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I am trying to make 120 (4 rows and 30 columns) LED's go on and off alternatively. (At every given point in time 60 LED's are turned on and 60 off)

Components and Parameters : Function Generator : AFG 300; Switch Dg202b http://www.vishay.com/docs/70037/dg201b.pdf; Op amp-741Cn; transistor (Using 8 transistors) : MJH6284 http://www.onsemi.com/pub_link/Collateral/MJH6284-D.PDF; LED: BXRA-C0402 (Need a minimum of 7V to light up) http://www.leds.de/out/media/67920.pdf

I am using: A square pulse from a function generator (1Hz and 5V) A switch to control each row An op-amp to invert the square pulse (resulting in the alternating On and Off) The rows are connected in parallel (L1 to L3 to L5 to L7....L29 AND L2 to L4 To L6 To L8...L30). The columns are connected in series (L1 to L2 to L3 to L4) and are controlled by switches as shown in the image

The switches are placed in so that either a row(s) or column(s) can be turned on and off

I am supplying 5V at the base of the transistor through the switch (and op-amp depending on the light, L2 to L4 To L6 To L8...L30), +10 at the collector and -10 at the emitter.

The Problem: when I connect a single transistor to my circuit, the square pulse is 5V, as i increase the number of transistors the voltage given by the function generator reduces, which in turn reduces the luminescence of my LED's (some don't even light up).

I am using a 10 amp AC-DC power supply

I am not at liberty to change the components and the LED's are pre-wired, how do I get the maximum luminescence for all the LED's (at 9V and 15 Amp for the rows, and accordingly for the columns)

Where am I going wrong?? (NEED HELP ASAP)

  • \$\begingroup\$ Ballast resistors. \$\endgroup\$ Commented Feb 12, 2014 at 4:07
  • \$\begingroup\$ I assume the rectangle with a "pin 2" output is the signal generator. What is the other other connection from it you have going to the top of the column switches? And separately, have you tried resistors in series with the transistor bases on the odd rows? \$\endgroup\$ Commented Feb 12, 2014 at 4:14
  • \$\begingroup\$ @GeorgeWhite: The switches that control the turning on an off of a whole row or column are inturn controlled by arduino... Which through LABVIEW sends out +5V or grounds the switch based on what I choose.(ON or OFF). I did try the resistors on the base of the transistor. \$\endgroup\$
    – ShP
    Commented Feb 12, 2014 at 5:15
  • \$\begingroup\$ Also, at this point in time, I'm guessing the function generator I am using isnt providing enough current to drive the circuit, hence the voltage drop, I was wondering would connecting the Function generator to the base of a new Transistor and then using this transistor to drive the other 8 Transistors make any difference? (i.e, emitter of the new transistor is connected to the base of all the other transistors individually?) \$\endgroup\$
    – ShP
    Commented Feb 12, 2014 at 5:20
  • 2
    \$\begingroup\$ Good schematics make life a lot easier. \$\endgroup\$ Commented Feb 12, 2014 at 10:05

2 Answers 2



There are too many things wrong to be worth a blow by blow how to fix - it's not hard to do but you have several misunderstandings and several errors. Understand and fix and you'll be able to make it work.

What is the switching speed.
What model opamps are used.

The way you number the IC is non standard. One could just think that this was on the diagram only BUT the fact that you wrongly have V+ on pin 12 and NC on pin 13 BUT they would be correct IF you numbered the IC in the proper way suggests you misunderstand the pin numbering system (used on LL ICs). I was going to add to this diagram but see it would be wasted until the fundamentals are correct.

The physical IC numbers as shown here when viewed from above. Pin 1 will have a dot or indent or similar mark. (Connections on diagram below are correct but incomplete. Obviously drive on eg IN2 is needed to connect S2 to D2 etc. )

enter image description here

You misunderstand how the level translator switch works (and it is in fact unnecessary). You also do not need the opamps - you may be able to replace switch and opamps with 2 or 3 small transistors. BUT you can use switch and opamps if you must.

Switch is driven by the signal generator on pins 1 8 9 and 16 as required.

V+ to drive transistors goes into "Sx" 3 6 11 14.
Drive to transistors or opamps is taken out of "Dx" pins 2 7 10 15

You are trying to use the switch as a half-H-bridge or a SPDT switch - which it is not. It is a SPST switch and polarity conscious.

You are trying to drive the LED-drive transistor bases to -10V when they are off. This is not only unnecessary but wrong and could 'cause problems' quite apart from not being correct.

