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Strange high frequency push pull transformer action

My bet would be this (asterisk is winding orientation dot):

       .------------->
       |         .--->
       |*        |
       '-'-'-'-'-'
      =============
  (A) .-.-.-.-.-.-. (B)
      |*    |    *|
      |     |     |
  ||--'    +5v    '--||
  ||<-.           .->|| Q2
__||--+ Q1        +--||__
      |           |
     ===         ===

Could be of course, but Vds(Q2) would go to zero when Q1 was on, not
+5v/2.
 
M

mook johnson

Jan 1, 1970
0
.------------->
| .--->
| |
'-'-'-'-'-'
=============
(A) .-.-.-.-.-.-. (B)
| | |
||--' +5v '--||
||<-. .->|| Q2
_||--+ Q1 +--||__
| |
=== ===

Not possible. If the transformer's right, the drain of the "off" FET
has to rise when the other turns on, there's simply no other choice.

If the "off" FET's drain didn't rise to 10v, or spiked then decayed,
etc.--that would be different. For Vds not to rise at all, but fall
instead--that's not possible.

The transformer's wrong. Has to be.


That is strongly my suspicion as well. Just catching hell trying to
prove it to the transformer designer/vender.
 
T

Tim Williams

Jan 1, 1970
0
Could be of course, but Vds(Q2) would go to zero when Q1 was on, not
+5v/2.

And the "baseline" would go up to some flyback voltage, probably with an
exponential L/R decay, not just +5.

The asymmetrical waveform makes sense for the unbal (one driving CT) case,
but only on the middle transistor. A dual trace scope shot, monitoring
both drains, would be illuminating.

Tim
 
M

mook johnson

Jan 1, 1970
0
Shot in the dark:

It could be miller effect due to the capacitance
between the drain and gate..
The opposite side is generating a pulse in an instance and thus the
cap located there is pushing a pulse over to the gate and forces it to
bias on. Of course this is short and shouldn't last long however, at the
frequency you are operating at, it maybe showing the effects if you are
not driving the gate low and hard enough in the off state.

You should probe at the gate of Q1 when Q2 turns on.

Jamie

I considered that as well but the miller cap with a falling drain
voltage would turn off the Fet not turn it on.

Gate drive resistance is a bit high at 200 ohms so I considered that as
well. Driver output resistance is in the 10 ohm range
 
M

mook johnson

Jan 1, 1970
0
Mook seems to have gone very quiet. It would be good to hear the
outcome even if he has to swallow a little humble pie.

Still poking around. You guys thoughts and responses are great and
confirming that I'm not crazy. :)

The transformer guy is saying it is some fancy RF stuff and I don't have
the background to say he's wrong but my gut says its something 1st order
that is the problem.
 
M

mook johnson

Jan 1, 1970
0
Betcha he finds this:

.------------->
| .--->
| |
'-'-'-'-'-'
=============
(A) .-.-.-.-.-.-. (B)
| | |
+5v-' ||--' '--||
||<-. .->|| Q2
__||--+ Q1 +--||__
| |
=== ===

That would explain the first waveforms, anyhow.

Nope,
Tested for that and when Q1 would turn on Q2 would go negative ( at
least until the body diode turned on. Then all hell would break loos on
Q1s current.

That was my first thought as well.
Ready to swing my big hammer and see if that fixes it. :X

I a haven't had much time in the lab to poke around with it. its
"kinda" working as it is (good communication) but the SNR is much lower
and I was after. This has the priority lower than it should be.
 
J

Jamie

Jan 1, 1970
0
mook said:
That is strongly my suspicion as well. Just catching hell trying to
prove it to the transformer designer/vender.
I still think it's miller effects, only because of the voltage you have
indicated, which is ~ 2.5 volts, and the Vgs(th) is around 1.5 or so for
a 700x and putting that in context the way it is, it's making a nice
clamp at said gate voltage minus a little due to R(on). The fast raise
from the opposite side is forcing this condition.

