PopularFX
Home Help Search Login Register
Welcome,Guest. Please login or register.
2024-06-20, 01:21:06
News: If you have a suggestion or need for a new board title, please PM the Admins.
Please remember to keep topics and posts of the FE or casual nature. :)

Pages: 1 2 [3] 4 5 6 7 8 9
Author Topic: DSRD pulse generator  (Read 196445 times)
Group: Guest
@Peterae

It is important to switch off as quick as possible the last stage. The switch on is not critical.
I have also tested many NFET. The NFETs in the range 200-600v 10A give the best results. The more powerful FET able to handle bigger currents and lower voltages give poorer results. I suppose it's due to the higher output capacitance and predominatingly to the inside protecting diode which cancels the back emf while the interesting voltage for blocking the DSR diode is not yet reached.

   

Group: Administrator
Hero Member
*****

Posts: 3894


Buy me some coffee
Quote
It is important to switch off as quick as possible the last stage.
OK Thanks, i see.

I think that lower amp devices have faster on and off times generally as well for the reasons you explained.
   

Group: Professor
Hero Member
*****

Posts: 3400
It is important to switch off as quick as possible the last stage. The switch on is not critical.
I think you are confused.
The DSR Diode interrupts the current in the secondary winding - not the power switch (e.g. MOSFET). This current interruption is responsible for the generation of the short pulse via the inductive effect of the secondary circuit.

The power switch (e.g. MOSFET) on the primary side should switch OFF when the voltage in the capacitor on the secondary side is at its maximum and the current in the secondary winding is zero (or just crossing zero).  This happens after the time that is equal to half of the self-oscillation period of the RLC circuit formed on the secondary side.
« Last Edit: 2012-11-10, 13:35:28 by verpies »
   
Group: Guest
I think you are confused.
The DSR Diode interrupts the secondary current - not the power switch (e.g. MOSFET) on the primary side.

The power switch (e.g. MOSFET) on the primary side should switch OFF when the voltage in the capacitor on the secondary side is at its maximum and the current in the secondary winding is zero (or just crossing zero).  This happens afer the time that is equal to half of the self-oscillation period of the RLC circuit formed on the secondary side.


You missed the point of the back emf which motivated my previous reply. From what I observed, the DSR pulse is a lot increased by the back emf pulse when they coincide. The back emf pulse is rather wide in comparison with the DSR pulse. A quick switch off narrows the back emf pulse and so improves the effect. I adjust the synchronization thanks to the "pumping" pulse width.
I wonder why most of the DSR circuits that I see here and there, possess a diode that prevents the back emf to occur.

It is also to be emphasized that L in the RLC circuit formed on the secondary is (theoretically) 100% coupled to the primary coil. If the primary is not switched off quickly enough, the NFET which is not yet fully open presents a not negligible load for the DSR pulse induced in the primary coil and so reduces the efficiency.

   

Group: Professor
Hero Member
*****

Posts: 3400
I wonder why most of the DSR circuits that I see here and there, possess a diode that prevents the back emf to occur.
The flyback diode is there to ensure that the currents keeps flowing through the primary winding after the power switch (e.g. MOSFET) stops conducting.
The current flowing in the flyback diode keeps the core of this transformer saturated, which has the effect of decreasing the inductance of the secondary winding. The latter increases the reverse current through the DSR diode which results in a larger nanosecond inductive spike when that diode abruptly stops conductiong that high current.

Remember, that the current in the primary winding flows in the same direction after the power switch turns off.
This current does not reverse direction when the power switch transitions from the ON state to the OFF state.  Many people think that the current reverses because of the word "back" in the phrase "back EMF" - wrong!

The core must remain saturated between t1 and t2. See the Belkin paper and my waveform diagram.
   
Group: Professor
Hero Member
*****

Posts: 2996
Welcome, verpies!

Would you please comment on the MrClean device which you have diagrammed at EF? -- here's the thread

http://www.overunityresearch.com/index.php?topic=1566.msg26388#msg26388

   
Group: Guest
The flyback diode is there to ensure that the currents keeps flowing through the primary winding after the power switch (e.g. MOSFET) stops conducting.
The current flowing in the flyback diode keeps the core of this transformer saturated, which has the effect of decreasing the inductance of the secondary winding. The latter increases the reverse current through the DSR diode which results in a larger nanosecond inductive spike when that diode abruptly stops conductiong that high current.

