discharging

D

default

Jan 1, 1970
0
I tried the buz21 mosfet , it work good, but heats up real quickly.
thats a reason why I wanted to use the 3055. . The buz21 only work when I
apply at least 5v to the base. Itried a new 3055 with only 1v at the base as
Jamies suggested, but it did not work.
I really want to get it going with the 3055 transistor

k
Buz21 spec is 100 volts and point one ohm on resistance. It shouldn't
be getting hot. Same application? source to ground - drain connected
to the coil and coil connected to positive?

If that got hot, something is wrong. Possibly it needed more gate
voltage - I don't have the specs in front of me - as a rule of thumb I
figure an N channel mosfet should turn on with about +4 volts on the
gate- and that is close to its linear region (not turned on fully)

All Mosfets require a fairly high gate voltage to turn on (that is an
intrinsic characteristic of the part) - but a very low gate current
(Pico amperes - compare that to the 3.3 amps a 3055 wants to switch
ten amps.)

Another thing you can encounter with a mosfet is the gate drive
current is so low that a sloppy layout can pick up stray power line
hum, or an on board ground loop causing oscillations, or some dirt
that is conductive - get one of them toggling at a mega hertz or so
and they do get warm.

The only time they become hard to drive is when you try switching them
fast - then the relatively high gate capacitance will take some
current . In a low frequency or DC application that doesn't apply.
High frequency to a power mosfet would be ~20 kHz or higher - In that
case the power goes to charging and discharging the gate capacitance
rapidly - reactive current, does no work, produces no heat, but you
still have to be able to supply it..

A 2N3055 with a volt of positive polarity on the base with respect to
the emitter (emitter grounded) should be turned on very hard. Most
bipolar transistors turn on at point six volts and are over driven at
..7 volts. One volt base drive is a lot. At that current the
collector should be very close to ground potential.

It should be working . . . Bad batch of parts, wrong part entirely,
bad or just destroyed part? (they can't have an unlimited current
pumped into the base - over seven amps will destroy them) You must be
overlooking something.
 
J

Jamie

Jan 1, 1970
0
Ken said:
HI

I did you a voltage divider. , I have about 4 volt to the base.
The battery supplying the coil is a 12 v battery and I am barely getting 5
volt impulses from the coil.
i am not doing any Em kick back as you mentionned.

ken
if you are truly applying 4 volts to the base
with the Emitter to common then you have already
damaged your 2N3055. you should be reading around
0.5 ... 1.0 max on that base .
is it possible you do not have your battery
ground connected to the 555 common ? if so, if
your worried about isolation then use an opto-coupler.
 
K

Ken O

Jan 1, 1970
0
A 2N3055 with a volt of positive polarity on the base with respect to
the emitter (emitter grounded) should be turned on very hard. Most
bipolar transistors turn on at point six volts and are over driven at
.7 volts. One volt base drive is a lot. At that current the
collector should be very close to ground potential.

It should be working . . . Bad batch of parts, wrong part entirely,
bad or just destroyed part? (they can't have an unlimited current
pumped into the base - over seven amps will destroy them) You must be
overlooking something.


After some more testing, the BUZ21 do not heat,.. thats good.
I ordered ten 2N3055 from digikey. They cant be all scrap... could they?
I will do some more test on them tonight or tomorrow. I worked on this all
day. I used 0.5 volt to the base of the 3055, but I maybe there is not
enough current to drive it. My ammeter cant detect anything.
from the specs: http://www.ortodoxism.ro/datasheets/motorola/2N3055.pdf
Ib max is 7amps and Veb is 7v so emitter is grounded leaving a max of 7v
to base.
I'll work on this a bit later or maybe choose another high power transistor.

ken
 
K

Ken O

Jan 1, 1970
0
you better put a diode test on the base to emitter of those transistors
to make sure they are still alive.
in any case.., you don't apply that much current in the base.

try a 60 ohm or near it from the 555 output driving the base.
place a cap of lets say around .01 from the collector to emitter.

