Connect with us

Project Problems! Need guidance.

Discussion in 'General Electronics Discussion' started by Bharanidharan, Oct 7, 2014.

Scroll to continue with content
  1. Bharanidharan

    Bharanidharan

    25
    1
    Jul 22, 2014
    Hi guys,

    Summary:

    I am trying to make a product that can identify defects in a metal sample using Eddy current.
    For this, i am using inductor coil transducers (8 of them) of value 135uH. 4 of them are transmitters and the remaining four are receivers.

    The transducers are being switched by two multiplexers ADG508. one of them for transmitters and another for receivers. I am using this so that i can increase the number of coils.

    These multiplexers are controlled using Arduino UNO. Inbetween the multiplexers and the transmitters I added a Voltage to current converter circuit to increase the current flow which is upto 1 Amp. This is done using OPA552 in parallel connection.

    The received signal is then precision rectified and filtered and sent to arduino.

    Then the signal is processed using a software.

    Problems:

    1. Since I started using Voltage to current converter (transmitter) the load resistor keeps on burning out. So, i changed it to a 5W resistor of value 22 ohm (previously 22ohm with 0.2W) which now gets incredibly hot. how do i reduce this heat?

    2. I am using a tank circuit with cap 470nF in parallel with the inductor. This capacitor is in series connection with the Resistor of 22 ohms. The voltage capability of this capacitor is 50V. But this also gets burned. I am not sure why?
    My power supply is +/-15V and the output voltage is about 10 V or less. please some one explain why it is burning.

    3. After i changed to 5W resistor, my power supply (-15V is dropping to -13V) and the (+15V remains the same). If i change both of them to 13 V then it is not dropping anymore. Can anyone explain why is it dropping? is it because of the resistor?

    4. The received signals are of different voltages, after rectification and filtering i measured their voltages using the multimeter since the value will be a DC. This was working until I connected the ADC pin to the filter end which makes all the values to be constant.
    The filter is RC low pass, with 470nF cap and 15 ohm resistor. the values i obtain before connecting adc are 600mv, 300mv, 200mV, 500mV and after adc is 300mV constant for all the receivers. I tried to use a 1k resistor in parallel to the filter and then send the signal to the adc, but that failed as well. Please explain why it is acting like this.

    Thank you.
     
  2. Gryd3

    Gryd3

    4,098
    875
    Jun 25, 2014
    You will need to post a circuit diagram before anyone can help diagnose, or all we will be doing is guessing.
     
  3. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,271
    Nov 28, 2011
    Also post some photos of your coil assembly.
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,480
    2,828
    Jan 21, 2010
    I suspect that one of the following is happening:

    1. You have a shorted turn in your inductor(s)
    2. You are using too low a frequency (or DC)
    3. Your inductor is saturating
    4. The inductance is far lower than you believe
     
  5. hevans1944

    hevans1944 Hop - AC8NS

    4,581
    2,142
    Jun 21, 2012
    Yeah, post a schematic with labeled components (so we can make specific comments) and pictures are always appreciated. For specialty parts (the transducers maybe?) identify by manufacturer and model or part number.

    Is your tank circuit 470 nF capacitor resonant with the 135 μH inductor at the driving frequency? If so, there will a LOT of circulating current to heat up the 22 Ω resistor in series with the capacitor. What IS the driving frequency?

    This looks like a really FUN project. I hope you make a lot of money if you can get it working! NDT (non-destructive testing) is a "big thing" in aerospace applications.
     
  6. Bharanidharan

    Bharanidharan

    25
    1
    Jul 22, 2014
    Hi guys,

    uploaded a paper sketch of the block diagram and the circuit diagram and also the coil assembly.

    in the circuit, the inductances are of the same value of 135uH.

    V2I means Voltage to current converter.

    The input frequency is 20 KHz and the input voltage is 2 volts.

    Filter is LPF RC R = 16 ohms and C = 470nF and it is being controlled by arduino uno.

    Thanks.
     