The transistors should be driven from V+ when on and ground when off. Vout to LEDs will be V+ - darlington drop. ee datasheet but probably 1.5V +. At the currents you are using the power dissipation in the darlingtons will necessitate a significant heatsink.


Get IC pinout correct.

Drive IC on correct pin.

Feed V+ to switch inputs and take drive from switch outputs.

Remove -10V from logic.

-10V on opamps may be useful but should not be necessary.

Redraw as per hand drawn version with new understanding.
Show IC and transistor names.

ROUGH edit of diagram. DO NOT just try and copy - understand changes and draw your own and show us.
LM324 opamp allows single supply. -10V not needed. Certainly not on switch IC.

enter image description here

_________________________ OLDER BELOW HERE _________________


We can definitely make this work if you do your part properly.
The components used look capable. Just needs details sorting.

Taken at face value you describe a system with too little voltage swing to turn the LEDs on and off fully. A dig through the switch data sheet after I had spent too much time answering the question as asked shows that the "switch" is in fact a level shifter capable of switching +/- 22V DC with 0/2.4V inputs. You are attempting to supply 0V/+Vhi to the transistor bases - which is as it should be.

The system is almost certainly capable of doing what you want with minimal extra effort - possibly with buffering of the switch outputs to increase current drive.

Your transistors are 20A darlingtons with current gains of 1000+ so should work well.

However, you MUST show a clear diagram of what you are actually doing. You do not need the whole diagram and it can be hand drawn if neat. We need:

  • Pulse generator - voltage levels, "ground" connected where? (V-/gnd presumably.

  • connection to switch IC showing part number and pin numbers. Voltage levels of supply as seen at input and output are "helpful".

  • Clear connection of one string via direct drive.

  • Same via opamp.

I believe I could probably draw this up as it is intended to be BUT you ARE able to and should.


I wrote most of the following before I realised that your "switch" is a level shifter. I'll not change it at this stage - much still applies. Entries in [square brackets] have been added.

The transistor drivers are shown as common collector = emitter followers located above the LED columns. [Correct]

It is not obvious how you can get base drive to the emitter followers from a 5V pulse that will lift the bases to the required 9V + 1 Vbe drop. [You don't - the level shifters are meant to apply 10V+]

The diagram larger version here is messy but readable BUT does not appear to show a ground reference on the sig gen. [Now assume it is 0/+5 wrt true ground]

You do not show resistors between the transistor bases and either the opamps or function generator drive inputs. A single input will tend to clamp the sig gen somewhat above the LED drop for the driven string and other strings cannot be higher. [Still somewhat true but ignore until more fundamental issues sorted].

Column current is about 1A or about 2A depending on how I interpret your data. Your transistors are 20A rate (good) and have high gain as they are darlingtons (good).
Gain is 450 min but 18,000 max. So hopefully you'd get 1000x + gain so a few mA base drive. [yes]

As shown you do not seem to have any way to turn the LEDs on fully and and off fully with only 5V swing from the sig gen. It seems that you could achieve what you want simply by increasing the available voltage swing from 5V to 10V or more. This can be achieved with one transistor and 2 resistors. [level shifter doing this - or should be]

If you can show an uncluttered diagram of how the circuit is really connected ... [Yes - as at start].

  • \$\begingroup\$ I attached a hand drawing of a portion of a circuit, (with pin connections of switch, function generator, op-amp and transistors along wih LED'S). At this point I do not have any external resistor between the function generator (or op-amp) to the transistor base. \$\endgroup\$
    – ShP
    Commented Feb 12, 2014 at 15:14
  • \$\begingroup\$ Also, I did try increasing the signal generator voltage from 5 to 10V, but as I increased the number of transistors, I saw the same issue, the pulse started to drop. \$\endgroup\$
    – ShP
    Commented Feb 12, 2014 at 16:21
  • \$\begingroup\$ I have attached a portion of the circuit. (Stating pin connections; \$\endgroup\$
    – ShP
    Commented Feb 13, 2014 at 17:47
  • \$\begingroup\$ I have attached a portion of the circuit. (Stating pin connections); Connecting the Negative terminal of the LED to ground does not light it up when transistor is on. I tried the connections and as i increased the number of transistors (i.e the number of rows; the voltage of the square pulse decreased.) \$\endgroup\$
    – ShP
    Commented Feb 13, 2014 at 17:56

Honestly, if I was going to do this I'd go with LPD8086 or WS2812 strips. That will save a ton of work in wiring, and you will get pretty much identical brightness across all of them. Plus color.


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