Just to be on safe side, when you say: from the collector to emitter.
you mean positive on the collector ? I only have 0.1uF left 35 and 50 v
ones.. I assume electrolytics?
I just want to make sure I do exactly hat yuou want me too.
My ideas bank is practically exhausted. I hope to wake up with some new one
tomorrow :)

ken
 
D

default

Jan 1, 1970
0
After some more testing, the BUZ21 do not heat,.. thats good.
I ordered ten 2N3055 from digikey. They cant be all scrap... could they?
I will do some more test on them tonight or tomorrow. I worked on this all
day. I used 0.5 volt to the base of the 3055, but I maybe there is not
enough current to drive it. My ammeter cant detect anything.
from the specs: http://www.ortodoxism.ro/datasheets/motorola/2N3055.pdf
Ib max is 7amps and Veb is 7v so emitter is grounded leaving a max of 7v
to base.
I'll work on this a bit later or maybe choose another high power transistor.

ken

Your parts should be good. Digikey is a prime source and unlikely to
be trash.

The base emitter junction takes .6 volts to turn on .5 probably won't
turn it on at all. It is a diode junction - begins to conduct at
point 6. A hot transistor will conduct with a lower base voltage.
Heat it to 100 C and it would probably switch on at .5 volts.

The 7 volts you're quoting is the absolute maximum rating - that's the
point where it smokes - not part of anything you're likely to see
(except maybe with a scope in a high power dynamic application - like
a brief spike)

Look down at the "on" voltages graph that's where you should be
running and with a 50 ohm coil (if that's right) you should be
working with about point 65 volts.

2N3772 or 2N3773 are somewhat better parts and frequently used to
replace the 3055. But if the 2N3055 isn't working it is unlikely
another bipolar will.

In one of your posts you make it sound like you connected the 555
directly to the base of the transistor without a current limiting
resistor - if so it is probably dead.

Test the transistor to make sure it turns on when it should:

Put your transistor emitter to ground, collector to the coil and coil
to +12. Nothing should happen. Connect a 100 ohm resistor from +12
to the base and the coil should turn on - that tests that part of it.

Connect the 100 ohm to the output of the 555 (pin 3) and it should
turn on and off with the oscillator.

It wouldn't be a bad idea to put a damping diode across the coil (a
reversed biased diode across the coil (band lead, cathode to +). Then
when the transistor switches off the voltage spike won't eat the
transistor.

I have an induction coil that easily produced 450 volts when it
switches off from a 20 volt power supply. I use a horizontal drive
transistor (actually three of them in parallel) rated for 450 volts
and it still requires a zener to protect the transistor (can't use a
damping diode in this case because I want it to "ring" when it
switches - that 450 V spike produces 100,000 volts on the secondary)
 
J

Jamie

Jan 1, 1970
0
Ken said:
After some more testing, the BUZ21 do not heat,.. thats good.
I ordered ten 2N3055 from digikey. They cant be all scrap... could they?
I will do some more test on them tonight or tomorrow. I worked on this all
day. I used 0.5 volt to the base of the 3055, but I maybe there is not
enough current to drive it. My ammeter cant detect anything.
from the specs: http://www.ortodoxism.ro/datasheets/motorola/2N3055.pdf
Ib max is 7amps and Veb is 7v so emitter is grounded leaving a max of 7v
to base.
I'll work on this a bit later or maybe choose another high power transistor.

ken
you better put a diode test on the base to emitter of those transistors
to make sure they are still alive.
in any case.., you don't apply that much current in the base.

try a 60 ohm or near it from the 555 output driving the base.
place a cap of lets say around .01 from the collector to emitter.
 
P

Pooh Bear

Jan 1, 1970
0
Ken said:
I tried the buz21 mosfet , it work good, but heats up real quickly.
thats a reason why I wanted to use the 3055. . The buz21 only work when I
apply at least 5v to the base.

