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I am trying to make an IR transmitter, and have the following set up:

schematic

simulate this circuit – Schematic created using CircuitLab

I was thinking of this circuit, because I want to run through the LED ~700mA with 20mA output from the microcontroller. And putting R2 at the collector would mean a lot of power wasted by R2, doing this I am hoping to minimize that.

However the problem is I want to put the BJT into saturation mode of operation using this configuration. I found this table on wikipedia, that shows what voltages should be applied to what contant for different modes of operation:table

I can understand the table, but have problems calculating the resistor values. Here's what I do know (for saturation):

  • Q1 collector current \$I_c=700mA\$
  • D1 voltage drop \$V_f=1.4V\$
  • Q1 CE voltage drop \$V_{CE}=0.3V\$
  • Q1 BE voltage drop \$V_{BE}=0.95V\$

Using these values how can I calculate resistors needed to put the transistor into saturation mode of operation?

Could you provide more general details about the calculations, that could work for any mode of operation.

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You are using the wrong topology for what you are trying to achieve. Here is a better circuit:

First, we figure out R2. This is what sets the LED current. You want 700 mA. D1 drops 1.4 V, and let's say the transistor saturates to 300 mV. That leaves 3.3 V across the resistor. (3.3 V)/(700 mA) = 4.7 Ω. Note that this resistor will dissipate (3.3 V)(700 mA) = 2.3 W.

Now for the base resistor. You need 700 mA collector current, so you need at least that divided by the transistor gain as base current. Let's say the minimum guaranteed gain of the transistor at 700 mA is 50. (700 mA)/50 = 14 mA minimum required base current. Figure the B-E drop will be 750 mV, so that leaves 4.25 V across R1. (4.25 V)/(14 mA) = 304 Ω. So the largest common value that guarantees everything within spec is 270 Ω.

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  • \$\begingroup\$ 2.3W resistor was why I am trying to do this different. I was thnking of using R2 on the emmitter and getting smaller voltage acorss it? \$\endgroup\$ – Golaž Jan 16 '15 at 15:21
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    \$\begingroup\$ You're drawing 700 mA from a 5 V supply. The LED drops 1.4 V of that, so ultimately 2.5 W are going to get dissipated somehow somewhere besides in the LED. It will be cheaper and easier to dissipate that power in a resistor than a transistor. The circuit I show pushes most of the dissipation onto R2, which keeping the LED current reasonably predictable. \$\endgroup\$ – Olin Lathrop Jan 16 '15 at 16:09
  • \$\begingroup\$ With paralleling 1/4W resistors I could achieve 2.3W at specific resistance, yes? \$\endgroup\$ – Golaž Jan 16 '15 at 17:58
  • \$\begingroup\$ @Gol: Yes, but you'd need a lot of them. 2.3 W / 250 mW = 9.2, so at least 10. Ten 47 Ohm 1/4 W resistors in parallel would do it in theory, although I'd use a few more for some margin (and adjust the resistance of each accordingly, or course). However, a single higher power resistor will be easier. \$\endgroup\$ – Olin Lathrop Jan 16 '15 at 18:10
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I agree with Olin on the better transistor configuration, but having done exactly what you've done so many times, I'd encourage you to consider a MOSFET circuit instead this time, unless the BJT is an absolute requirement. Honestly, I've used BJTs for everything switch related forever out of simplicity, familiarity, and having a huge surplus of them. But I've come to see that MOSFETs are often a much better choice, (partly thanks to @OlinLathrop ) and they are usually dirt cheap too. Look at this example, which is similar to what you're doing.

simple Mosfet Switch

The reason I would do it with a MOSFET now is because like you, I wanted to put the BJT transistor into deep saturation to get it to maximally behave like a switch, to minimize dissipation. But the fact is, the transistor in this case is still going to have to dissipate at least 1/2 watt or more all tolled, and that assumes your square wave duty cycle is 50%. If heat or efficiency are of any concern to you, a MOSFET will be a much better choice. Here's a little N-channel MOSFET (actually called a hexfet) at mouser.com, an IRLML6246. It may be overkill for your app, but just consider how this little 50¢ device can switch up to 4 amps for you, with such low "on resistance" that it would dissipate only 28 milliwatts in your application.

Mouser Listing of IRLML6246

The other consideration, I don't see that you mentioned the drive frequency, which will be a consideration regardless of whether you use a FET or a BJT, and whether you care how harmonics beyond the fundamental frequency. For that particular FET I mentioned, the turn on delay time is about 4nS, and the turn off is 11nS. Adding those and taking the recipricol, that should make this FET good to at least 60kHz. (you might speed the off time with a lower resistance from gate-source).

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  • \$\begingroup\$ I agree a FET is better here. I was going to add a circuit for that but had to go do something else. The new IRFML8244 has even better specs and is even cheaper. That's my new jellybean part for low voltages, replacing the IRLML2502. \$\endgroup\$ – Olin Lathrop Jan 16 '15 at 16:30
  • \$\begingroup\$ I am switching at 40kHz with 50% duty cycle. Thanks for the suggestion I will be using BS270. \$\endgroup\$ – Golaž Jan 16 '15 at 17:59
  • \$\begingroup\$ Thanks Olin. Good tip! They are only like 15¢ at mouser if you buy 10, and the specs do look good. hash-tag 'sold on MOSFETS'. :-) \$\endgroup\$ – Randy Jan 16 '15 at 20:40

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