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I have a circuit in which I want the operation of one part to trigger operation of another part. I am using some adafruit sound boards, and I want that while the sound is being played, a set of lights comes on. Poking around showed the right thing for this would be transistors, being electrically operated switches essentially. Logically, everything seems to be working fine, but the light (LEDs) are coming on SUPER dim. It is noticeable that they are on, but they are not actually throwing any light out.

schematic

simulate this circuit – Schematic created using CircuitLab

This instagram post shows how I am expecting the 3xWhite LEDs to look. Nice and blindingly bright (the bright spot on the left breadboard). This is with them just hooked up directly to the circuit, before I tried introducing the switching transistor.

This one on the other hand (video) shows how it is actually working. They are functioning logically correct, in that they are turning on while the audio plays, and then turning off when it is done, however they are only just barely turning on, not bright at all.

My first thought was either the LED or transistor-base resistors were providing too much resistance, so I've stepped them both down (originally they were both 220Ω), and even tried taking them out completely. That didn't help at all.

I've also tried a number of different transistors, the TIP102, TO-92, MC16-0100, and an NTE-199, all with the same effect.

My current working theory is that the line the sound board uses to drive the amplifier does not provide enough amps to fully switch the transistor, so it is kind of "partially" closing. I think that may be the case based on this datasheet segment from the sound board (if I'm reading it right). I think I had seen in my google travels that most of these small transistors switch at around 100mA.

enter image description here (source: https://cdn-shop.adafruit.com/datasheets/vs1000.pdf )

I don't even know if that is actually possible though. Can a simple BJT actually be "partially" triggered or is it a strictly binary ON/OFF?

The white LEDs I don't know a whole lot about. They are the original LEDs from the toy I am upgrading. I'm hoping to be able to just use them for the new circuit. Using my multimeter's resistance mode, they come on at 200 and 2000 ohms, but nothing above that.

Any other ideas or feedback? I could ultimately move the 3xWhite LEDs onto the trigger switch circuit without the transistor, but then they are just on/off depending on the trigger. I am really hoping to get them to stay on as long as the sound is playing, even if the trigger is released.

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    \$\begingroup\$ A transistor can be "partially" on. This is how we make analog electronics such as amplifiers etc. \$\endgroup\$ – Joren Vaes Oct 28 '17 at 16:52
  • \$\begingroup\$ @JorenVaes - that is good to know. I tried googling it, but that simple question didn't seem to come up with any answers. So maybe my theory as to why this isn't working is correct then. \$\endgroup\$ – eidylon Oct 28 '17 at 16:53
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    \$\begingroup\$ YOur issue is not the transistor but the LEDs. You do not have enough voltage to drive three LEDs at 5V. Post the spec for the LED if you have it. \$\endgroup\$ – Trevor_G Oct 28 '17 at 16:57
  • \$\begingroup\$ @Trevor The circuit itself has no problem driving the LEDs. Like I said, if I take the whole transistor bit out of the circuit and just wire them directly into the power/ground, they show up beautifully, blindingly bright as expected. So it is definitely something to do with the transistor bit; they are clearly getting voltage, but not enough. That is why I was sub-questioning whether transistors can be partially closed. \$\endgroup\$ – eidylon Oct 28 '17 at 17:00
  • \$\begingroup\$ @JorenVaes - thanks for that link. I am new to schematics, though I did try to do some basics like not overlapping text and grouping pins by function (not board layout) etc. I'll give that a read over though too to work on improving them in the future. 😎 \$\endgroup\$ – eidylon Oct 28 '17 at 17:02
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Your problem is very simple: you need to provide a higher LED voltage or go to something other than a Darlington transistor.

Look at the data sheet for the TIP102. Vce(sat), the collector emitter voltage when fully on, can be as much as 2 volts at 6 mA. The forward operating voltage for white LEDs is nominally in the 3 - 3.5 volt range. This means that even without limiting resistors at all there is unlikely to be much current through the LEDs, since with a 5 volt supply there is simply not enough voltage available.

Your best bet is to replace the TIP102 with logic-level NMOS FET. With only 5 volts available for gate drive, "regular" FETs may work, or they may not. Try something like an IRZL44N. They are a buck or two from places like Digikey and Mouser.

Alternatively, drive the LEDs directly from the battery, although if you do you should increase the LED resistors to reflect a resistor voltage of 3 to 4 volts.

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  • \$\begingroup\$ This answer makes the most sense to me, because it explains both why the LEDs aren't working, as well as why they do work when I take the transistor out of the circuit. Also explains which number I need to look at for the flow - Vce(sat). To go with that, what is the usual nomenclature/terminology for the number which indicates the value necessary to actually switch the transistor on? One other Q, I tried looking up the IRZL44N in google and on mouser, and couldn't find it. Just want to double check there wasn't a spelling error there at all. \$\endgroup\$ – eidylon Oct 30 '17 at 16:53
  • \$\begingroup\$ @eidylon - Oops - IRLZ44N. Sorry. And the number you want is hfe, also known as beta, and generally gain. This varies with operating conditions, and if you want Vce to be small (like a switch turned on) the usual rule of thumb is between 10 and 20. For Darlingtons this isn't true - their gain is much higher, at the expense of minimum Vce. Look at the data sheet for the TIP102,and you'll see the base current specified for various collector currents. \$\endgroup\$ – WhatRoughBeast Oct 31 '17 at 5:52
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Really need to know the LED current required but I have a feeling you are running into the current output limit of the micro. If so there will not be enough base current, even without a base resistor, and hence not enough LED current.

Try hooking it up as a Darlington instead. You do not need a base resistor in this configuration since the 47R is reflected back to the micro times Hfe squared.

Or use a n N-MOSFET.

schematic

simulate this circuit – Schematic created using CircuitLab


EDIT

Now I can see your video!

Now I am sure you are overdrawing from the micro.

You would be better to hook up the LEDs this way.

schematic

simulate this circuit

Run them in series direct from the 9V supply.

You need to calculate a new value for R1. Benefit is, less currents, and less strain on your regulator.

Also, With this configuration you may not need the Darlington.

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  • \$\begingroup\$ Really? Admittedly I don't have a lot of knowledge yet, but everywhere I've read about LEDs always seems to say its a bad idea to run them in series; and additionally each successive one then gets dimmer, no? \$\endgroup\$ – eidylon Oct 28 '17 at 17:25
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    \$\begingroup\$ @eidylon No. LEDs are current driven animals. Having them in series, they all have the same current so they should all be about the same brightness. Putting them in parallel they are never the same. \$\endgroup\$ – Trevor_G Oct 28 '17 at 17:28
  • \$\begingroup\$ @eidylon please note I changed the second schematic. Having the R below the emitter in this version may put the emitter voltage too high for the micro to turn it on. I'd also try it without the extra transistor first. I think it will likely work fine without it wired this way, but put a base R bin. \$\endgroup\$ – Trevor_G Oct 28 '17 at 17:31
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    \$\begingroup\$ @eidylon, I suggest that you keep reading and learning. I suggest that you don't trust anyone that says that is a bad idea to run leds in series. Most of times the series connection is really the better option. \$\endgroup\$ – mguima Oct 28 '17 at 20:03

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