Re: Joule Thief - still not working....

On Sat, 25 Jul 2009 21:10:42 GMT, Jon Kirwan
<jonk@xxxxxxxxxxxxxxxxxxx> wrote:

On Sat, 25 Jul 2009 13:43:14 -0700, John Larkin
<jjlarkin@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:

On Sat, 25 Jul 2009 19:41:33 GMT, Jon Kirwan
<jonk@xxxxxxxxxxxxxxxxxxx> wrote:

On Sat, 25 Jul 2009 12:01:01 -0700, John Larkin
<jjlarkin@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:

On Sat, 25 Jul 2009 18:38:34 GMT, Jon Kirwan
<jonk@xxxxxxxxxxxxxxxxxxx> wrote:

On Sat, 25 Jul 2009 15:40:36 GMT, ehsjr <ehsjr@xxxxxxxxxxxxxxxxx>
wrote:

Jon Kirwan wrote:
On Fri, 24 Jul 2009 08:51:39 -0700, John Larkin
<jjlarkin@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:


<snip>

At any
rate, the value of R influences both the ON time and the OFF time.

<snip>

The OFF time is NOT determined by R1, so far as I can tell.

John Larkin is right.

In the sense that Ic_peak is higher. The off time equation is:

t_off = Ic_peak * Lprimary / (Vout + Vschottky - Vbattery)
t_on = Ic_peak * Lprimary / (Vbattery - Vcesat)

I suppose I failed to follow John's point because I was looking at
these as separate processes and the fact that while frequency is
indeed influenced by Rbase, that it isn't important because it is also
influenced by Lprimary and that can be adjusted, as needed.

So I like to look at the design process here as choosing the necessary
Ic_peak from what greg and I were talking about earlier:

N = (Vout+Vd-Vcesat)/(Vin-Vcesat)
Ic_peak = 2 * Iout * (N + 1)

And then using Lprimary by selecting a desired 'f':

Lprimary = (Vin-Vcesat)*(Vout+Vd-Vin)/(Ic_peak*(Vout+Vd-Vcesat)*f)

Rbase isn't selected due to off time, on time, etc., but instead based
upon the BJT and Ic_peak:

Rbase = 2*beta*(Vin-Vbeon)/Ic_peak

That's the thing I don't like about this circuit, the large value of
Rb, which wastes power and seriously slows down transistor switching
time, wasting more power. All this gets worse as Vin increases.

I'll look at that aspect. But on first glance it doesn't seem that
much. For example, at 2mA during the ON time and figuring 4.5k,
that's 18mW even assuming that the ON time is 100% of the total, which
it isn't. Meanwhile, the circuit is delivering more than 600mW to the
load, according to a spice simulation here. Frequency is 100kHz.

Transistor turnoff is especially ugly.

The reverse transit time seems a significant problem as the frequency
goes up.

To turn a transistor off fast, you have to suck current out of the
base. This circuit doesn't do that. At best is applies a wimpy
negative current into the base at turnoff time.

Let me try and think about this more closely. Assume for a moment
that we are driving six 3.3V LEDs. Output voltage is 20V. Battery
voltage is three 1.5V's for a total of 4.5V. When the transformer
flies back, there is about (20-4.5+.35) across the collector winding,
or 15.85V... call it 16V. This is reflected back by flux changes to
the base winding as ... 16V to oppose the 4.5V battery. 11V or so,
net. That's a fair push to sweep out charge, isn't it, even with Rb?

Are you assuming a 1:1 transformer? Back-biasing the base by 11 volts
will zener the b-e junction. That degrades the transistor beta over
time.

But it's charge we're trying to pull out of the base, and that takes
current, not just voltage.


Okay. So maybe the point arrives to me, now. What you are saying is
that the base drive current during the ON phase is of a similar size
to the current available during the reverse transit time, which really
needs to be a LOT higher to be much faster. But then I'd argue that
it isn't all that important because _with_ only that current, the OFF
transition is still in the area of the reverse transit time of the BJT
and that with a 2N2222, that's only about 100ns or so. Making that
faster is fine, but if the frequency is only 100kHz or so, it's not
going to make a huge difference.


But then... okay. Maybe I'm not getting the point, at all.

The base only needs 0.7 volts or so; it's silly to get that from a 5
volt supply, or more like 10 when the transformer feedback winding
signal is added.

Actually, that helps, doesn't it? First, doubling the voltage means
the battery can be usefully used down to lower values. Second, a
higher voltage implies more of a current source behavior via Rb, as
Vbe droop variations are a lower percent of the total.

The "current source bahavior" is the enemy of speed. One more resistor
can help a lot.

To ground from the base?

No. Make a voltage divider from Vbat to ground. Its output voltage
should be +0.8 or so, enough to ensure powerup turn-on. Connect the
feedback winding between that and the base. Scale the turns ratio to
give maybe 2 volts p-p across the feedback winding. Scale the divider
resistors so that the thevenin impedance of the divider results in the
desired turn-on base current. Now we have a low-voltage, low-impedance
base drive circuit. When the transistor is on, most of the drive
current is coming from ground, not Vbatt. And there's plenty of stiff
drive to turn the transistor off.

More complex circuits can do even better.

John

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