Yes, that's normal. It would work better still ( less heat ) if you applied more
volts to the gate.
Itried a new 3055 with only 1v at the base as
Jamies suggested, but it did not work.
I really want to get it going with the 3055 transistor

Bipolar transistors are not *voltage driven* they are current driven.

Your immediate problem seems to be a lack of understanding of basic electronics.

Graham
 
K

Ken O

Jan 1, 1970
0
Bipolar transistors are not *voltage driven* they are current driven.

Your immediate problem seems to be a lack of understanding of basic
electronics.


yes, and to create current I have to add a resistor in the output of the
555, I tried 5kohm ( little current) and 10ohm (lots of current) but
nothing.

Ken
 
P

Pooh Bear

Jan 1, 1970
0
Ken said:
yes, and to create current

No, it doesn't 'create' current. That's to *limit* current so the 555 isn't
driving a near short circuit ( the 2N3055's base ) , or more accurately - to
define the base current.
I have to add a resistor in the output of the
555, I tried 5kohm ( little current) and 10ohm (lots of current) but
nothing.

Well you're clearly doing something basic very wrong.

I note you haven't posted a schematic yet.

Graham
 
D

default

Jan 1, 1970
0
As for the electromagnet effect: with the resistor you cut the

huh??? should be half as strong.

You're right - ampere turns, cut the current in half with the same
current and the ampere turns is 1/2 not 1/4 as I stated.

Ken is getting me confused.
 
D

default

Jan 1, 1970
0
the cmos ones can't source much.

Good observation. I've been taking it for granted he's using a
standard NE555. If he was using a cmos part and trying to drive a
workhorse like the 2N3055 that would explain some of the problems.

The other thing I tend to doubt <at this point> is that the coil has a
resistance of 50 ohms.
 
K

Ken O

Jan 1, 1970
0
I note you haven't posted a schematic yet.


you are right,
I made a smallpage for this effect:
http://www3.sympatico.ca/lerameur/

the resistance of the coil is 10.2 Ohms
I am using the chip NE555N from

I am getting a good effect with the Buz21 chip (mosfet). I just want to do
it with the 2n3055.

ken
 
D

default

Jan 1, 1970
0
you are right,
I made a smallpage for this effect:
http://www3.sympatico.ca/lerameur/

the resistance of the coil is 10.2 Ohms
I am using the chip NE555N from

I am getting a good effect with the Buz21 chip (mosfet). I just want to do
it with the 2n3055.

ken
I can believe the ten ohms. I rewound the rotor for my motorcycle and
it was 4 ohms and, surprisingly, uses a 3055 for a regulator.

Ten ohms shouldn't take a lot of base drive it is only just over an
amp.

What is that thingee between what appears to be the base and emitter
of the 3055 - a diode? How is it in the circuit - forward or reverse
biased? can't see the cathode band . . . that has to be reverse
biased to allow the 3055 to have enough current to work - and it is
totally unnecessary no matter which direction it is in - might be of
some use with a mosfet to protect the gate, but not with a bipolar.

Air core coil? Iron has 800+ times the permeability of air, so an air
core electromagnet is weak indeed.

There's nothing to be gained from using two power supplies - one for
the 555 and one for the load coil. The 555 is happy working from 12
volts and will provide some more base drive to the 3055.

And that may account for the heating in the mosfet you saw earlier
(two batteries). 6 volts is too close to the 4-5 volts the mosfet
needs to turn on completely. The 555 can drop a volt or two and that
puts the mosfet in its linear region (not turned on fully - drops
voltage across it - generates heat)
 
K

Ken O

Jan 1, 1970
0
What is that thingee between what appears to be the base and emitter
of the 3055 - a diode? How is it in the circuit - forward or reverse
biased? can't see the cathode band . . . that has to be reverse
biased to allow the 3055 to have enough current to work - and it is
totally unnecessary no matter which direction it is in - might be of
some use with a mosfet to protect the gate, but not with a bipolar.

