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  3. We know the output voltage is related to the ratio of the number of turns. And the output current related to the inverse of the number of turns. The more voltage gives one more current at the primary so in an ideal transformer the power is the same on both sides of the transformer. But in a pulse transformer is it not the rate of change of current in the primary that gives rise to the change in magnetic flux that produces the voltage? I need to learn more about pulse transformers. I had in mind trying out different frequencies on the automobile coil/transformer but had a problem when
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  6. Harry , I need to unpick what you have found. The highest output across the secondary of your coil was 159.8V, when I presume one would expect several kV? (I don't know what the turns ratio is in such a coil but ignition systems are designed to produce 5-15kV) Am I correct in saying that to get a higher secondary output one will need a bigger primary current - to store more energy in the primary coil? Also are you saying that the current in the primary is a function of the Gate frequency - perhaps a lower frequency will allow more current into the primary? What's DUT? Regards,
  7. Using the 100 watt bulb and two different ignition coils this is what I got: AC Output(VOM) 1 meg load Input sig. 204 Hz Peak voltage across 7ohms coil type 67.4 v 6.0v 1.84 v std auto ignt. coil 159.8 v 9.68 v 5.44 v 17.4 v 8.0 4.0 - 4.2 v Capacitor discharge coil The std auto coil is a 12 volt
  8. PRESENTO EL MISMO PROBLEMA, SI NOS PUDIERAN AYUDAR PORFAVOR GRACIAS SALUDOSSS
  9. RTC helps to reveal the real time to the machine thus it's critical to some applications. Using MicroPython on Ameba RTL8722, we can get RTC time in just 2 lines, here is how, Materials Ameba x 1 Steps RTC module help microcontroller to keep track of time and is essential to our time module. Here we an example to demonstrate how to get local time and update the time. Copy and paste the following code line by line into REPL to see its effect. rtc = RTC() rtc.datetime() # get date and time
  10. The student should always respect everyone and be respectful to everyone. A famous institute is working to provide accounting assignment help australia to each and every student for their help. Appreciation to our CBD team he reached on time. The man always is within his limits. anyway, thanks for this blog.
  11. Harry, that's very industrious of you to build a physical version of the circuit. Have you have substituted my upper 530 for a light bulb because it is just acting as a variable resistor? Also could you clarify if you got a high voltage on the secondary of the ignition coil? If you did, even with the small currents you measured in the primary, what does this all mean for my current design 'HV Supply 1B' shown a few posts back? Julian
  12. Don't laugh that is an emulation of your circuit. I was thinking that there is not enough load in the circuit with only the coil (the mosfets are acting like switches only off or on) so I started out with 7 watt incandescent lamps but could not get two mosfet to work so I switched to only one; that works well. In the photo at the lower right is an automobile ignition coil and the big black thing is an 80 volt battery (borrowed from my chainsaw). Working up to 13 bulbs (normally 7 watt bulbs) I got 0.571 ma through a 7 ohm resistor. Switching to a 100 watt bulb got 0.594 ma, the IR
  13. Hi Harry, I'm not clear where your two variable resistors are and in the attached Pic 1 I have put where I think one is. My supply is set at 50V and the gate pulses at 200Hz, 50% so 'hoping' to be able to get 750-1kV out at the secondary when connected to the electrolysis cell. As I come from a Physics background and not electronics I'm trying to understand why increasing the current in the primary will increase the voltage across the cell. Is the problem that the 530 is throttling both the voltage and the current and hence the multiplication by the secondary? From my limited ex
  14. Hi Marta, Thanks for the pictures. Looks good. I already received some more details from our Product Manager concerning updating the embedded FW of the devices. I will discuss this later today with him. Is it oke for you to further talk directly via e-mail? This is more convenient and I can also cc some of my colleques in the discussion? If oke, please contact me on [email protected] Regards, Peter
