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What resistor value do I need?

andywest

I have an 8 ohm, 0.5 watt speaker that I want to drive directly from an IO pin of an FPGA. I just want to generate simple square wave beeps, so I think I can connect like this:

 

FPGA IO pin -> resistor -> speaker -> ground

 

The IO pin is set to 3.3V LVTTL, and I can choose between 4mA and 8mA.

 

My question is what resistor values would I need for 4mA and 8mA to get the loudest volume without drawing too much current?

 

Thanks!

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  • Hi Andy,

     

    If you are going to drive the speaker with a current between 4mA and 8mA, I suspect you will struggle to ear anything at all. For a 0.5W 8 Ohm speaker, nominally you get the loudest sound when you drive the speaker at maximum current. Doing a little calculation, the max current would be 250mA. Assuming you will be able to drive the speaker with 8mA current, this would only give 0.5mW.

     

    To make things worse, maximum power transfer between the source and the load requires the two to have matching impedance (8 Ohm in this case). I don't know what is the resistance of the FPGA output pin, but a typical termination resistance value is 50 Ohm, in which case, even connecting the output pin directly to the speaker (not a good idea!), there is already quite a mismatch between the source and the load. Adding 400 Ohm resistor to limit the output current will only make things worse, and most of the power will end up being dissipated there rather than reach the speaker.

     

    I think the easiest way to interface the speaker with the FPGA output is to do what John suggested: use a MOSFET to drive the speaker, and drive the MOSFET with the FPGA output pin.

     

    One note on the use of 2 output pins to get a differential output with doubled range (i.e. -3.3V to +3.3V): I would check your FPGA datasheet if negative output voltages are supported first, as most FPGAs only support positive output voltages (i.e. 0V to +3.3V). If you really want/need to increase the output range, you would need to use a level shifter (you can build one using an OPAMP, and use it as as comparator), but even doubling the range, you will still have the limitation on the output current imposed by the FPGA output pins.

     

    Fabio

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  • Hi Andy,

     

    If you are going to drive the speaker with a current between 4mA and 8mA, I suspect you will struggle to ear anything at all. For a 0.5W 8 Ohm speaker, nominally you get the loudest sound when you drive the speaker at maximum current. Doing a little calculation, the max current would be 250mA. Assuming you will be able to drive the speaker with 8mA current, this would only give 0.5mW.

     

    To make things worse, maximum power transfer between the source and the load requires the two to have matching impedance (8 Ohm in this case). I don't know what is the resistance of the FPGA output pin, but a typical termination resistance value is 50 Ohm, in which case, even connecting the output pin directly to the speaker (not a good idea!), there is already quite a mismatch between the source and the load. Adding 400 Ohm resistor to limit the output current will only make things worse, and most of the power will end up being dissipated there rather than reach the speaker.

     

    I think the easiest way to interface the speaker with the FPGA output is to do what John suggested: use a MOSFET to drive the speaker, and drive the MOSFET with the FPGA output pin.

     

    One note on the use of 2 output pins to get a differential output with doubled range (i.e. -3.3V to +3.3V): I would check your FPGA datasheet if negative output voltages are supported first, as most FPGAs only support positive output voltages (i.e. 0V to +3.3V). If you really want/need to increase the output range, you would need to use a level shifter (you can build one using an OPAMP, and use it as as comparator), but even doubling the range, you will still have the limitation on the output current imposed by the FPGA output pins.

     

    Fabio

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