Diode boost converter

410

08:02:14

An inductor is connected in series with a diode and capacitor (a resistor is shown parallel to the capacitor, representing the circuit's load). A relay is used to momentarily shunt the inductor, so that it stores up energy. When the relay is switched OFF, the inductor must dump its energy through the diode and into the capacitor. The capacitor cannot discharge back through the diode, so it accumulates an increasing voltage.

published 9 years ago

Karsten

9 years ago

Nice, but the battery (at least in EC) produces current spikes of 2.3 kA.

Imrul

9 years ago

This can be solved by increasing PWM frequency. Butt it may not switch the relay. So need to use BJT or Mosfet

hurz

9 years ago

100n impedance is already lower then 1mOhm relais contact. So most power is going into heat at this contact. Efficeny <<50% increase freq and or inductivity

thebugger

9 years ago

Not to mention that there are no relays that switch at 1Khz, even worse at spiked switching. The coil hysteresis will not allow you to switch this fast. Furthermore, no mechanical device can switch at such speeds without wearing out the contacts

cjwinstead

9 years ago

Thanks for the comments, but this example is just a circuit theory demo for an introductory chapter on diodes. The goal is to demonstrate basic theory, not multiply the practicalities of implementation.

add comment in editor

EveryCircuit is an easy to use, highly interactive circuit simulator and schematic capture tool. Real-time circuit simulation, interactivity, and dynamic visualization make it a must have application for professionals and academia. EveryCircuit user community has collaboratively created the largest searchable library of circuit designs. EveryCircuit app runs online in popular browsers and on mobile phones and tablets, enabling you to capture design ideas and learn electronics on the go.