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Breadboard is likely to have higher parasitic capacitance between tracks than either of your filter capacitors (10 pF and 5 pF) - this means trouble The 100 uH inductor is quite possibly self-resonating below the frequency you are aiming for and this means you won't get sustained oscillations. For both of the above try lowering the inductor value (L1) by ...


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JFET exponential amplifier, sinus generator, OTA. The output signal of the OTA is a current which is proportional to the transconductance of the OTA established by the amplifier bias current and the differential input voltage.


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Emitter-coupled Franklin oscillator with switched power supply.


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Oscillators work because the output gets fed back to the input with just the right phase shift to be the same phase coming out again. In the case of the that circuit, that means that the output is -- roughly, because the chip has delay -- in phase with the input. It's close to being in your "acts like a resistor" mode. The weird thing about that ...


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The following oscillator should work with precision op amps (low offset voltage), as long as the values of \$r\$ and \$T\$ do not need to be very precise. Defining \$\omega=2\pi/T\$, I get \$r=\frac{R_4/R_3}{2 R_2 C_2}\$ and \$\omega^2=\frac{(R_4/R_3) (R_5/R_6)+(R_4/R_6)+(R_5/R_6)}{R_1 C_1 R_2 C_2}-r^2\$. SPICE simulation seems to give the expected ...


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There is negative_feedback for the bias_point. Thus the circuit will self_bias into linear_region.


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Bias a WEIN_BRIDGE oscillator with charge on each of the two capacitors. Then enable the opamp output. A CA3080 has this Enable ability.


0

I have found out that the method to get the amplitude is actually quite simple. It has to do with damping due to the resistance in the circuit. - Case 1: If G1 > G: Here, the Amplitude is Imax/G, since if it's greater than that, there is a positive resistance (1/G) which will dampen it back to Imax/G and if it's less than that, there is a negative ...


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You are correct - leaving it open will enable the oscillator. One way of covering your bases is to put a resistor in your PCB but don't stuff it if not needed. In this case, it would be between the tri-state pin and the positive power supply. Putting such a component in the PCB also allows you to wire something to that point if you need to test it or to ...


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Depending on the application of your crystal, if its connected to a microcontroller with configurable clocks, you might be able to divide the clock with the MCU with a large divisor and then have it toggle a testpoint-gpio. Say divide by 128, you get a 62.5KHz signal that would be pretty easy to measure with almost any ADC/frequency counter during factory ...


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At the moment, all I can say is the following: The circuit is an electronic model of a pendulum clock - that means: Each period it receives (from the comparator) an additional "kick" which can compensate the losses of the analog loop. In the literature, this principle is called "restoration of initial conditions". The transfer function ...


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sometimes you need to enable the Chip (the oscillator) before it oscillates, to save energy. Try to enable it over I2C. Page 11 on the Datasheet you sent: Note 1:The ST bit must be set to enable the crystal oscillator circuit.


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The LM324 has not got enough open loop bandwidth to sustain a gain of 3.3 at 500 kHz. Try running at a lower frequency to prove this. If you made the inductor 1000 uH you would get oscillation: - And, if you made C1 = 10 nF (inductor still at 1000 uH) you would get a tad more amplitude: - But see how the op-amps's slew rate limiting is converting this ...


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Assuming the slope of your current source is higher than the LC tank conductance and steady state amplitude is much higher than the linear range of your current source, the fundamental component of current will be: $$I_{f_o} = \frac{4I_{max}}{\pi}$$ This current will flow only through the conductance since LC impedance is infinite at resonance, so the ...


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Energy method Assume oscillation is \$v_o = A \sin (\omega t)\$. Current and instantaneous power going into each element is (respectively) Capacitor \$i_C(t) = C \frac{d v_o}{dt} = C\omega A \cos(\cdot)\$ \$P_C(t) = C A^2 \omega\sin(\cdot) \cos(\cdot)\$ Inductor \$i_L(t) = \frac{-A \cos(\cdot)}{\omega L}\$ \$P_I(t) = \frac{-A^2 \sin(\cdot) \cos(\cdot)}{\...


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Your best approach for determining the frequency is to use the earlier suggestion of having some software on the MCU produce a low frequency output and measure that with a (calibrated) frequency counter. However, your most likely issue is the magnitude of the oscillation -- this is determined by the quality of the crystal and the parasitics and components ...


