I have an electronic toy that lights up an LED accessory. The LED accessory consists of two banks of four LEDs each. The accessory plugs into the toy via a 1/8" stereo jack. I am trying to build a simple distribution amplifier that will allow the toy to light up multiple banks of LEDs instead of just one.
Currently, the main thing that is confusing me is that, when I measure voltage coming from the 1/8" stereo jack, I see about -4 volts between ground and the left or right output. This is making a big problem for me. I wired up a transistor to a power source, but the transistor never opens, because the negative voltage never exceeds its threshold. Do I need to build a circuit to invert the voltage? What is the right way to handle this?
Are you saying the toy is outputting 4 volts to turn on the LEDs in the accessory?
You now want to add an interface between the toy and LED accessory to drive more LEDs?
How are the LEDs in the accessory wired?
LarryD:
Are you saying the toy is outputting 4 volts to turn on the LEDs in the accessory?
Technically, I think the toy is outputting -4 volts, but yes. This is my problem. I think that I need a transistor that opens up when the base/gate leg goes negative, not positive. Or maybe I need to build a 1:1 inverting amplifier?
You now want to add an interface between the toy and LED accessory to drive more LEDs?
I intend to replace the LED accessory entirely. The LED accessory is a set of glasses that you wear, and the toy flashes LEDs over your closed eyes to make patterns of light. I want to hook up more than one set of glasses (which I will build) so that multiple people can use the toy at the same time. The situation would be analagous to a headphone distribution amplifier--in fact, I considered using just that, but my understanding is that audio amps are made for AC signals, and I'm not sure whether they would work properly with a DC signal.
How are the LEDs in the accessory wired?
Without disassembling the glasses, I can't say for sure. I know that there are two banks of four LEDs--one on the left and one on the right channel of the stereo jack. I don't know whether the LEDs are wired in series, parallel or what. They are white-ish LEDs, so they may have a voltage drop around 3-3.5 volts. If the supply is 4 volts, this would suggest that they must be wired in parallel.
The jack is on the toy and you are plugging a stereo plug in it where you measure -4 volts between 1 & 3 or 2 & 3 with negative lead (on the voltmeter) on terminal 3 (ring)?
Do all the LEDs flash at the same time for a given side?
Can you tell if there are electronics in the glasses or resistors and LEDs?
EDIT: Will you be using the old glasses in the new configuration?
LarryD:
The jack is on the toy and you are plugging a stereo plug in it where you measure -4 volts between 1 & 3 or 2 & 3 with negative lead (on the voltmeter) on terminal 3 (ring)?
I measure -4 volts with the black lead on 3 (ring) and the red lead on 1. I also measure -4 volts between 3 and 2. Technically, it is not a constant -4 volts, since the lights are flashing and the voltage is constantly going on and off. In reality, the meter usually reads around -3.2 to -3.8 volts, with -4 being an absolute maximum.
Do all the LEDs flash at the same time for a given side?
Correct.
Can you tell if there are electronics in the glasses or resistors and LEDs?
I am pretty certain that there is not. The LEDs are mounted to a piece of plastic that has traces printed on it. The plastic is stuck to the inside of the glasses with double-sided tape. There are no resistors that I can see or feel. I don't want to disassemble the glasses to be 100% sure.
I have previously said that the LEDs are white, but upon closer inspection, I think they are actually amber. I looked it up, and amber LEDs typically have 2 volts of drop.
EDIT: Will you be using the old glasses in the new configuration?
LarryD:
Is this your best guess how things are now?
Correct. If we assume that the LEDs drop between 2-3 volts, then they cannot be wired in serial off a supply that is running around 4 volts or less. The only quibble I would make with your drawing is that I don't detect any resistors in the glasses. I know that best practice with parallel LEDs is to give each one its own resistor, but this isn't exactly Bang and Olufsen, you know
No resistors implies the resistors may be built into the LEDs.
There may be surface mount resistors though.
Are the LEDs on a circuit board?
Can you describe the LEDs ?
Well, one approach is to simply connect the jack sleeve to your 5V positive supply, and use the tip and ring voltages to control the bases of two PNP transistors (whose emitters connect to the 5V) via 2k2 resistors. the only risk of this is that the sleeve of the jack is now connected to your 5V and at some risk of contacting a ground of your 5V supply. A 220 ohm resistor in series with that connection will protect that.
