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i want to build a bluetooth speaker with an additional audio-input jack.

When the 3.5mm audio-plug is inserted, i want to switch off the bluetooth-module, so that i can switch between these two audio input signals. The bluetooth-module is powered by 5V via USB-Connector.

The audio-jack has a detection-pin which is connected to GND when the plug is inserted. I am searching for a circuit which turns off the bluetooth-module when the detection pin goes to GND.

I am very new to electronics, but nevertheless i tried to create a little schematic and i dont know if it'll work. Here pin 1 of the switch is the detection-pin and pin 3 is the GND-pin of the audio jack. So when the plug is inserted the "switch is pressed" (i did not found a usable audio jack for me in eagle)

enter image description here

I dont know if i have to use a transistor or if some pullup/pulldown resistors only should do the job.

Thanks in advance !

*EDIT*

Just for clarification: The BT Module isn't transmitting anything. It is a receiver for the second audio-input-signal. Anything i want to do is to switch to the jack as audio-input when a plug is inserted. The audio-signal then goes to a LM386 Amplifier driving an 8-Ohm Speaker.

PS.: Since the circiut will be soldered to a perfboard, i cannot use any SMD-Packages.

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This is a better circuit, because the BT module can draw quite a bit of current when transmitting and you would need a lot of base current to guarantee the voltage does not drop too much:

schematic

simulate this circuit – Schematic created using CircuitLab

You could use a dual MOSFET such as the SiA519EDJ, which includes gate protection, for this application, or a single P-channel MOSFET M1 for the switch and a BJT or N-channel for the driving transistor M2. Another advantage of this circuit is that the current is quite low when the BT module is off- only 50uA.

To make it even more bulletproof against ESD on the jack, use a MOSFET with gate protection (as suggested) and put a resistor such as 2K in series with the switch.

Edit: Functionally, M2 turns off when SW1 is closed. That allows R1 to pull the gate of M1 up to +5, turning off M1. When SW1 opens, R2 pulls the gate of M2 up to 5V, turning it on, that pulls the gate of M1 down to 0V, turning M1 on and energizing the load.

You require two transistors in order to have the switch grounded and have the load 'off' when the switch is closed. If you could connect the switch to +5 then you could just have a resistor from the gate of M1 to ground and short gate of M1 to source (+5) to turn it off.

Use a breakout board to incorporate SMD packages. MOSFETs in through-hole packages don't tend to be very good at the low voltage/high current/logic level gate end of things.

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  • \$\begingroup\$ i dont get it :( why should i use two mosfets ? When i simulate this circuit on CircuitLab, i measure 5V when the switch is closed or open. Can you go more in Detail explaining your schematic please ? Thank you! \$\endgroup\$ – steffka Feb 26 '16 at 15:24
  • \$\begingroup\$ You need to add a reasonable load to ground to simulate the BT module, otherwise the tiny leakage in M1 will make it appear as if it was on. The MOSFETs models in the schematic won't be the same as the one I suggested, but it should be functional. See my edit above for a description of how it works. \$\endgroup\$ – Spehro Pefhany Feb 26 '16 at 16:12
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Yes, this should work. It may be better to use a FET in this circuit. Check out this question for a more detailed answer in a very similar situation.

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  • \$\begingroup\$ Can you recommand a specific FET for this application ? I think the maximum current draw is 500mA as the bluetooth-module usually is powered by a normal USB-connection \$\endgroup\$ – steffka Feb 26 '16 at 10:29
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    \$\begingroup\$ An emitter follower is not a good solution \$\endgroup\$ – EM Fields Feb 26 '16 at 13:18

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