Good point Rick,
I should have explained that even though the larger inductance reduces
the rms current in the primary significantly, the positive and
negative peak currents are highly asymmetric. Simulating with a
sinewave input, the positive peak current is about 110mA whilst the
negative is
It's also hard to see how the circuit could work with C1 in series with
the transformer current. Why is a capacitor needed if you use the
transformer?
Maybe there is some effect that will balance things out, but if the two
currents are unequal, actually it would be the integral of the two
The transformer allows a DC path to exist on the secondary side, but
you still have the capacitor on the primary side of the circuit. If
the positive and negative pulse currents are not equal, you will still
have a problem on the primary side. You need to remove the cap C1.
I
The most recent circuit you posted is not the same as your original
and as Gene pointed out, you have now made a series resonant circuit
between the 220nF cap and the 200uH primary inductance.
In the simulation, the source resistance is zero, the ESR of the cap
is zero and there is only 0.25R
Dnia piątek 24 czerwca 2011 o 13:10:35 myken napisał(a):
This is strange in my simulation the attached circuit works fine. In
real life it kinda works but the signals are distorted like you can see.
I think that has something to do with the fact we used a pulse
transformer to try the circuit.
On 06/24/2011 07:10 AM, myken wrote:
This is strange in my simulation the attached circuit works fine. In
real life it kinda works but the signals are distorted like you can
see. I think that has something to do with the fact we used a pulse
transformer to try the circuit. If we
Vcc and Vss are still sensitive to load. So if the design requires
both Vss and Vee be equal and opposite, then it needs regulation -
zener, for example.
Your absolutely right, but that's what comes next. It is the regulators
how create the difference in load balance.
On 06/24/2011 06:10 AM, myken wrote:
I think that has something to do with the fact we used a pulse
transformer to try the circuit.
Looks like a diode drop to me, not transformer problems.
You've got a series chain of reactances. You're gonna get oscillations
and lagging, leading voltage
On 06/22/2011 04:39 PM, Andy Fierman wrote:
Vcc and Vss are still sensitive to load. So if the design requires both
Vss and Vee be equal and opposite, then it needs regulation - zener, for
example.
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Hello all,
I would appreciate some expert advice.
I have a system which rectifies a sine wave input signal of 20Khz after
a LC filter (see Rectifier_sim.jpeg)
Everything works fine if LOAD_1 and LOAD_2 are equal. Vx is then
(almost) the same as Vin. And Vcc and Vss are equal to the
Ooops,
Just missed the Undo Send window ...
Typo in (i):
if the source has a peak to peak swing of x volts but a dc offset of y
then (neglecting the diode drops) vcc = x/2+y and vss = x/2-y.
:)
Andy
On 19 June 2011 11:01, Andy Fierman andyfier...@signality.co.uk wrote:
Rick is
Hello all,
As a remark on your observation Andy I would like to say in my defence
that I simplified and reduced the problem/information simply to avoid
wasting anyones time with details and (in my opinion) irrelevant side
effects. That the two load resistors are in fact a representation of two
What is the purpose of C1 and L1? If you want to filter anything, it
should be AFTER you rectify the signal to DC. A series cap is going to
remove low frequencies... like DC which is attenuated very highly. So
much in fact that you can't draw a DC signal through a capacitor. That
is why
Yeap, it should be a very low power power supply. Vx is not important
Vcc and Vss are.
Vin can be anything from 15Khz to 28Khz so a transformer is not the most
desired option.
I have designed two SMPS for Vcc and Vss but there load to the rectifier
are not the same, with the described result.
Simply reproducing the filter twice, one for each polarity of
rectifier will not work.
If you can float the load or the source then splitting the circuit
into two and using a bridge rectifier in each will work OK.
The attached shows what I mean.
Cheers,
Andy.
www.signality.co.uk
On 06/16/2011 02:30 PM, myken wrote:
Hello all,
I would appreciate some expert advice.
Are you trying to make a low current power supply?
I agree with DJ - the unequal loading on + and - cycle will average to
something other than zero (unequal capacitors, unequal diodes, etc) If
Vx must
When you put two capacitors in series, there's no way to know what the
voltage between them will be. You have three with a common central
connection Vx. V1 acts to charge the node, the loads act to discharge
it, so an unequal load means unequal discharging and thus nonzero
average node voltage.
On 06/16/2011 01:30 PM, myken wrote:
see Rectifier_sim.jpeg)
Everything works fine if LOAD_1 and LOAD_2 are equal.
Lose C1. Lose R1. Add L2 feeding D1.
Separate D1 D2.
Add some rc filter R between
D2, C2. Or try moving L1 between D2, C2
John
___
Thanks DJ,
I had the same thought that Vx was floating somewhere unwanted, that's
why I added the resistor (which didn't work).
Gazing at this problem for a couple of days make me miss the obvious,
just split the filter. Brilliant.
I'll give it a try.
Robert.
On 16/06/11 20:48, DJ Delorie
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