If you want your jack sleeve grounded to the negative of your new 5V supply, you can have NPN transistors with their bases to ground, their emitters to the tip and ring connections via 1k resistors, and their collectors going to the 2k2 resistors described above.
The obvious thing would be to have small chip resistors on the flexible PCB, but if the LEDs are from the same batch (as they would be in manufacture), they would have matched forward voltages anyway.
LarryD:
I was thinking something like this but it's 11:50 and time dream on it.
Nope, logic is inverted.
Need P-channel FETs driving from +5V
Note by the way, that the "new" glasses will be wired with the opposite polarity to the "old" ones. Unless you implement double logic inversion.
Of course, if you are going to power this arrangement from, say, a 2 Amp "plug pack" whose output has no ground connection, there is no reason why you cannot construct the whole arrangement with a positive "ground" so that both the toy and your "new" glasses have the original positive sleeve on the jack and your "distribution" device uses a PNP transistor to detect each input voltage and control an NPN transistor to switch the LEDs.
It only really gets more complex if you wish to involve an Arduino.
I'm studying last night's responses now. In the mean time, I wanted to share with you this approach, which seems to be working. I think it is not the most elegant approach, but it involved components that I had on hand (an opto-isolator). I'm not sure if there are any hidden gotcha's that I'm missing.
Shame on me for not having a ready supply of op-amps; I think a unity gain inverting amplifier would neatly solve this problem, with no shenanigans like putting 5v on the ground rail, and such.
Paul__B:
Well, one approach is to simply connect the jack sleeve to your 5V positive supply, and use the tip and ring voltages to control the bases of two PNP transistors (whose emitters connect to the 5V) via 2k2 resistors. the only risk of this is that the sleeve of the jack is now connected to your 5V and at some risk of contacting a ground of your 5V supply. A 220 ohm resistor in series with that connection will protect that.
I am trying to get my head around this approach. I have always worked with circuits with a common negative ground, so the idea of connecting the 5v supply to anything that is nominally "ground", like the sleeve of the jack, raises alarm bells for me. I think what you are getting at is that the sleeve is not actually "ground" in the sense that I'm thinking of it, because it is sourcing current. So connecting the sleeve to the power supply is basically just connecting two power sources together, which of course is fine. As long as the two power sources are similar in voltage, no harm will come of it. Is that right?
If you want your jack sleeve grounded to the negative of your new 5V supply, you can have NPN transistors with their bases to ground, their emitters to the tip and ring connections via 1k resistors, and their collectors going to the 2k2 resistors described above.
Would it be too much to ask for a schematic? I'm having trouble picturing this.
The obvious thing would be to have small chip resistors on the flexible PCB, but if the LEDs are from the same batch (as they would be in manufacture), they would have matched forward voltages anyway.
This is my guess as to what is happening. I would be surprised, based on the overall construction of the glasses, if any "fancy" SMD components were used. But I can't be sure.
@Paul__B yes it would invert.
@ joshuabardwell
I like that.
A 4.7K resistor from emitter to ground is needed.
A 4N35 with >100% CTR would be recommended.
If you need to sharpen the edges for some reason, turn on timing for FET, then select an appropriate base to ground resistor (pin 6) can be added.
LarryD:
A 4.7K resistor from emitter to ground is needed.
I realize this is electronics 101, but could you explain why? I thought that the current-limiting resistor associated with the LED would limit current through the transistor. Also: do I have my load on the correct side of the transistor?
The opto isolator transistor turns on/off as it's led goes on/off.
With +5 volts on the opto's transistor collector you need a ground path for current to flow.
The input resistance of the FET is ideally infinite hence there is no ground path through the gate to source to ground.
Adding a 4.7K from the emitter to 0 volts gives you the needed ground path.
Hence, when the transistor is on the emitter will be at +5 volt turning the FET on (O/P LEDs will be on).
When the transistor is off, the emitter is at 0 volts through the 4.7K and the FET will be off (O/P LEDs will be off).
If the original glasses LEDs were on with a negative o/p on the plug, using your opto isolator circuit will do the same thing.
