3
\$\begingroup\$

How can we measure the voltage at the coil of a Slayer Exciter?

We have built a simple 9V Slayer Exciter for a school project. We are examining the conservation of energy at the transformer. The Slayer Exciter works fine: when held in front of it, a TL-light gives light. We use a BD-139 transistor. But we haven’t succeeded at measuring the voltage of the coils.

Measuring with a multimeter give incorrect voltages, like 0V or less than 1V. So, we thought that maybe a voltage divider would work. We connected it to the top of the larger coil and to the ground of the battery. We connect an oscilloscope to the smallest resistor, at the start and at the end of a resistor. It shows a sinus. However, If we put the resistor out of the voltage divider, the oscilloscope shows the same sinus. Without the resistor! If we put one cable from the oscilloscope to the voltage divider out, it also gives the same result. If we measure with two unconnected cables at the oscilloscope it gives a weaker signal. We believe that the electromagnetic field is causing a problem to the oscilloscope and the multimeter.

Our questions are:

  1. How can we measure the voltage at the coil of a Slayer Exciter, the primary and the secondary?
  2. Why is there no current going through the top of the bigger coil to the voltage divider?
  3. How can we stop the Slayer Exciter from creating an Electromagnetic field? The transistor creates an AC current, but how can we know what the frequency is of that current? Is it a specific property of the transistor?
  4. We have tested different transistors, all of them NPN. But the BD-139 was the only transistor working. The others got too hot. What are the reasons why they don’t work?

We have built the Slayer Exciter through the following scheme:

The Scheme of the Slayer Exciter

\$\endgroup\$
  • \$\begingroup\$ See electronics.stackexchange.com/questions/214954/… , although that doesn't really explain it very much either; a lot of power is unavoidably dissipated in the transistor. \$\endgroup\$ – pjc50 Nov 25 '17 at 19:08
  • \$\begingroup\$ A series chain of 1N4007 diodes, plus a pith ball electroscope comes to mind. You might try that anyway. Calibrating it would be the remaining problem. \$\endgroup\$ – jonk Nov 25 '17 at 19:45
2
\$\begingroup\$
  1. To measure voltage at the secondary coil correctly you need a device with very high resistance, about 100 MOhm or even 1 GOhm. So, when you create a divider use 100 MOhm resistor for oscilloscope input withput divider (it will probably have 1 MOhm input resistance), or 1 GOhm resistor for oscilloscope with 1:10 probe (it will have 10 MOhm input resistance). In both cases you'll get 1:100 divider with proper input resistance. Using AC voltmeter is not a good idea because most cheap AC voltmeters work correctly only with low frequency signals (less than 1 kHz) and frequency of your generator is much higher, I guess. Voltage at primay coil may be measured by direct connection of oscilloscope to it's ends.
  2. I don't think that there is no current flow to the divider. You can calculate it according the Ohm's law, using R1 value of divider.
  3. You can measure the frequency using oscilloscope. Measure the period T and calculate frequency as F=1/T. This freq is a property of transformer - resonance frequency of tank, consisting of secondary coil and it's stary capacitance. You can't stop this circuit to generate electro-magnetic field, because it is it's primary function. Take into acount, that any alternating current produces alternating megnetic field, and alternating magnetic field produces alternating electric field.
  4. In this circuit transistor works in active mode and have to dissipate considerable power, so 2222A is bad choice and BD139 is much better. You can try to tune base resistor, trading output power versus transistor heating.

schematic

simulate this circuit – Schematic created using CircuitLab

EDIT: The divider circuit should meet at least two criteria: 1.It should attenuate input signal by factor of about 100 (because the transformer have about 1:100 turn ratio). 2. It should have at least 100 MOhm input resistance (to minmize it's influence on secondary voltage level). Divider, consisting of 100 MOhm and 1 MOhm resistors meets both criteria. Ideal oscilloscope has input with infinite impedance, so when you connect it to the divider where will be no influence. But typical real oscilloscope has input, which equivalent circuit presented on the drawing (parallel connection of 1 MOhm and 10 pF, this values probably printed near scope input connector). The input Rosc plays role of second divider resistor and you needn't add another one. Oscilloscope input IS the second resistor of divider. If you want to get 1:1000 attenuation you may either use 1 GOhm at divider 'high side' and continue to use scope input as 1 MOhm at 'low side', or continue to use 100 MOhm at 'high side' with 100 kOhm at 'low side'. In the last case you neet to place 100 kOhm 'low' resistor in parallel to oscilloscope input.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Thank you for your answer, we really appreciate your help. Do I have to place the oscilloscope in series? We made our voltage divider with 2 resistors. One has 1KOhm and the other one has 10Gohm. We connected the oscilloscope to the 1KOhm resistor in parallel. Why do we need to place the oscilloscope in series and not parallel? \$\endgroup\$ – user3505506 Nov 26 '17 at 18:24
  • \$\begingroup\$ You are right connecting oscilloscope in parallel to 1 KOhm resistor. in my sheme there is no 1 kOhm resistor since it's role plays input resistance of the oscilloscope (1 MOhm or 10 MOhm depending on 1х or 10x probe setting). Also a think 10 GOhm/1 kOhm ratio is too much (it reduces voltage 10 millon times). I guess 10 GOhm/(1..10 MOhm) ratio will be sufficient. \$\endgroup\$ – Eugene K Nov 27 '17 at 0:25
  • \$\begingroup\$ Thank you for your answer. I made a typo in my first comment, we had a voltage divider of 1KOhm and 10MOhm. The GOhm has to be MOhm. I don't understand your scheme fully. Our oscilloscope has a 1:1 probe. So we need the resisors 10GOhm and 10 MOhm and place the oscilloscope parallel on the smallest resistor. Or according to your drawing, do we need to place the oscilloscope in series after the 100MOhm resistor? \$\endgroup\$ – user3505506 Nov 27 '17 at 16:22
  • \$\begingroup\$ I edited the answer to clarify the role of oscilloscope input resistance in the divider circuit. \$\endgroup\$ – Eugene K Nov 27 '17 at 18:05
  • \$\begingroup\$ Thank you very much for your help. Your Edit makes it very clear. We will test it tomorrow. I will let you know if we succeeded. I showed your answer to my teacher, he was very excited. Our oscilloscope has a 25pF. He had a question: Can we add another capacitor of 2500pF parallel to the 100MOhm to eliminate the influence of the capacitor in the oscilloscope (Because the ratio 1:100 between the resistors)? \$\endgroup\$ – user3505506 Nov 28 '17 at 17:54
0
\$\begingroup\$