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If you want to see whether the PIC oscillator runs with the crystal, then turn it on and see whether it runs. Design is the time to be doing margin testing, to see whether a spread of crystals and a spread of PICs at a spread of voltages all start up and run reliably, not production. By production, you've established a crystal supplier and a PIC supplier ...


1

Since you have an external crystal, not an external crystal oscillator, it is extremely sensitive for connecting any measurement equipment to it. External crystal oscillator outputs logic level square wave and it would be much easier to measure. The frequency is too high to measure with your DAQ directly, it would be necessary to write a test program to PIC ...


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You probably should verify the actual frequency as well as the presence of the oscillator, at least approximately, in order to detect the situation where, say, a 4MHz or 10MHz crystal (or external oscillator) is populated. In general to do that you can use an adequately sensitive input circuit that it doesn’t unduly affect the oscillator output node, and a ...


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The TCA0372 is a dual 1A power Op Amp with reasonable low saturation and Class AB low Idle current. This might be suitable for your use or low power VFD control but better with PWM and current sensing. But for micro-stepper-mode they use DRV8844 dual H or Quad 1/2 H bridge FET drivers with PWM control with adjustable current limit to simulate analog ...


0

The resonant frequency of 10pF and 0.1 henry is Fres = 1/(2 * PI * sqrt (L * C)) = 0.159/sqrt(0.1 * 10e-12) Fres = 0.159 / sqrt(1e-12) = 0.159 * 1,000,000 = 159,000 Hertz. Does the opamp have much usable gain, at 160KHz?


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Your 384 ohm resistor is in the wrong place: - You may also need to lower R1 to allow the circuit to initiate and sustain oscillations. The OP-07 is going to struggle with the operating frequency - try making C1 ten to one hundred times higher initially. Don't be too surprised if sine purity isn't outstanding. Here's a simulation result: - Because the OP-...


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As a kid, having been given a 100,000 Hertz crystal of size 1" by 1" by 2" from some WW2 military radio, I tried repeatedly to get that beast to oscillate, listening at the 5MHz WWV channel for some interference or beat note that would indicate successful oscillation. I never could get oscillation. The challenge with a crystal is the interface ...


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People who have hard times when they try to understand feedback amplifier oscillators probably see the light if they can handle an audio system which has a mic, an amplifier and a speaker. Very soon they learn that the system starts to whistle or roar if there's too much gain or the mic is too near the speaker. The audio system has mathematically the same ...


0

The Barkhausen criterion is necessary but not sufficient to sustain a constant oscillation. e.g. There must some nonlinear mechanism to regulate the loop gain to be 1 for a sine signal. When 1-AB goes 0, Vs (the input) also go to 0, meaning no input is necessary to maintain a constant oscillating output. However, some initial condition and a non-linear ...


0

if nothing else, the PI_matching capacitors on either end of the resonator will demand power. Power (capacitor) === F * C * V^2 Notice the power scales linearly with the frequency. Example: Power = 10MHz * (20 + 10pF) * 1volt_rms^2 Power = 1e+7 * 30e-12 * 1 = 30e-5 = 0.3 milliWatts, just to charge/discharge the capacitance. Then the amplifier has quiescent ...


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Source: https://www.nxp.com/docs/en/data-sheet/S32K-DS.pdf In short the chip will consume power depending on: What peripherals are enabled Current consumption of peripherals (use high impedance if possible) What clock speed chip is running at The clock speed pf the S32L142 is determined by the PLL (and if you select an internal or external oscillator, it ...


1

Yes, you can use a square wave LO. It makes the multiplication by the data trivial. With digital data, it can all be done with logic, and with analogue IQ data you can do it with 4053-type analogue muxes. It sounds like you're not expecting the Arduino to produce the LO. Use a 4x LO digital clock, clocking a /4 Johnson counter. That's most easily made with ...


2

If you are able to get this working on a proto-board, it is by accident and not design. The leakage capacitance on a proto-board is greater than some of the capacitors you are using. Proper use of de-coupling capacitors on the power supply is neigh impossible due to inductance in the proto-board conductors. Its upper limit is often < 1 MHZ. In general, ...


1

Remove 5pF cap. was the solution add >= 0.01 uF on Vcc to gnd near IC and use short leads made the signal cleaner. Everything as instructed by Tony Stewart.


