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Arduino controlled boost converter
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pigrew:

--- Quote from: Arznei on June 28, 2018, 09:52:22 pm ---Thanks for all your comments!

@pigrew
I put the transistor driving circuit in place exactly to increase the gate voltage. From my measurements it does seem to pull the gate to a few hundred mV on LOW and nearly to the full 9V on HIGH. May I ask how you got to your values of 1.2V or 3.8V? I don't want to find out my circuit just so happens to work out of pure luck.


--- End quote ---

I am assuming that you use 5V logic from your micro. Since they are Darlington pairs, you should get two Vbe drops over the transistors. When the base is 5 V, I assumed 0.6 V for each junction, so 5-2*0.6 = 3.8 V. Checking the simulation, the gate voltage is getting much closer to 0 and 5V. So, maybe I'm wrong...
not1xor1:

--- Quote from: Arznei on June 28, 2018, 09:52:22 pm ---Thanks for all your comments!

@pigrew
I put the transistor driving circuit in place exactly to increase the gate voltage. From my measurements it does seem to pull the gate to a few hundred mV on LOW and nearly to the full 9V on HIGH. May I ask how you got to your values of 1.2V or 3.8V? I don't want to find out my circuit just so happens to work out of pure luck.
The frequency is the PWM frequency I talked about. It is currently at 62.5kHz and using the standard Arduino methods I can't really go higher than that and the next lower frequency would be around 7kHz, which i think may be a little low.
I do sense the output via an analog input and regulate the duty cycle on that input. It does oscillate but it stays decently around the target voltage.

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I guess something is wrong with your measurements.
There is no way an emitter follower (single BJT or darlington) can increase the gate voltage.
That configuration boosts just the current.
Arznei:
@pigrew @not1xor1
Oh I see, maybe I didn't clarify that properly. The V+ symbol in the schematic is supposed to be the +9V rail, that is also used to drive the inductor. The micro uses 0-5V to drive the UC pin, but with the transistor stage it should then be increased nearly to V+ wich is 9V in my case, right?
gcewing:

--- Quote from: Arznei on June 29, 2018, 05:22:33 am ---with the transistor stage it should then be increased nearly to V+ wich is 9V in my case, right?

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No. Each base-emitter junction drops 0.6V, so with the UC output at 5V, the emitter of the upper darlington will only be at 5 - 1.2 = 3.8V.

There are a number of ways you could get a bigger output voltage swing. One way would be to connect a resistor from the bases of the darlingtons to +9V, and an NPN transistor with its emitter to ground and its collector to the darlington bases, and drive the base of that transistor from the UC through a resistor. That should give you a 0 to 9V swing at the darlington bases, and a 1.2 to 7.8V swing at the output.
T3sl4co1l:
There's no current sense!

Tim
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