These two photos in comment are very informative!

The circuit generates at about 2 MHz frequency. It's quite high, so there is the need to take into account electro-magnetic radiation and reception. Main rule: keep area of any loop with high frequency current as small as possible. Magnetic field radiation is proportional to that area. The same is true for oscilloscope connection circuit. It have to be shielded (to be protected from electric-field) and must have as low loop area as possible. For your setup at least two points must be fixed: 1. Use coaxial cable to connect oscilloscope to the circuit. Two separated wires (black and red, with that huge distance between them) – is inappropriate oscilloscope connection method. To test amplitude of signal induced directly on oscilloscope/cable, measure signal at both disconnected cable input (as you do on second picture) and short-connected cable input. This amplitude must be as small as possible. 2. Reduce area of primary coil loop. Twist first winding wires on all their path from coil to breadboard. Use decoupling capacitor (parallel ceramic C2+ electrolytic C3) to bypass HF current near T1. Try to keep wire lengths from these caps as small as possible.

Some other ideas: 1. Try to insert ferrite rod into the coil. This will increase coupling between windings and reduce both working frequency and magnetic field radiation. Try to increase first winding turn amount (for example to 10...30). 2. Try to measure voltage across first winding, connecting OSC ground to '+power' wire.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Thank you very much for your answer. We decided to solder our circuit. This is how it looks like now: 1drv.ms/f/s!AtjkJBr7In_ogrlBSgPYZW5pxNYMSA, do you have tips? We now also use a coax cable. As a result the voltage is less. The voltage is less at the voltage divider, but also less at an unconnected cable by the coil. Why is the voltage at the divider less? If we connect the red cable to the voltage divider and the black unconnected. We see a similiair pictures when the black cable connected to the ground. In the next days we will solder the capacitors. (1/2) \$\endgroup\$ – user3505506 Dec 5 '17 at 20:33
  • \$\begingroup\$ But what is the function of the capacitor in this circuit? What do you mean to with: to bypass the HF current near T1. Is there a cominbation of AC and DC current on the same cable? And how did you chose the values of the capacitor? If we put a ferrite rod in the coil. The period is twice as big as original. The voltage is a little bit less. If we put Iron in the core the period is the same, but the voltage halves. Is this because the formula of the inducter of the coil? Do you have other tips or ideas to measure the voltage of the coil? The coax cable has still influence of the coil. (2/2) \$\endgroup\$ – user3505506 Dec 5 '17 at 20:40
  • \$\begingroup\$ Can I ask you another question? Do we need to ground the oscilloscope itself to the ground of the battery? Could that be the problem? And we dont have shorter coax cable, is that maybe the problem. Thanks in advance. \$\endgroup\$ – user3505506 Dec 8 '17 at 20:35
0
\$\begingroup\$

It's right decision to solder the circuit! It should be much reliable now.

According to you experiments, I think that electro-magnetic coupling between circuit and probe still exists. To estimate it's value, try two experiments.

  1. Leave both ends of oscilloscope cable unconnected and observe amplitude. Move the cable around circuit and watch amplitude change, try different geometric configurations.
  2. Connect both ends of cable to each other (make short circuit). Again, move the cable and observe amplitude. Try to make loop of maximum area. At the opposite, try to twist both ends before connecting them.

Consider each wire in the probe and in the circuit as receiving and transmitting antennas. After these experiments you can get estimation of non-conductive coupling strength, which interfere with conductive coupling while you are trying to measure secondary coil voltage.

To minimize magnetic field radiation, all loops of significant current must have as small area as possible. Te only such loop in the circuit is primary coil loop, which consists of collector-emitter path of Q1 and of battery. The function of capacitors is to exclude battery from this loop. High frequency (HF) current will flow through the capacitors, not through battery. To further reduce area of HF loop, twist the ends of primary coil (red wires) on their path to Q1. Place them and C2, C3, Q1 as close to each other, as possible. You can measure radiation level using closed loop of oscilloscope cable, as mentioned above.

When you measure secondary voltage, connect GND clip to 'minus' pole of battery to make conductive path of scope's input current. The length of coax cable is not so important (coax configuration is intrinsically immune to radiation). But the length and configuration of non-coax part of cable (red and black wires) is important. So, the best way is to make your own specific coax cable with very short non-coax part (about 1 cm). At least, twist non-coax part of existing cable to maximize it's HF radiation immunity.

The goal is to significantly reduce 'radiation' path of signal, to make possible accurate measurement of signal via conductive path.

|improve this answer|||||
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.