Putting a much slower signal in there as a test could verify that and
you could then watch the.

Oh well, what a world!>..


Jamie
 
J

Joerg

Jan 1, 1970
0
Could be of course, but Vds(Q2) would go to zero when Q1 was on, not
+5v/2.

Not really. Those 2N7002 have several ohms Rdson, then there is the wire
resistance in the transformer, all sprinkled with a pinch of leakage
inductance.
 
Not really. Those 2N7002 have several ohms Rdson, then there is the wire
resistance in the transformer, all sprinkled with a pinch of leakage
inductance.

True, but none of that should matter, right? The Q2 winding section
is supposed to be unloaded when Q1 turns on. Essentially no current
flows in Q2, so the Vd(Q2) should match Vd(Q1) nearly exactly.
 
I had considered that except the OP said,
   "If I probe Q1 drain instead of Q2, I get the same result. where Q1
goes from Vin to ground and goes from Vin to ~ 1/2Vin when Q2 turns on."

Which implies there isn't a circuit asymmetry.

Agreed, it just didn't seem that all these statements can be true, so
I took the most likely guess.
 
T

Tim Williams

Jan 1, 1970
0
Michael A. Terrell said:
Would you rather pass it through a team of Clydesdales? ;-)

Probably taste better!

Tim
 
J

JW

Jan 1, 1970
0
Joerg wrote:
[...]
If it was saturation the current intake would be high and the 2N7002
FETs would attempt to unsolder themselves.
yeah well, that would help out in the replacement process :)

Not always, I've seen device leads that have gotten so hot that they seem
to weld to the pad or via. A soldering iron won't melt the lead (or
whatever it's become).
 
Joerg wrote:
[...]

If it was saturation the current intake would be high and the 2N7002
FETs would attempt to unsolder themselves.

yeah well, that would help out in the replacement process :)

Not always, I've seen device leads that have gotten so hot that they seem
to weld to the pad or via. A soldering iron won't melt the lead (or
whatever it's become).


The old solder is oxidized. I've found that adding some fresh solder
(With a mild RMA flux) helps disolve the old crap.

Our processes are all RoHS, so I do the same. The first thing I do
when soldering a part is add leaded solder to the socialist solder. I
usually take the excess off with some solderwick.
 
F

Fred Abse

Jan 1, 1970
0
My money is still on the centre tap isn't really the centre!! Measure
the inductance of each primary, then across the drain connections and
come back here with the result.

Quick and dirty: Take the transformer out of circuit, excite the secondary
at, say, 5MHz. With respect to the terminal you think is the center tap,
look at what you think are both primary ends with a scope. They should be
antiphase, and reasonably equal. If not, you've found it.
 
B

Bill Sloman

Jan 1, 1970
0
JW wrote:
@newsfe20.iad>:
Joerg wrote:
[...]
If it was saturation the current intake would be high and the 2N7002
FETs would attempt to unsolder themselves.
yeah well, that would help out in the replacement process :)
Not always, I've seen device leads that have gotten so hot that they seem
to weld to the pad or via. A soldering iron won't melt the lead (or
whatever it's become).
  The old solder is oxidized.  I've found that adding some fresh solder
(With a mild RMA flux) helps disolve the old crap.

Our processes are all RoHS, so I do the same.  The first thing I do
when soldering a part is add leaded solder to the socialist  solder. I
usually take the excess off with some solderwick.

It's not socialist solder, it's "we-don't-want-to-make-children-dumber
solder".

krw is a classic example of why this is probably a good idea.
 
JW wrote:

Joerg wrote:

[...]

If it was saturation the current intake would be high and the 2N7002
FETs would attempt to unsolder themselves.

yeah well, that would help out in the replacement process :)

Not always, I've seen device leads that have gotten so hot that they seem
to weld to the pad or via. A soldering iron won't melt the lead (or
whatever it's become).