    Attached Files:

  7. Bharanidharan

    Bharanidharan

    25
    1
    Jul 22, 2014
    1. I used it before on a different circuit it worked well and also tested it by directly connecting it to the FG. still had no problems.
    2. Using a frequency of 20 KHz. its not that low.
    3. might be possible when connected to the circuit. How to check and rectify it?
    4. Also may be the case, but this worked with my previous circuit.


    I have attached the coil assembly and the block diagram and the circuit diagram in the thread. Please find it.

    Thanks .
     
  8. Bharanidharan

    Bharanidharan

    25
    1
    Jul 22, 2014
     
  9. Bharanidharan

    Bharanidharan

    25
    1
    Jul 22, 2014
    Posted it. Please see the thread.

    Thanks
     
  10. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,271
    Nov 28, 2011
    Interesting. What op-amp are you using for the output driver in the voltage-to-current converter?

    What happens to the voltage-to-current converter inputs when not driven by the multiplexer? Do they just float?

    Do you switch the multiplexer only on zero-crossings of the input signal?

    Do you fade the input signal to zero before switching the multiplexer, then fade it back up?

    How many cycles of input signal go through on each multiplex selection?
     
  11. Bharanidharan

    Bharanidharan

    25
    1
    Jul 22, 2014
    Mux is always driving the Voltage to current converter until external switch off.

    The opamp i am using is opa552.

    I do not fade my signal to zero, I am using a sinusoidal signal and I don't think i am switching at zero crossing. how do i do that?

    There are four receivers, each one cycle that makes it four cycles. Not sure if i am right. because I am actively sending the transmitter signal and the receiver just picks up what it has for the moment and send it to the measurement side.
     
  12. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,271
    Nov 28, 2011
    According to the schematic, there are four voltage-to-current converters, and the mux only drives one at a time. The other three will be undriven. Their input voltages will be undefined unless you add a resistor to pull them to 0V.
    You need to add a zero crossing detector that monitors the 20 kHz signal, and only change the multiplexer setting at the instant of the zero crossing. This isn't enough to give clean transitions with no disturbances; the best way would be to fade the signal in and out for each magnet over a period of several cycles. The quicker the fade-up or fade-out, the more modulation products are generated. Look up amplitude modulation sidebands.

    You could do this using four VCAs (voltage controlled amplifiers or attenuators) instead of a multiplexer. That would also remove the need to synchronise with zero crossings. You can make VCAs with the LM13700. They require an analogue control signal.
     
    Bharanidharan likes this.
  13. hevans1944

    hevans1944 Hop - AC8NS

    4,581
    2,142
    Jun 21, 2012
    Changing the capacitor value will indeed affect the resonance frequency, f = 1 / (2pi√(LC)). For the values you listed (0.47 μF and 135 μH) resonance occurs at about 19,980 Hz. Close enough to 20 kHz, but small variations in frequency around the resonant tank frequency can have a large effect on the circulating current in the tank, depending of the Q of the tank. The coil resistance broadens the resonance response by lowering the Q.

    Since you are pulse-exciting the coil by switching its excitation through the CMOS analog multiplexer, things get more complicated. It would be interesting to know how much the coil "rings" between excitation pulses. Is there a particular pulse repetition rate that the Arduino UNO produces when stepping through the addresses of the multiplexor? Just curious. I don't think this is related to your resistor and capacitor overheating.

    Capacitors in parallel-resonant tank circuits can have large circulating currents at resonance. With an oscilloscope you should measure the voltage waveform across the 22 Ω resistor, or with a clamp-on oscilloscope current probe measure the current through the resistor in series with the LC parallel-connected tank circuit. Do this just to get some idea of what is going on. The 22 Ω resistor should be temporarily moved from its present location to the ground end of the coil for a voltage measurement. Do this to avoid having to make a differential voltage measurement across the resistor in the presence of a large common-mode AC potential.

    There are specialized non-polarized capacitors, called "snubber" capacitors, that are specifically designed to handle low frequency, high-amperage, currents without overheating. Check an on-line distributor for part numbers and available values. These caps generally have at least 600 VDC rating. A spec sheet for a typical snubber can be found here. I really doubt the capacitor is heating from circulating currents at the power levels you are providing. Any good-quality polyester or polypropylene capacitor should be okay. Make sure its voltage rating is at least twice the peak-to-peak voltage you measure across the tank.

