3 replies to this topic

[cce1911]

So I have gotten my shift register to work with my N+2, but the lasers I'm trying to drive require 300mA of current each. So to drive 9 of these, I need almost 3A at 2.5 - 4.5 vdc.  This is way too much for the N+2 and the 74HC595. I've been looking at some transistor arrays, but I'm not sure how to wire them. Can anyone point me to a circuit that uses this 8ch Darlington Sink driver (or something similar)?

[Paul Newton]

Hi,

 

The package contains 8 NPN "darlington pairs".

Each pair is two NPN transistors wired up so that one drives the other giving a higher current gain than with just one transistor.

To turn ON a pair, the Netduino or shift register output drives current into the input (base of the transistor), and the current returns to the Netduino via the ground.

When a pair is ON, it allows a high current to flow INTO the output pin and down to ground.

This is not obvious when the output is called "output".

 

To understand how to wire it up, imagine putting a manual switch on the negative terminal of your laser. e.g. the switch controls the current out of the laser to ground.

This switch is the darlington.

 

So permanently wire the positive pin of the laser to the positive laser supply, and connect the negative pin of the lasr to one of the "output" pins.

Connect the ground to the ground of the laser power supply, AND to the ground of the Netduino / shift register.

Connect the "common" to the positive of the laser power supply (the common is linked to an array of protection diodes that prevent damage to the transistors if the load is inductive).

Finaly connect the Netduino or shift register outputs to the "inputs" of the array.

 

When the Netduino or shift register output a high voltage, current flows into the input, and this allows a larger current to flow into the output from your laser.

 

BUT.

 

Looking at the ratings in the data, I don't think you can drive many lasers from one of these chips.

The current is OK, but the power is not.

 

In the table on page 4 of the data sheet, there is a parameter V_CEsat "collector emitter saturation voltage" which gives the voltage that develops accross the transistor when it is ON.

A transistor is not a perfect switch and when it is ON, it still has some resistance which means a voltage will develop.

 

Volts x Amps = Watts

The more current that flows through the darlington, the hotter it is going to get.

 

In the table, it says that V_CEsat = 1.3V for a load current of 350mA.

This means that with your 300mA laser, there will be about 1.3 x 0.3 = 0.39 = 390mW

For eight lasers the power adds up in the package to 0.39 x 8 = 3.12Watts

Unfortunately, on page 2, the absolute maximum ratings table shows that even for the Dual In-line version, the maximum power disipation is 1.47W

 

This means that you could only really drive three lasers from one chip.

 

You might want to consider using inidividual darlinton devices rather than the array.

 

Hope this all helps - Paul

 

EDIT

I made some measurements with diagrams on using a darlington on this Wiki page.

[cce1911]

Paul,

Thanks for this very informative post. I was trying to avoid adding 9 components to my system, but based on your observations, that may be unavoidable. Given this, would MOSFETs be a better choice than Darlington pairs?

-Capel

[Paul Newton]

Lots of people report using MOSFETs on this forum, I prefer to use Darlingtons - mostly because I can remember what the circuit symbols mean.

The issue here is not whether its a MOSFET or a Darlington, but that the array is packing lots of transistors into a small space and so there is a limit to how much power (waste heat) they can generate in total.

 

To be honest, for me, it came down to what I could get hold of off the shelf at the local electronics store.

I chose something that was in stock and could handle the load I wanted to use!

 

Have fun - Paul