ringing inductor in boost circuit
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From: Michael Robinson on 28 Jun 2010 11:00 "Jamie" <jamie_ka1lpa_not_valid_after_ka1lpa_(a)charter.net> wrote in message news:gjRUn.84$3%3.65(a)newsfe23.iad... > Michael Robinson wrote: > >> I breadboarded a little voltage booster. >> Here's the business end of the circuit: >> >> +5v >> | >> ,------------+ >> | | >> | | >> | ) >> | )220uH >> /e ) >> -1k--| | >> \c ,----+--+-->|--+--out >> | | | | | >> | 68pF = | | | >> | | /c | | >> '--470--+--| - | >> | \e ^ = >> 4k7 | | | >> | | | | >> '----+--+------' >> | >> gnd >> >> The pnp is driven by a square wave. I also have feedback from the >> output, which I didn't show in this drawing, controlling the duty cycle >> to keep the output voltage steady at 16 volts. I used jelly bean >> transistors, Pn2222 and PN2907. The inductor has a fraction of an ohm >> resistance. The circuit is running at about 15kHz with light load. I >> only needed a few tenths of a watt for my purposes. >> I scoped the circuit. When the npn switch turns off the voltage at the >> collector jumps to 16 volts and clamps there, as it should. But when the >> voltage at the collector drops below the voltage on the filter cap, the >> inductor rings. >> >> The diode clamp across the emitter-collector and the 68pF cap are there >> to adress this. The diode clamps the initial negative excursion at the >> collector. The base-collector cap damps the ringing. >> Without that cap in place, ringing is quite pronounced. It starts out >> with a magnitude of ten volts, decays exponentially and is still ringing >> when the npn switch turns on in the next cycle. The cap cuts the intial >> amplitude of the ringing from about ten volts to about two, and it dies >> out much faster. >> >> Is the ringing of the inductor unavoidable in a circuit like this? > have you tried putting a small R in series with the inductor to lower the > Q ? > The ringing occurs because the interwinding capacitance of the inductor forms a tank circuit with its inductance. Attaching a resistor to one of the leads won't change the Q of this tank circuit.
From: dwight on 28 Jun 2010 16:52 On Jun 23, 2:42 pm, "Michael Robinson" <kellrobin...(a)yahoo.com> wrote: > I breadboarded a little voltage booster. > Here's the business end of the circuit: > > +5v > | > ,------------+ > | | > | | > | ) > | )220uH > /e ) > -1k--| | > \c ,----+--+-->|--+--out > | | | | | > | 68pF = | | | > | | /c | | > '--470--+--| - | > | \e ^ = > 4k7 | | | > | | | | > '----+--+------' > | > gnd > > The pnp is driven by a square wave. I also have feedback from the output, > which I didn't show in this drawing, controlling the duty cycle to keep the > output voltage steady at 16 volts. I used jelly bean transistors, Pn2222 > and PN2907. The inductor has a fraction of an ohm resistance. The circuit > is running at about 15kHz with light load. I only needed a few tenths of a > watt for my purposes. > I scoped the circuit. When the npn switch turns off the voltage at the > collector jumps to 16 volts and clamps there, as it should. But when the > voltage at the collector drops below the voltage on the filter cap, the > inductor rings. > > The diode clamp across the emitter-collector and the 68pF cap are there to > adress this. The diode clamps the initial negative excursion at the > collector. The base-collector cap damps the ringing. > Without that cap in place, ringing is quite pronounced. It starts out with > a magnitude of ten volts, decays exponentially and is still ringing when the > npn switch turns on in the next cycle. The cap cuts the intial amplitude of > the ringing from about ten volts to about two, and it dies out much faster. > > Is the ringing of the inductor unavoidable in a circuit like this? Hi You could make a voltage doubler that would transfer more of the lost energy to the load. The ring would swing around the 5v line so a diode from the 5v line and two diodes would knock down the amplitude of the ring by at least 50%. One diode from the other end of the coil to the capacitor and the other diode from this junction to the load. Maybe make it a tripler or quadrupler and really knock it down. It would improve efficiency as well. You'd want to put a negative blocking diode in series with transistor instead of across it to take advantage of the ring rather then dumping it. The idea would be to take advantage of the coils deficiencies. Don't know if it would work but just thinking. Dwight