The old solder is oxidized. I've found that adding some fresh solder
(With a mild RMA flux) helps disolve the old crap.

Our processes are all RoHS, so I do the same. The first thing I do
when soldering a part is add leaded solder to the socialist solder. I
usually take the excess off with some solderwick.

I would hardly call adding leaded solder to a joint a RoHS process. How
certain are you of only leaving the allowable 0.1% lead in a repaired
solder joint?

Idiot. I'm an engineer. Nothing I touch with an iron goes to
customers. I can do anything I want with it.
 
T

Tim Williams

Jan 1, 1970
0
Bill Sloman said:
It's not socialist solder, it's "we-don't-want-to-make-children-dumber
solder".

krw is a classic example of why this is probably a good idea.

So you have the peer-reviewed journal articles illustrating a medically
and economically important improvement in life quality and expectancy
resulting from the reduction or elimination of lead from electronic
circuitry, which is traditionally disposed of in landfills, where the lead
content has been proven to leach out into the nearby aquifer?

Socialist Solder, I like it.

Tim
 
T

Tim Williams

Jan 1, 1970
0
JW said:
Not always, I've seen device leads that have gotten so hot that they
seem
to weld to the pad or via. A soldering iron won't melt the lead (or
whatever it's become).

Intermetallics. Copper and tin form hard, higher-melting-point compounds
when heated sufficiently for diffusion to take over.

It's more noticable when attempting to dismantle copper plumbing*, where
tin solder is used (usually with antimony, rather than copper or silver,
to harden it), the joints are tight, and the temperature control is poor.
If the age alone doesn't do it, overheating a sticky joint can make it set
up that much harder!

*Gee, we [humanity] haven't used plumbum (lead) for "plumbing" in over a
century. Maybe we should call it cupring (cuprum = copper) or something?
Heck, even copper is falling out of favor these days for chincy plastic
snap-together things. Go figure...

Tim
 
Mike said:
On 06/11/2012 17:47, [email protected] wrote:
On Tue, 06 Nov 2012 11:22:36 -0500, "Michael A. Terrell"


JW wrote:

Joerg wrote:

[...]

If it was saturation the current intake would be high and the 2N7002
FETs would attempt to unsolder themselves.

yeah well, that would help out in the replacement process :)

Not always, I've seen device leads that have gotten so hot that they seem
to weld to the pad or via. A soldering iron won't melt the lead (or
whatever it's become).


The old solder is oxidized. I've found that adding some fresh solder
(With a mild RMA flux) helps disolve the old crap.

Our processes are all RoHS, so I do the same. The first thing I do
when soldering a part is add leaded solder to the socialist solder. I
usually take the excess off with some solderwick.


I would hardly call adding leaded solder to a joint a RoHS process. How
certain are you of only leaving the allowable 0.1% lead in a repaired
solder joint?


He's not in Europe.

I was aware of that, but its not unknown for US and other countries to
adopt RoHS in order to export to Europe.

It's also done because some non-RoHS parts are impossible to come by.
Its a smart move to broaden your market.

Maybe. Some have *no* interest in the europeon market. We do, but
that's a different discussion.
Therefore any work ought to use lead free solder except in
specialist areas.

Complete bullshit. We don't even *have* RoHS solder in our lab.
Engineering said:
It's also a good idea to prototype using RoHS
materials just in case there are any potential manufacturing issues and
reliability.

Utter nonsense.
There is the contradiction that on the one hand where he says "Our
processes are all RoHS, so I do the same" and then says "The first thing
I do when soldering a part is add leaded solder to the socialist solder".

I do the same, meaning pollute the connections, first, with leaded
solder. Are you illiterate? I suppose so...
It hardly sounds the "same" and it's a shame that anyone who might point
out this contradiction is apparently an idiot engineer.

It really isn't my problem if you can't read. What a moron!
 
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