Well it was originally remove, as in none, Someone told me to trythat but
i left if for the picture. It has been removed.
Air core coil? Iron has 800+ times the permeability of air, so an air
core electromagnet is weak indeed.

No there is iron in there, there is duct tale over it because there are
small rods

There's nothing to be gained from using two power supplies - one for
the 555 and one for the load coil. The 555 is happy working from 12
volts and will provide some more base drive to the 3055.

And that may account for the heating in the mosfet you saw earlier
(two batteries). 6 volts is too close to the 4-5 volts the mosfet
needs to turn on completely. The 555 can drop a volt or two and that
puts the mosfet in its linear region (not turned on fully - drops
voltage across it - generates heat)

would the batteries affect from working or not?
I will try it.

I will also put a voltage divider to i get about 0.5 volt into the base as I
did before. (Not in the picture)

ken
 
D

default

Jan 1, 1970
0
No there is iron in there, there is duct tale over it because there are
small rods
Oh. OK sounds like my induction coil. Lots of small wires.
would the batteries affect from working or not?
I will try it.

Just run the 555 from the 12 volt supply
I will also put a voltage divider to i get about 0.5 volt into the base as I
did before. (Not in the picture)

What is this about a voltage divider? The 555 outputs a source
(connection to the + supply - internal to the 555). You put a current
limiting resistor to keep the 555 from dying in an effort to supply
the (effective short) base emitter junction.

A "voltage divider" implies, to me, that you put two or three
resistors to set the base voltage - that isn't how it is done=== the
555 can output just so much current. The 3055 will eat that current
and more (killing the 555 if you don't limit it)

The way it is normally done - you calculate the current that the 3055
needs to turn on with the load you have. (from the data sheet on the
3055) That will be about 100 milliamps with a 10 ohm coil working from
a 12 volt supply. The 555 can source over 100 milliamps (source=puts
out a positive voltage and current, "sink" is the opposite - takes the
current to ground). 555's do both - sink and/or source.

That's plenty to turn on the 3055. You figure your power supply
voltage, the drop inherent to the 555, the drop inherent to the 3055
(the six tenths of a volt or so) and calculate a current limiting
resistor so that the 3055 turns on and the 555 doesn't die.

The short answer: There should be a 100 or 200 ohm resistor between
the 555 and transistor base. It should be capable of dissipating a
watt of power - not the 1/4 watt resistor we all work with - bigger.

There should be no diode between the base and emitter.

There should be a reverse biased diode (cathode (side opposite the
arrow) or the band on the diode) to the plus 12
V and coil where it connects to the 3055 (that will keep the 3055 from
shorting when it tries to interrupt the current and creates a big
voltage spike. - the diode is reversed biased so doesn't eat any
current. When the transistor switches off it absorbs the reverse EMF
and saves the transistor from the high voltage spike.

(common practice with coils driven by transistors)

Can you watch the alt.binaries.schematics.electronics Usenet group? I
can draw and post a jpg to show it.
 
K

Ken O

Jan 1, 1970
0
Hi
I did as you told, put everything on the 12v battery. butmy 555 did not like
that, one reason why i was putting it on 6v.
It blew up in flames.. do they put gun powder in these thing?
I,ll have to wait and buy some more tomorrow.

ken
 
P

Pooh Bear

Jan 1, 1970
0
Ken said:
Hi
I did as you told, put everything on the 12v battery. butmy 555 did not like
that, one reason why i was putting it on 6v.
It blew up in flames.. do they put gun powder in these thing?

This means you're *definitely* doing something wrong !

Graham
 
D

default

Jan 1, 1970
0
Hi
I did as you told, put everything on the 12v battery. butmy 555 did not like
that, one reason why i was putting it on 6v.
It blew up in flames.. do they put gun powder in these thing?
I,ll have to wait and buy some more tomorrow.

ken
The 555 works to a power supply of 16 volts absolute maximum and
outputs 200 ma max and can dissipate 600 milliwatts max before it
overheats.

