Hi all,
In short, I am working on a device that is to take a pulse train from a microcontroller output (Duty Cycle < 0.5% @ 5V, 30 Hz), give it a current boost using a Darlington pair, and increase the signal of the voltage by a factor of 12 with a step-up transformer.
In order to get an output on-time current pulse of around 40 mA it would mean that my Darlington pair current through the transformer primary windows would need to have an on-time value of around 500 mA.
Ideally I would like to have this unit battery powered, and am considering using series AAs or a 9V, but not sure if they can handle the load of the Darlington pairs for a significant duration of use. The device is only operating for around 3 seconds.
From wikipedia it mentions that the RMS value of a pulse train is:
\(
I_{DPrms} = I_{pk}\sqrt{Duty Cycle} = 500mA\sqrt{0.005} ~= 35 mA
\)
That seems like a relatively small amount for the battery to provide, but I feel as though the 500 mA draw might be too much for the battery to take.
Hope that someone might be able to lend a hand, for deciding on a power supply method.
Thanks,
JP
In short, I am working on a device that is to take a pulse train from a microcontroller output (Duty Cycle < 0.5% @ 5V, 30 Hz), give it a current boost using a Darlington pair, and increase the signal of the voltage by a factor of 12 with a step-up transformer.
In order to get an output on-time current pulse of around 40 mA it would mean that my Darlington pair current through the transformer primary windows would need to have an on-time value of around 500 mA.
Ideally I would like to have this unit battery powered, and am considering using series AAs or a 9V, but not sure if they can handle the load of the Darlington pairs for a significant duration of use. The device is only operating for around 3 seconds.
From wikipedia it mentions that the RMS value of a pulse train is:
\(
I_{DPrms} = I_{pk}\sqrt{Duty Cycle} = 500mA\sqrt{0.005} ~= 35 mA
\)
That seems like a relatively small amount for the battery to provide, but I feel as though the 500 mA draw might be too much for the battery to take.
Hope that someone might be able to lend a hand, for deciding on a power supply method.
Thanks,
JP