.7V threshold, so needs again, a series resistor to determine the current.

And I wonder what your "3.3" battery actually is?

Since you are going to power it with an Arduino UNO, you will actually be running it at 5V, so the voltage drop of the transistors will not be a particular problem. Now for multiplexing, you have to decide whether you are going to "strobe" by rows or columns. Whichever grouping you will use for the strobe, selected one after the other, will need to provide the current for four LEDs, so you would want something like 1k resistors in series with the bases of those transistors. The other direction grouping you will use to place the pattern for each strobe, needs 220 ohm resistors in series with the LED connections to the collector, and a 3k3 resistor in series with the base of each of those transistors.

Now actually, with modern efficient LEDs (which I decline to refer to as "super-high-brightness" or such as they are no more or less than efficient in comparison to early model LEDs which were simply awfully poor - I essentially need to throw out my stock of the older LEDs as the are simply too poor to consider using for any project ), 20 mA even for a quarter of the time as they are strobed one-in-four, they will be very bright so in fact, you could just as well use 1k resistors in series with the side of the array that is not the "strobe" lines, and use no transistors.

7

Sorry for not replying, but it is a very busy period at uni atm.

@Paul__B, thanks for the info. I only started working with electronics this year, so Im very new to everything. I always thought resistors were there to dissipate power to reduce the load on the LED in this case...

I was -completely wrong- thinking that it would simplify the circuit scheme. I probably will use a 5V regulator if the project becomes standalone, or the arduino's 5v output if the matrix is limited.

And about the resistors on the BJTs, how do you determine those? Also the drop in voltage, I still didnt find the specs in the datasheet.

I will try to upload a new circuit today, to check if I got it...

Thanks everyone for the info so far!

8

I always thought resistors were there to dissipate power to reduce the load on the LED in this case

No it is there to control the CURRENT through the diode.

And about the resistors on the BJTs, how do you determine those?

There is not much need to calculate the exact value it is not helpful. Use a 1K.

Also the drop in voltage, I still didnt find the specs in the datasheet.

It is called the saturation voltage and sometimes called Vsat or Vce ( Voltage collector / emitter )

9

Grumpy_Mike:
It is called the saturation voltage and sometimes called Vsat or Vce ( Voltage collector / emitter )

Thanks!

But still, it is 'only about 0.4V for the 2N3906...

So here is the promised update, sorry it took so long :confused:
With this circuit, I would multiplex row by row, right?

10

If you take that whole circuit and filp it top to bottom and left / right it will then approach a circuit that people could understand.

Rules for a good schematic.

  1. +ve at the top ground at the bottom
  2. LEDs anodes above cathodes
  3. signals go from left to right

With this circuit, I would multiplex row by row, right?

As drawn yes

11

Grumpy_Mike:
If you take that whole circuit and filp it top to bottom and left / right it will then approach a circuit that people could understand.

Rules for a good schematic.

  1. +ve at the top ground at the bottom
  2. LEDs anodes above cathodes
  3. signals go from left to right
    As drawn yes

As mentioned, Im new to circuit drawings. Thanks for the tips