Remember, that the current in the primary winding flows in the same direction after the power switch turns off.
This current does not reverse direction when the power switch transitions from the ON state to the OFF state.  Many people think that the current reverses because of the word "back" in the phrase "back EMF" - wrong!

I know your last point and I agree. It is the ordinary functionning of the back emf, following the simple rule of the direction of the current passing either from + to -  through a load , or from - to + through a generator.
  • During the back emf, the voltage reverses direction but not the current because the coil is now a current generator releasing its current which continues its way as before, while before, it was the power supply that provided the current to the coil acting as a load.
  • The equivalent effect for a capacitor is to charge it and then to shortcut it. Thus the capacitor status changes from load to generator but now, as we have a voltage generator, it's the voltage that remains in the same direction while the current reverses its direction. The current peak of the capacitor discharging in a shortcut is the equivalent of the voltage pulse of a coil whose we open the circuit.
With our DSR diode, we have a capacitor, and so, as you say, the current is reversed. Nevertheless the Belkin's paper is not the best explanation of the pure DSR effect because he mixes the problem with the question of the core saturation. The core saturation is not an obligatory step of the process but just a possible improvement. The pulse can occur without core saturation.

It's clearly explained in wikipedia:
"The principle of DSRD operation can be explained as follows: Short pulse of current is applied in the forward direction of the DSRD effectively "pumping" the P-N junction, or in other words, “charging” the P-N junction capacitively. When the current direction reverses, the accumulated charges are removed from the base region. As soon as the accumulated charge decreases to zero, the diode opens rapidly. A high voltage spike can appear due to the self-induction of the diode circuit. The larger the commutation current and the shorter the transition from forward to reverse conduction, the higher the pulse amplitude and efficiency of the pulse generator (Kardo-Sysoev et al., 1997)."

So we understand that the DSR pulse is not mainly due to the release of the diode charge but to the opening of the inductive circuit by the diode. As the inductance of the secundary circuit is coupled to the primary, both have to be taken into account for the final result, using the coefficient of mutual induction. If the primary is looped by a flyback diode, then the inductance viewed from the secundary is reduced. I agree that a current can be increased by reducing an inductance but if the inductance is already reduced at the moment of the DSR pulse, the  current stored inside the inductance and available for the impulse, is less because it is proportionnal to the inductance. Without flyback diode, the full current stored in the inductance is available for the pulse, including that one of the primary: the DSR diode is using the current that the flyback diode could waste. It's my option.


   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4103

It's clearly explained in wikipedia:


Funny,  please take a look at the (co)author of that wikipedia article.

Regards Itsu
   

Group: Professor
Hero Member
*****

Posts: 3400
Would you please comment on the MrClean device which you have diagrammed at EF?
Please take a look at my comments about this schematic diagram here.
   

Group: Professor
Hero Member
*****

Posts: 3400
I know your last point and I agree.
I am glad, but I was not writing about the current direction for your benefit.  I always think of beginners and lurkers that also read my replies.

Nevertheless the Belkin's paper is not the best explanation of the pure DSR effect because he mixes the problem with the question of the core saturation. The core saturation is not an obligatory step of the process but just a possible improvement. The pulse can occur without core saturation.
That is true - the saturation effect is not mandatory.  However, if the saturable transformer is used, then the flyback diode prolongs the time when the core is saturated. 
Actually, I am a proponent of the transformerless method because there are less variables to coordinate in a circuit without a saturable transformer. That's why I described such circuit here

So we understand that the DSR pulse is not mainly due to the release of the diode charge but to the opening of the inductive circuit by the diode.
Correct. That's why the reverse current must be maximized and di/dt @ t2 must be the greatest when the DSR diode interrupts the current abruptly.

As the inductance of the secondary circuit is coupled to the primary, both have to be taken into account for the final result, using the coefficient of mutual induction.
That's where you go wrong, because as soon as the transformer's core becomes saturated, the primary and secondary windings become uncoupled.  In other words, their coefficient of mutual inductance decreases drastically.  That's why Mr. Belkin states, that after t1, the primary charging circuit can be neglected.
« Last Edit: 2012-11-11, 04:05:36 by verpies »
   
Group: Guest

Funny,  please take a look at the (co)author of that wikipedia article.

