# step up converter

Discussion in 'Electronic Basics' started by lerameur, May 28, 2007.

1. ### lerameurGuest

Hello,

I am building a step converter, the circuit works and is based on the
following web site:
http://services.eng.uts.edu.au/~venkat/pe_html/ch07s3/ch07s3p1.htm
My circuit boost from 5v to 150v, bit it takes about 45 seconds to
reach the maximum voltage with the proper duty cycle. I found out by
using a bigger coil I get a faster result. Is there a way to reach the
150v within 10 seconds or less? or is this how it usually works and we
cannot do anything about it..

Thanks

ken

2. ### CharlesGuest

Use a push-pull converter configuration for a much faster response time.

4. ### Guest

Strewth!. I didn't think these things could get that horribly
complicated. Any chance of you posting somewhere your actual circuit?.

5. ### lerameurGuest

Figure 1 on the link above.
http://services.eng.uts.edu.au/~venkat/pe_html/ch07s3/ch07s3p1.htm
with a 2n3055 and a 3300UF 350v elec. cap, no resistor at the load.
That simple. The coil I wound myself with 300 feet of magnetic wire.

ken

6. ### Guest

(at some point you'll need to know your inductor value)

You've complete control over how fast the capacitor charges up.
At 150V your capacitor stores a whopping 40 Joules of energy. A one
second recharge time could be actioned by storing 40J in the inductor
and switching across once per second (an impractical 0.3H and 17amps).
Or storing 4mJ and switching across 10000 times a second. A 10 second
recharge would get you down to a more reasonable couple of amps.
Joules energy stored in the inductor is equal to 1/2 x L x current^2.
(it's the inductor 'peak' 'current at the point where the transistor
is switched OFF, that counts.)

Basically it's how many amps you can get through the inductor when the
transistor is ON and the number of charging/switching cycles you can
put up with each second.
More power implies a lower inductor value as ...
Amps per second rise through inductor = Voltage across inductor / L.
I.e low L values allow a high current (hence energy) build up in a
short time.
Somewhere there is a happy balance. You've got to look for the sweet
spot.

With the 300' of wire and old 2N3055 transistor, your lazy charge
problem could be down to voltage loss across the ON transistor and
high wire resistance.
You've only 5V to play with and at the needed couple of amps or so, an
odd volt and a few ohms means the inductor sees only a fraction of the
full voltage and the design is lost.

7. ### lerameurGuest

Hi,

I did tried different inductor values, I actually get better
performance using a higher value coil.
In one case i used a 12v supply battery and a 220uF inductor (Dn4516-
ND at digikey). I get increment just about the same as before. With a
high value inductor, the current might be less, but the potential peak
is higher.

Ken

8. ### Guest

The lower the coil resistance the better. A high R value can quite
easily dominate the circuit performance. Dependant on quality you can
have a higher inductance but lower R. Commercial inductor makers
strive constantly to reduce their R values.
I can't believe what I've just gone and done but have downloaded and
actually read a 2N3055 datasheet. .
Looks like you'll be losing 1V across it when the inductor is charging
but more to the point, you'll need to be driving the base with
something like 1/2 amp, which is a biblical quantity of current .
Would suggest it's far, far better to get hold of a cheap power FET
instead. You'll get that volt back and have far simpler switching
needs.

9. ### lerameurGuest

Hi,
I have been driving the 2N3055 with a microcontroller, its far from
driving 0.5 amp and it works. I am also looking into driving a high
speed step up transformer. I cannot even find that at digikey. Is it
me or does somebody knows a part number at digikey ?

ken