boost circuit

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

 

[Bob Monsen]

<lilipot@ymail.com> wrote in message
news:f646a429-7782-4fd2-9d35-e92eee9b32f5@79g2000hsk.googlegroups.com...

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