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I'm currently converting an old MP3-player (a Creative Zen Stone) to a standalone music player for my children. One of the things I'd like to do is to add an LED which shows whether the device is powered on or off.

Due to how the MP3-player is built, the only place where I can easily modify the circuit is between the battery and the device itself. Therefore, my idea was to place the LED in parallel to the device and use a transistor to switch it on whenever there's a current flowing through the player.

schematic

simulate this circuit – Schematic created using CircuitLab

The battery says that it delivers 3.7V. The green LED needs 20mA at a forward voltage of 2.1V, so given the 3.7V from the battery it needs an 80Ω resistor. I'm currently using a 100Ω one, because that's the closest that I got.

This design somewhat works, in the sense that when I power on the MP3-player the LED lights up. However, it seems the player doesn't get enough voltage now, because it doesn't work properly (which it does without the additional elements attached). The base-emitter saturation voltage of my transistor is around 0.7V (accordingly, I'm measuring that the player only gets 3.0V).

How can I resolve this problem? Is there a different circuit design that I can use or should I start looking for a transistor with a very low base-emitter saturation voltage instead (assuming those exist)?

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  • \$\begingroup\$ That's an interesting approach. I've redrawn your schematic in the conventional manner with positive on top and negative on the bottom. You have the battery polarity reversed on your diagram. \$\endgroup\$ – Transistor Aug 12 '17 at 15:57
  • \$\begingroup\$ @Transistor Thanks! I'm farely new to electronics, so I haven't picked up all the conventions, yet. \$\endgroup\$ – Florian Brucker Aug 12 '17 at 15:59
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    \$\begingroup\$ @Reinderien I want to keep the design "wireless" with regard to power supply. Using another battery might be an option, however, I'm afraid that this might interfere with the charging logic of the MP3-player. \$\endgroup\$ – Florian Brucker Aug 12 '17 at 16:28
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    \$\begingroup\$ @Reinderien that actually sounds pretty nice, but I never tried to build a light pipe myself. But the idea of using the original led has something: Florian could use a phototransistor to "read" the LED! \$\endgroup\$ – Marcus Müller Aug 12 '17 at 16:46
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    \$\begingroup\$ electronicdesign.com/components/fundamentals-led-light-pipes , fictiv.com/hwg/design/… , for example \$\endgroup\$ – Reinderien Aug 12 '17 at 16:55
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Have a look at the LM124 comparator.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. The LM124 can compare inputs down to 0 V. They feature an open-collector output suitable for driving the LED.

How it works:

  • The comparator, as the name suggests, switches its output based on the comparison of its inputs. If IN+ > IN- output switches high. If IN+ < IN- output switches low.
  • We'll assume that you can tolerate a 0.1 V drop on your power supply and that your MP3 device draws > 50 mA when on.
  • We need to set R1 to a value that will generate > 0.1 V at 50 mA. \$ R1 = \frac {V}{I} = \frac {0.1}{0.05} = 2 \; \Omega \$. 2.2 Ω is the nearest standard value above 2 and will generate a slightly higher voltage which will make the system more reliable.
  • R2 and R3 create a voltage divider to hold IN+ at about 0.1 V.
  • C1 holds this voltage steady in case voltage fluctuates with volume of the player.
  • When the device is turned on IN- voltage will be greater than IN+, the output will switch low and D1 will light.

Measure the current drawn by your device in standby and recalculate for R1.

The LM124 is available in 8-pin 0.1" DIL chip which you will be able to solder or insert in a DIL socket.

Welcome to the world of electronics. This is an interesting little project. If you take my approach you'll read through the datasheet, learn how to pronounce the words, study the graphs, wonder what it all means but start to figure it out from context and studying other circuits and application notes.

