<lilipot@[EMAIL PROTECTED]
> wrote in message
news:f646a429-7782-4fd2-9d35-e92eee9b32f5@[EMAIL PROTECTED]
> Hi,
>
> I have a question concerning a boost circuit Vs a dc motor.
> here is a link to a boost circuit:
> www.intersil.com/engineeringtools/tools/buckandboostcircuit.pdf
> In the step-up circuit, when Q1 is on, L1 is grounded. I feel that
> this circuit resembles a lot like an electric motor and its coil.
> The analogy is when Q1 is on, it would be like when the brushes makes
> contact with the coils. The duty cycles with the time the brushes has
> contact with the coils.
> I tried implementing this circuit to a dc motor pole and I was not
> getting any boosted voltage.The circuit was the same as shown in the
> link, but I did not add a load resistance, just a big capacitor. Does
> somebody knows why I am not getting a boosted voltage?
> There was not filter or smoothing capacitor on the poles of the
> motor. I used a small electric motor from a remote control car.
>
> lili
These things work by storing energy in the coil when the pass transistor
is
on, and then dumping that energy into the output when the transistor is
off.
The amount of boost depends on how quickly you can turn off the
transistor,
since
V = L * I'
Where I' is the rate of change of current with respect to time.
So, the faster you can change the current across the inductor, the more
voltage you will get (for a given inductor).
To get a boost effect, the V in the equation above must be greater than
the
output voltage plus the forward voltage of the diode.
So, to get 10V, for example, given a 1mH inductor, you would need to
switch
the current from 1A to 0 in 0.1ms.
So, the question is, how are you turning off Q1 above? Also, how much
inductance does the motor coil have?
BTW, you would probably have more luck if you bought an inductor to play
around with. They are pretty cheap.
Regards,
Bob Monsen
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