Regards Itsu

I don't care who writes but what is written.

   
Group: Guest
Quote
As the inductance of the secondary circuit is coupled to the primary, both have to be taken into account for the final result, using the coefficient of mutual induction.

That's where you go wrong, because as soon as the transformer's core becomes saturated, the primary and secondary windings become uncoupled.  In other words, their coefficient of mutual inductance decreases drastically.  That's why Mr. Belkin states, that after t1, the primary charging circuit can be neglected.

Replace "uncoupled" by "much less coupled" and I agree.
Now you speak of a setup which is not mine. In mine, the coupling is maintained, this can be observed by the correlation of the signal of each coil, and so the back emf is available at the secondary winding and used differently than in Belkin's setup. So I'm perplex because my previous explanations concern essentially my setup but your objections consist in considerations about the functioning of another circuit. They don't apply to mine.

   

Group: Professor
Hero Member
*****

Posts: 3400
Yes, less coupled is more accurate because mutual inductance does not fall down all the way to zero, even if the core saturates and behaves like air.

If your device does not take the core of the transformer into saturation, then this circuit might as well be built without a transformer, e.g. like this one, with inductors only.
   
Group: Guest
...
If your device does not take the core of the transformer into saturation, then this circuit might as well be built without a transformer, e.g. like this one, with inductors only.

Agreed. I have tried with a single coil, a sliding core and a connection point to the capacitor that I could move along the coil, but I got a bit inferior result whatever the position of the core, the position of the contact, and the input pulse width. Nevertheless I'm not convinced that it is a question of principle, because the output pulse was disturbing the functioning of the monostable and the quality of the input pulse. With the transformer, the output is more isolated from the electronics, it is floating, and the reaction problem is weaker. More experiments are needed and also to redesign the circuit according to the rules of engineering in matter of HF amplifiers, for example it is needed to shield the stages from each other, to filter the connection to the supply power with choke inductances and so on...

The pulse tests systematically crashed my sat receiver which was powered and in stand-by, not at the same floor. Then it was impossible to start it again from the remote. This is the clear sign that the pulses travel everywhere along the mains lines and that I am far from good experimental conditions  >:(.

   

Group: Administrator
Hero Member
*****

Posts: 3894


Buy me some coffee
Quote
With the transformer, the output is more isolated from the electronics
This is what i really like about this pulsar, the transformer gives electrical isolation, i want to be able to build 2 of these and delay the pulse between each stage firing by a variable nS, i have the electronics already built to do this, i have used fet stages before to do this, but always had lack of isolation and the resulting pulses always managed to feedback to the driving circuit to cause false triggering and eventually self destruction, using the transformer hopefully will give me that isolation  O0
   
Group: Guest

The pulse tests systematically crashed my sat receiver which was powered and in stand-by, not at the same floor. Then it was impossible to start it again from the remote. This is the clear sign that the pulses travel everywhere along the mains lines and that I am far from good experimental conditions  >:(.


Have you considered placing a flyback 'Zener' with series resistor directly across the offending inductance? This will waste less energy and provide some protection for devices susceptible to the flyback spike but not the intended result spike.
   

Group: Professor
Hero Member
*****

Posts: 3400
With the transformer, the output is more isolated from the electronics, it is floating, and the reaction problem is weaker.
The additional isolation is a good point. Extra isolation is more forgiving to non-optimally laid out circuits.


R1 and R2 are usually tailored to the individual characteristics of Q1 (in the neighborhood of 4.7Ω).
« Last Edit: 2012-11-16, 18:20:00 by verpies »
   
Group: Guest
@ww
The perturbation is proportional to the back emf. If I limited it with one way or another, I would also reduce the effect that I want to use to enhance the DSR effect.

@verpies
I will try later. For the moment I defer the experiments until I receive the mosfet drivers that I ordered today (TC4429).

   

Group: Administrator
Hero Member
*****

Posts: 3217
It's not as complicated as it may seem...
verpies,

Won't there be a problem with that scope ground?


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   

Group: Professor
Hero Member
*****

Posts: 3400
Won't there be a problem with that scope ground?
No, the +150V supply rail is galvanicaly isolated from the scope's power supply.