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  • \$\begingroup\$ You'll also learn how to spell "comparator" :) \$\endgroup\$ – Reinderien Aug 12 '17 at 16:38
  • \$\begingroup\$ Nice. Another note: R2 and R3 have (probably intentionally) been designed "strong", i.e. the ca. 100 µA that flow through these at all times have been chosen so that they are in the same order of magnitude as the Opamp/comparators supply current, and sufficiently larger than the input leakage currents. If you can find an Opamp that uses less quiescent current than the LM124 (that is certainly possible, but finding one that can compare voltages ~ 0.1V over negative supply isn't easy), then increasing the values of R2 and R3 (and decreasing C1) would result in longer battery life. \$\endgroup\$ – Marcus Müller Aug 12 '17 at 16:38
  • \$\begingroup\$ And: it's very possible that under full load (i.e. high volume into low-impedance earpieces), the MP3 player might draw much more than 50 mA, so that it'll sense weak battery when things get loud. In that case, you'd need to reduce R1 – but that would conflict with reliable detection; not sure how Transistor would solve that problem :) \$\endgroup\$ – Marcus Müller Aug 12 '17 at 16:43
  • \$\begingroup\$ Idea: maybe replace R1 with a Schottky diode, which has a nonlinear Voltage-over-current curve? Schottkys with Vf of 0.1 to 0.2V at ~50 mA do exist and aren't that expensive. That way, you'd never get a "large" voltage drop \$\endgroup\$ – Marcus Müller Aug 12 '17 at 16:44
  • \$\begingroup\$ @Reinderien: Thanks. I've fixed my spill chick. Everytink is spelt properley know. \$\endgroup\$ – Transistor Aug 12 '17 at 16:52
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For power indication a good super-bright 5000mcd (cheap) Red LED only needs 2mA or ~ 760 Ohms across MP3 player input DC to get 500mcd offered in poor LEDs at 20mA

Then keep switch to MP3 wired as before. This way there is no voltage drop and minimal current load. Get the right LED and R.

https://www.digikey.com/product-detail/en/wurth-electronics-inc/151054RS03000/732-11408-ND/7315780. $0.23 (1)

You dont need a transistor when you have a mechanical switch to do the job.

nowgeterdun

added

The best LED feature is the one built in which is better and could be brighter by careful selection in a larger 5mm case without drawing more current and more useful than anything else suggested.

enter image description here My measuring the voltage on each pin, one can determine which is common cathode to ground or common anode to V+.

You can replace it with a 2 colour R/G 5mm case mounted LED readily available from Mouser/Digikey etc and use AWG30 magnet wire used in many wallworts for the primary winding if you dont have any surplus.

When both R + G are active it may appear yellow to orange. such as <<

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  • \$\begingroup\$ Thanks for your input! However, I'm not sure I understand: in my current design, the LED is not the problem, the voltage drop between base and emitter is. And the LED you mention has a forward voltage of 1.9 V, so putting it in series with the player will get even less voltage to the player than my current design. In addition, the player is switched on by holding the "play / pause" button for several seconds, so I can't simply hook into that without duplicating some logic (that's why I'm trying to get at it via the battery). \$\endgroup\$ – Florian Brucker Aug 13 '17 at 13:28
  • \$\begingroup\$ Don't change anything in your MP3 player power connections ok? No drop voltage, no transistor ,nothing in series. Just share voltage in parallel with LED which uses a series R as I said above. \$\endgroup\$ – Sunnyskyguy EE75 Aug 14 '17 at 4:57
  • \$\begingroup\$ That won't work: if I add the LED in parallel to the player then it will be always on, since it's directly attached to the battery. I need to take into account whether a current is flowing through the player. The options I know of are to put the LED in series with the player (not enough voltage), use a transistor as in my original post (not enough voltage) or use an comparator as @Transistor has suggested. If you can think of another option then I'd be grateful for a more detailed explanation, since I seem to be missing your point. \$\endgroup\$ – Florian Brucker Aug 14 '17 at 12:07
  • \$\begingroup\$ doh .. last time I explain this. it goes across mp3 after the switch, not across the battery so it is only On when switch is On. \$\endgroup\$ – Sunnyskyguy EE75 Aug 15 '17 at 2:10
  • \$\begingroup\$ Sorry, but I still don't understand. As I mentioned before, there is no mechanical switch, so I'm not sure what you mean by "across mp3 after the switch". \$\endgroup\$ – Florian Brucker Aug 20 '17 at 16:48

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