1

Consider building an oscillator that can be swept over a range from 80 MHz to 160 MHz (typically a Colpitts oscillator) and mix (using a 4 quadrant mixer) the output with a fixed frequency of 80 MHz. The difference frequency (wanted) can be extracted from the mix and will range from DC to 80 MHz. The sum product (unwanted) will range from 160 MHz to 240 MHz ...


6

There are some MCUs that do have a low frequency (like 10kHz or 32kHz nominal) oscillator so that the MCU can continue to do things like scan a keyboard while it is drawing little power. For example, the MSP430FR2311 has a VLO function. For such purposes, the actual frequency of the oscillator is not very important, +/-25% is fine, just accurate enough that ...


7

To get < 50 ppm accuracy the Q of the resonator must be > 10k. This is done either with MEMs Ceramic crystals acting like a tuning fork or a Quartz Crystal ( too big) This technology is not compatible with silicon chip materials or adds cost to a custom fabrication where it might not be needed. However, if the 32kHz is just for low power and not high ...


0

From personal research I have found that the ohmic resistance of a diode is approximately 25Ohms. The total resistance will bw 2x25 = 50Ohms. The initial energy stored in the capacitor is 1J.The power dissipated on the resistor is 10^5x50= 5x10^4=10^3/2 W. L = 1H. C = 20mF. The frequency of the RLC oscillation is 1/sqrt(LC) = 1/sqrt(1x10^3x20) = 1/0.15 = 6....


0

here is a simulation that I just quickly did and here are the results: from the simulation, it can be seen that it took ~670,000 seconds (=7.75462963 days) to discharge the capacitor. I have chosen capacitor value to be 2 micro farads and initial voltage of 1000 volts, which means it will store the energy of 0.5 * 2 * (10^-6) * (1000) * (1000) = 1 J I have ...


1

You are using ideal tank, add a resistor in parallel with the LC tank, should work then. Also, you have a floating node between the drains of the MOS. Split the inductor as a series combination of two 0.04nH inductors and connect the node in between to VDD/2 = 1.25V. Will definitely work then.


0

Are electromagnetic waves produced during the oscillation of charge in a LC oscillator? Yes, and this is a fascinating niche-topic in antenna-design, called "electrically-short resonant antennas." If a coil/capacitor is far smaller than quarter-wavelength, it's still able to become a "virtual antenna" which behaves as if it was greatly ...


0

Let's assume you connect a charged capacitor and an inductor in parallel the circuit starts to oscillate. There's a sinusoidal electric field between the ends of the coil and that's enough for creating electromagnetic radiation. The radiation power can be quite small in practical LC circuits when compared to the power of the energy transfer between the ...


2

Consider a disk capacitor "shorted" with an external loop. Even if the external loop has no resistance, there is some inductance associated with the loop, and therefore the discharge can lead to an oscillation at a frequency determined by the inductance and capacitance of this structure. If this loop is sufficiently large compared to the free ...


1

Actually, an LC tank (as part of a functioning LC oscillator) will produce a near field electromagnetic field around the inductor as alternating currents move through it. However this won’t be suitable for long distance transmission of electromagnetic waves to a receiver. It can be used for close transmission as similar to transformer operation. For long ...


-1

Maddy --- an antenna produces 3 different emissions. There is the radiative field, and two reactive fields. This per Corson and Larraine book. And wikipedia "near and far fields" Close to the antenna, the two reactive emissions are very strong. But their rate of dropoff is very fast, so eventually (about a wavelength away) the third form of ...


0

If you abruptly turn on the power, imbalances in charges and currents and voltages will turn one transistor fully on and turn the other transistor full off ---- with that state continuing as the base of the Off transistor slowly rises from -VDD or similar levels (unless the emitter-base breakdown serves as a clamp on the negative_going excursion). And that ...


3

Maddy, you have a good idea. However, there is a QUALITY FACTOR we can compute for even simple L+C circuitry And at a Q of 5 (ratio of XL/R = Xc/R = 5 at the resonant frequency), the voltage drops by 50% over each cycle. That also means the stored power drops by (0.5)^2 or 4:1 drop. Now let us suppose your L+C (closed_circuit_loop, so the energy can slosh ...


2

No, you can't construct this with real components. If nothing else, you'll never get rid of the effective series resistance of the capacitor. There is no 100% efficient circuit in real life.


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