If a rectifier was used to produce +V2 and it was directly connected to the mains line, then it would be a problem if the scope's ground was also connected to the main's neutral - but nobody in their right mind would do such a thing.

P.S.
When using N-Channel MOSFETs it's good to keep current sensing resistors (CSR) on the high side in order to avoid making a Source Follower, inadvertently.
Also, it is convenient to keep the scope probe's ground wire on the +V2 rail when obtaining the trigger signal for the scope. See diagram below:



Before you suggest that it would be better to use a 1:1 isolation transformer to get at the trigger signal at pin 4/8 while scoping the CSR on the high side of the power supply, watch this video and this video.
« Last Edit: 2012-11-16, 14:50:36 by verpies »
   
Group: Guest
First tests with my mosfet driver, received today (TC4429).

I'm not familiar with this kind of component but I'm very impressed. It is given to drive 6A peak onto a 10nF load in about 30ns and seems really do its job, without heating.
I am using a quadruple OR NAND HEF4011 to générate the pulse adjusted around 400ns. The weak output from the HEF4011 is connected through a 1K resistance to the TC4429. Both are powered by the same 8v power supply. The output of the TC4429 is connected to the IRF640 gate. The IRF640 is powered with 30v.

This works perfectly. I easily get 500v pulses on 50Ω

The bad news is that the 1N5408 diodes that I ordered also with the TC4429 (7 for only 1€  :)) don't work at all. They are reputed for their DSR effect but none of mines work. I conclude that the effect depends on the fabrication process which can depend on the maker or the factory. So I advise everybody rather to try diodes from old electronics scraps until finding one.

« Last Edit: 2012-11-14, 19:48:17 by exnihiloest »
   

Group: Professor
Hero Member
*****

Posts: 3400
The weak output from the HEF4011 is connected through a 1K resistance to the TC4429. Both are powered by the same 8v power supply. The output of the TC4429 is connected to the IRF640 gate. The IRF640 is powered with 30v.
I connect the output of my NOR gates directly to the UCC27511 MOSFET driver and I supply my NOR Gates with +5V and my MOSFET Driver with +16V.  Power MOSFETs' gates benefit from higher voltage (as long as it is below VGS Max - apx.20V)
   
Group: Guest
I connect the output of my NOR gates directly to the UCC27511 MOSFET driver and I supply my NOR Gates with +5V and my MOSFET Driver with +16V.  Power MOSFETs' gates benefit from higher voltage (as long as it is below VGS Max - apx.20V)

I can vary the 8v power supply. I didn't note any difference above 8v, up to 15v. Now my IRF640 works under its possibilities. Under 150v, maybe I would need more gate voltage, and it's right now possible. To be continued...

These MOSFET drivers are so easy to use... The electronics industry has never disappointed me since the time of the germanium and the AF116 from Philips.  O0

   
Group: Guest
I have replaced the IRF640 by a 2SK2545.

Here are the main differences:
IRF640 -  200V max - 18A max - 125W - Rds 0.18Ω - Input cap: 1.3nF
2SK2545 - 500V max - 8A max -  30W - Rds 0.8Ω - Input cap: 1.3nF

The pulses have increased from 500V up to 850V / 50Ω (14.4 KW), with only 30v power supply. I think that the difference is due to the stronger back emf, probably due to the higher voltage that can handle the 2SK2545. The DSR pulse is around 6-7 ns. It is maximum when synchronized with the falling edge of the back emf. The radio EMI have increased in intensity and frequency up to 400 Mhz.
So the good choice for the transistor when one wants use the back emf voltage, seems to be a high voltage MOSFET. We don't need a big power, because the peak power of a MOSFET can be near 16 times the mean power (4 times the current), which is enough taking into account that the duty cycle of the pulses is very low.

Verpies was right: to increase the voltage of the driver from 8 to 11v now increases the output pulse.

Here is the shematic, only the 4011 used for generating the input pulse is not represented:


Next step: to increase the power supply voltage to 60V.

   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4103

Hi exnihiloest,

very impressive, one question concerning the 4429.
This is an "inverting" MOSFET driver.
So this means that the output of it (and the gate of the MOSFET) is always positive except for the pulse duration, meaning it keeps the MOSFET conducting all the time except for the pulse time, Right?


Regards Itsu
   
Pages: 1 2 [3] 4 5 6 7 8 9
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2024-06-20, 01:21:06