From: dwight on 28 Jun 2010 17:43 On Jun 28, 1:52 pm, dwight <dkel...(a)hotmail.com> wrote: > On Jun 23, 2:42 pm, "Michael Robinson" <kellrobin...(a)yahoo.com> wrote: > > > > > I breadboarded a little voltage booster. > > Here's the business end of the circuit: > > > +5v > > | > > ,------------+ > > | | > > | | > > | ) > > | )220uH > > /e ) > > -1k--| | > > \c ,----+--+-->|--+--out > > | | | | | > > | 68pF = | | | > > | | /c | | > > '--470--+--| - | > > | \e ^ = > > 4k7 | | | > > | | | | > > '----+--+------' > > | > > gnd > > > The pnp is driven by a square wave. I also have feedback from the output, > > which I didn't show in this drawing, controlling the duty cycle to keep the > > output voltage steady at 16 volts. I used jelly bean transistors, Pn2222 > > and PN2907. The inductor has a fraction of an ohm resistance. The circuit > > is running at about 15kHz with light load. I only needed a few tenths of a > > watt for my purposes. > > I scoped the circuit. When the npn switch turns off the voltage at the > > collector jumps to 16 volts and clamps there, as it should. But when the > > voltage at the collector drops below the voltage on the filter cap, the > > inductor rings. > > > The diode clamp across the emitter-collector and the 68pF cap are there to > > adress this. The diode clamps the initial negative excursion at the > > collector. The base-collector cap damps the ringing. > > Without that cap in place, ringing is quite pronounced. It starts out with > > a magnitude of ten volts, decays exponentially and is still ringing when the > > npn switch turns on in the next cycle. The cap cuts the intial amplitude of > > the ringing from about ten volts to about two, and it dies out much faster. > > > Is the ringing of the inductor unavoidable in a circuit like this? > > Hi > You could make a voltage doubler that would transfer more of > the lost energy to the load. The ring would swing around the 5v > line so a diode from the 5v line and two diodes would knock > down the amplitude of the ring by at least 50%. One diode from > the other end of the coil to the capacitor and the other diode from > this junction to the load. Maybe make it a tripler or quadrupler > and really knock it down. It would improve efficiency as well. > You'd want to put a negative blocking diode in series with transistor > instead > of across it to take advantage of the ring rather then dumping it. > The idea would be to take advantage of the coils deficiencies. > Don't know if it would work but just thinking. > Dwight Hi I gave it a little more thought. You'd need at least a voltage quadrupler. You'd use relatively small capacitors since it is not intended to be the primary voltage increasing circuit, just the dump for excess coil power. Dwight
From: Hammy on 28 Jun 2010 19:07
On Mon, 28 Jun 2010 08:00:40 -0700, "Michael Robinson" <kellrobinson(a)yahoo.com> wrote: snip >The ringing occurs because the interwinding capacitance of the inductor >forms a tank circuit with its inductance. Attaching a resistor to one of >the leads won't change the Q of this tank circuit. > Your saying its ringing when the diode switching from conducting to blocking or at least that is what it I gathered from >>But when the voltage at the collector drops below the voltage on the filter cap, the >> inductor rings. Here is an excert from Basso's book. Figure 7-10 showed some ringing appearing when the diode abruptly blocks. These oscillations find their roots in the presence of stray elements such as the leakage inductance, the lump capacitor, and all associated parasitic elements. Damping the network consists of artificially increasing the ohmic losses in the oscillating path.<------- The damping resistor value can be found through a few simple equations pertinent to RLC circuits. The quality coefficient of a series RLC network is defined by Q= Wo * L/ R |