Has the value of inductor been selected properly? The rule of thumb for a boost converter in continuous conduction mode is L = (V * D)/(R*I_l*f), where the parameters are evaluated at the _lowest_ specified (worst case for a boost) input voltage. V is the input voltage, I_l is the DC inductor current (I_o/(1-D)), D is the duty cycle, f is the switching frequency. R is the "ripple current ratio" and has an optimum value of 0.4 for continuous conduction mode in most cases.
At low voltages the inductor can be sized solely based on the DC load current requirements - if the current limiting is fast enough it doesn't matter if the inductor saturates under abnormal conditions.
If the inductor is still too big maybe use discontinuous mode instead? It'll be harder to stabilize, unfortunately.
But 4 ohms d.c.r. means it'll drop 600mV @ 150mA. Increasing f by a factor of 3 reduces i^2*r loss by that same factor. Or, keep the loss and trade for a smaller L and C.
Switching inductors without magnetic shields make RFI, magnetic nasties, etc. Shielded are better.
Ian _and_ a notable other poster on this topic are severely ignorant about the real world.
PWM BUCK is trivial. I chose BURST for the hub dynamo because it would be a freebie around the 555, and I was intent on rubbing the 555 up some snobbish noses ;-)
PWM BOOST is NOT trivial, though I'm well on my way to an off-the-wall STABLE solution... found by running the math instead of my mouth. ...Jim Thompson
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| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
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