    Your 11X-gain non-inverting op-amp circuit (10 kΩ feedback to 1 kΩ inverting input to ground) driving the paralleled OPA522 voltage-to-current converters is probably going into rail-to-rail saturation with a 2 V RMS sinusoidal input. The peak voltage of 2 V RMS is 2.82 V which becomes 31 V peak (62 volts peak-to-peak) after an 11X gain. The op-amp is powered from +/- 15 VDC supply rails, so assuming it is even capable of linear rail-to-rail output, that still leaves you lacking enough headroom to pass a 2V sinusoidal drive signal. When this op-amp saturates you will be essentially applying a DC pulse instead of a sinusoid to the coil. The current will rise appreciably. I would suggest decreasing the feedback resistor from 10 kΩ to 3.9 kΩ, thereby decreasing the gain of this driver stage to 4.9.

    Why are you not exciting the coil as a series resonant circuit instead of a parallel resonant tank? Parallel resonant tanks are high impedance to external exicitation, and it is difficult to control the circulating current in a parallel resonant tank. Just asking. I know you said the design worked fine before you added the voltage-to-current conversion.

    .
     
    Bharanidharan and KrisBlueNZ like this.
  14. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    Attached pdf analyzes the load current through the 22Ω resistor and parallel LC tank from DC to 100 KHz. Note that the power dissipation through the resistor at 60 KHZ is almost the same as at DC. As hevans1944 observed, due to the amplifier gain there would be a highly clipped waveform driving the load, rich in harmonics. With a 15V signal amplitude the third harmonic could be as high as 5V which equates to approximately 1W power dissipation in the resistor. That is certainly enough to burn a quarter watt resistor, and enough to warm up a 5W resistor. Also, the accuracy of the LC values may cause additional power dissipation in the resistor if resonance is not exactly at the driving frequency. Does the metal sample change the inductance of the coil?

    EP55.png
     

    Attached Files:

    • EP55.pdf
      File size:
      27.1 KB
      Views:
      118
    Last edited: Oct 9, 2014
    Bharanidharan likes this.
  15. Bharanidharan

    Bharanidharan

    25
    1
    Jul 22, 2014
    Thanks for the explanation.

    To be honest, i haven't even started experimenting with the metal samples, trying to get free space measurements is very much troublesome by itself.
     
  16. Bharanidharan

    Bharanidharan

    25
    1
    Jul 22, 2014
    I am providing an sinusoidal signal of 20 KHz to the CMOS analog Multiplexer. I don't understand what do you mean by "coil rings". Please provide an example or guide me to a web learning page? Not sure about the pulse repetition rate as well.

    Just did this, Checked the waveform is a square waveform. Does this mean the signal is clipped due to over amplification? The voltage is 10 Vpp across the resistor.

    I will have a look at the snubber capacitors. I used ceramic capacitors, electrolytic capacitors and polyster capacitors. Ceramic caused a small distortion in waveform, electrolytic was better and polyster causes a bigger distortion, to the point it looks like a combination of triangular signal and sinusoidal signal. I am currently using electrolytic for the voltage to current converters and polyster for the filter.

    if i change the gain won't my current rating be affected? i am trying to send a current of 1 A. I am assuming this value will drop proportionally to the gain reduced.

    i tried the series resonance circuit, it worked for a few seconds and the resistors got hot. The voltage waveform observed across the inductor seemed distorted. The IC's started to heat up ( hot enough for me to warm my cold hands in this cold) and I didn't want my IC to burn so i went back to parallel resonance circuit.


    .
     
  17. Bharanidharan

    Bharanidharan

    25
    1
    Jul 22, 2014
    Sorry, but I am unable to understand the part about undefined input voltages. I thought that until the mux control signal changes, the other switches are defined to be zero / to say no signals being sent through.
    For example, control signal is 00 and then S1 is switched sending the signal and the remaining will be zero/ no signal being switched. About adding a resistor to pull them to 0 V. The mux i am working with has a 300 ohm resistance +/-15V, does that mean it pulls the value to zero if not, how to use a resistor to pull to zero volt?