  15. I was using the fixed resistors as the simulator dos not have a concept of a potentiometer. Putting the 530 back in an using a combination of two resistors between 30k - 17k to 46k - 1k (where the lower value is to ground) I get good results with 40k - 7k and 30k - 17k but the peak current is 2 amperes for some reason. At 44k - 3k the current is 900ma but the output is down to 50v. I will continue looking at it. If you have a symmetrical pulse waveform say of 1 ampere I gather the rms value of the current is 1 * 0.707 and the rms voltage would be 100 * 0.707 so the equivalent pow
  16. Yes, the modules are in their original packaging, tightly packed. I am sending you pictures of the packaging. I will wait for a message from you
  17. Hi Harry, In your first reply at the start of this thread you suggested using a second MOSFET to control the current instead of a variable resistor, which is why I modified the circuit to include the IRF530. While it works to allow fine control of the current are you now suggesting I put a 50 Ohm resistor instead, based on your simulation? If so then this would surely need to be a 200W resistor (massive?) to be able to cope with 2A through it and I certainly don't have one of those - which is why I presume you suggested the FET in the first place. J
  18. Hi Marta, Still have not heard anything from our product manager concerning the update. I will be out of the office rest of the day, but will keep you posted. Another question: are all modules still in the original package container as can be seen on the picture? (question from our electronics supplier) 🙂 Regards, Peter
  19. Hi Marta, Thanks for your fast response. We need to upgrade the FW version. We are checking the possibilities. I will let you know asap. Best regards, Peter
  20. Hi Peter, Modules are still available. By answering your question: BLUEMOD+S42/AI FW 3.006 Manufacturers Part No 53346-04. Let me know if you have any questions. Best regards, Marta
  21. PWM is one of the crucial peripheral of an MCU, with MicroPython, you can control RTL8722DM to output PWM signal to fade and brighten an LED with ease, here is how, Materials Ameba x 1, LED x 1, Resistor(220ohm) x 1 Steps PWM use pulse width modulation to control output duty cycle and is widely used to control LED brightness and motor. Here we are using an LED to demonstrate how PWM works. Let us connect pin PA_26 to the anode leg of an LED which in series with a current limiting resistor and GND to cathode of the LED as shown below, Then,
  22. Dear Marta, Are these modules still available? What FW version do these modules have? Hope to hear from you soon. Best regards, Peter
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  25. I would think you are not hitting it with enough current. Using a resistor instead of the IRF530 as it is faster to use rather then dorking around with two resistors to simulate a pot. I am using a input pulses that 2.5 ms on and 2.5 ms off at 6.8 volts peak No bifilar coil nor diode in the output: resistor current output capacitor output resistor output voltage 100 1A 1.058 nf 1.16 meg 720 volts 100 1A 0 1.16 1.05 kv 50 2A
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  28. New section: For R1(max). Assuming a 8 volt transformer. Vin = 8.0 * 1.4 or 11.2 Taking the min and max to be plus or minus 10%. One would not need a voltage regulator if there where nothing to regulate. Vmin = (Vin - 10%*Vin) - (2 * voltage drop across the two diodes). Typically 0.7v each. And Vmax = Vin + 10%Vin - (2 * voltage drop). R1max = (Vmin - Vz)/(Izmin + Ib) Izmin is found from the zener data sheet; typically 10 ma. What happens at Vmax? R1max = (Vmax - Vz)/(Izmax + Ib) Where Izmax is the only unknown here. Izmax must be within the maximum curr
  29. Have an awesome project in mind using some LEDs. In that project I will be using some LED Fading Effect and few LED Chaser Circuits. But before jumping onto that, I thought I should create a short tutorial and show you guys how to fade a LED with or without an Arduino automatically or manually using a potentiometer. Video: https://youtu.be/IIUsdICycOw Sponsors This video is sponsored by PCBWay. PCBway: only $5 for 10 pcbs from https://www.pcbway.com/?from=CZcouple PCBWay specialize in manu
  30. Hi Harry, I inserted a 1.5 Ohm resistor into the Source line of the IRF840 as suggested (Pic 1) and calculated typical readings of the primary current of 30mA when the secondary was unloaded and about 160mA when the cell was connected (Pic 2), The value under load could have gone higher but not as high as 1A. This was with a square wave input of 200Hz from the pulse circuit and 12V from the PSU as before but with my bifilar coils in place (Pic 3). Once again the voltage across the cell drops to 3-4V under load and reads about 220V when the cell is not connected. I know to expect a vo
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