That 50 ohm resistor the picture shows going to the base of the
transistor (assuming too it really is the base) has to be larger - use
a 100 ohm and not lower than 75 ohms, or the 555 will die a hot death.
A 100 ohm should have enough current drive to turn on the 3055 with
a 10 ohm load.

I'm assuming the emitter of the 3055 is going to ground - right?

75 ohms should work up to absolute maximum of a 15 volt supply - a
lead acid battery puts out about 12.5 unless it is being charged then
it can go as high as 14 volts or so.
 
K

Ken O

Jan 1, 1970
0
Hi

Look at the message 4 from Jamie, he said that I should have about 0.5 v to
the base of the 3055. the output of the 555 is about 7v, so from this is why
I tried the voltage division thing you mentioned earlier

The 555 works to a power supply of 16 volts absolute maximum and
outputs 200 ma max and can dissipate 600 milliwatts max before it
overheats.

yes I saw 18v max , I did not change anything in this circuit, it was
working fine before.

That 50 ohm resistor the picture shows going to the base of the
transistor (assuming too it really is the base) has to be larger - use
a 100 ohm and not lower than 75 ohms, or the 555 will die a hot death.
A 100 ohm should have enough current drive to turn on the 3055 with
a 10 ohm load.

I'm assuming the emitter of the 3055 is going to ground - right?

yes it is.
75 ohms should work up to absolute maximum of a 15 volt supply - a
lead acid battery puts out about 12.5 unless it is being charged then
it can go as high as 14 volts or so.

yes its about 12.6v now

I still need to get some 555, I'll continue later tonight on it or tomorrow
at the latest
thanks for all the help.

ken
 
D

default

Jan 1, 1970
0
Look at the message 4 from Jamie, he said that I should have about 0.5 v to
the base of the 3055. the output of the 555 is about 7v, so from this is why
I tried the voltage division thing you mentioned earlier

He is correct it should be .5 to .7 if everything is working - what is
not correct is that you have to provide the .5 to .7 the transistor
will do that.

You provide current, but the transistor determines the voltage. You
limit the current you provide with a resistor (sized high enough to
keep the 555 from burning up and low enough to push enough current
into the 3055 to turn it on).

~100 ohms by my estimate, but as low as 75 should still work - lower
than that, and the 555 smokes. (will be too close or above the design
maximum for current)

One side of the resistor will have pretty close to the supply voltage
on it (when the 555 is outputting a high signal) the other side will
have .5 to .7 - that the transistor limits the voltage to.

You can check your transistor/coil drive simply by taking the side of
the resistor that goes to the 555 and connecting it to +12 - that
should turn on the transistor and energize the coil - disconnect it
and the coil should de energize.

That tells you that the resistor is low enough to turn on the 3055 and
it is hooked up correctly, and the 3055 is OK -

if it won't turn on and off as the resistor is connected to the +12 it
won't work with a 555 either.

Just a simple and easy way to check that part of the circuit
separately. "If'n that don' work ain't nothin' goin' ta work"

(that is disconnect the resistor from the 555 and connect it to the
+12 temporarily to check to see that the transistor turns on when it
should. DO NOT connect it AND pin 3 of the 555 to +12 - bad things
will happen.)

If your 555 still refuses to work on 12 volts you must be doing
something else wrong - for instance:

Pin 7 the discharge pin is the collector of a transistor inside the
555. In an astable multivibrator (the oscillator connection) it
discharges the capacitor that sets the timing. If the cap size is
very large or the resistor charging the cap is very small, that poor
little transistor gets hot and dies - and that symptom will be
exacerbated by raising the supply voltage. Exactly when the little
bugger is pushed close to its limit is also determined by frequency
and/or pulse width. (read the application notes - from national
semiconductor - on the 555 they should mention this)

All design is a balancing act and series of compromises. Until you
learn the basics, you'll be having trouble. You are learning . . .
that much already.
 
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