    So i need to add a zero crossing detector before the mux stage and find the time in microseconds where it actually crosses zero and use that time as a delay for next switching.
    Eg: time period of my signal will be 50 uS. and i need to use a detector to determine this zero crossing happens at 0, 25 and 50uS. While transmitting this first period i need to receive that signal and measure it. and then so on for four periods for four receivers?

    About this, I am also using a Waveform generator AD9830 to produce the other waveforms at different frequency. So only two pins are left for the arduino uno. I also need to find another microcontroller to make the system compact and also need to used a battery (similar to a mobile phone battery or a laptop battery) than a power supply and the power calculations seems to be dodgy as there are more linear components in the board.
     
  18. hevans1944

    hevans1944 Hop - AC8NS

    4,581
    2,142
    Jun 21, 2012
    Wow! What a can of worms we have opened up here!

    The first problem you need to solve is the non-linear performance of the transmitter circuit following the CMOS MUX, caused by the 11X gain stage going into saturation. As I mentioned in post #13 you MUST lower the gain so the amplified pulse output of the MUX channel is a sinusoidal waveform NOT a square pulse. At maximum output current of 200 mA, the OPA551 can only swing to within 3V of the supply rails. With ±15 V supplies, your sinusoid will be limited to ±12 V peak-to-peak amplitude, which means a maximum gain of only about 4 for a 2V RMS input can be accommodated without clipping the waveform.

    Note that your circuit configuration is NOT a voltage-to-current converter. It is a high input impedance voltage buffer with higher current capability to the load than a garden-variety op-amp. There is no negative feedback to make the output current proportional to the input voltage, which is what a voltage-to-current converter circuit does. The load current for any particular output voltage is a function of the load resistance, mostly 22 ohms in your circuit.

    Getting the current you need from the so-called voltage-to-current amplifiers (the paralleled OPA551 op-amps) is determined by their output current capability driving the 22 Ω resistor and the LC tank. The Texas Instruments datasheet says the OPA551 is limited to ±200 mA, so two in parallel will max out at ±400 mA. This is the peak-to-peak current; the RMS current from your sinusoidal pulses will be lower. The datasheet offers a solution capable of providing 1 A load current: a pair of external NPN and PNP power transistors connected in a classic "totem pole" circuit. There is a 100 Ω resistor from the op-amp output to the load to allow the op-amp to source current to the load when the output is less than the forward bias junction voltage (about 0.7 V), which occurs as the sinusoid waveform goes through zero. With this approach you only need one OPA551 per coil because most of the load current is provided by the totem-pole transistors. The load current will be determined by the load impedance and the output voltage per Ohm's law. Note that negative feedback causes the totem-pole output voltage to be proportional to the input voltage.

    Ringing occurs in a tank circuit when the excitation promoting oscillation in the tank is abruptly removed. The oscillations continue but at decreasing amplitude as the tank energy is absorbed by ohmic resistance losses. How long this occurs depends on the losses and is related to the Q of the tank circuit.
     
  19. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    It should be easy to drive a tank circuit with a sinusoid at the resonance frequency since the driving current will be zero. Of course, the current will be non-zero to initially energize the resonant tank, and to overcome any losses especially in a lower-Q tank. So why should a high-current driver circuit be necessary?

    However, if the driving signal is at a fixed frequency it will likely not be exactly the same as the resonance frequency of the tank and that will increase the drive current. Which leads to the question, what is the magnitude of the signal current in the coil? Would it be better to drive a simple coil directly or drive a tank off-resonance?

    With that in mind this analysis looks not just at the driving current through the load resistor but the inductive and capacitive currents as well, including several values of wire resistance in the coil. Note that in terms of the coil current there is nothing special about the resonance point, but including the capacitor is a way to decrease the driving current in the vicinity of resonance.

    EP55a.png
     

    Attached Files:

  20. Bharanidharan

    Bharanidharan

    25
    1
    Jul 22, 2014
    Hi,

    Thanks for the reply.

    I had to stop working on this for a while since my frequency generator is failing up. I am using AD9833 for a waveform generation and it stops working half of the time.

    I need to focus on that and then going into the circuit step by step.

    Thanks again
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-