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1

SoC ‘s that look like the two on your photo are ASIC’s and thus are an Arm and a Leg to buy. speaking of ARM’s .... TI has licensed CORTEX from ARM to put in their Open Multiple Architecture Platform called OMAP and may be developed on a BeagleBoard from DigiKey. Unless you have deep pockets, here some of what the big OEM’s use for SoC’s from a half a ...


0

1) NO 3mA is all that it uses max. 2) NO. Because it is to reduce ripple which reduces the internal signal/noise ratio (SNR) from supply noise. This is just a low pass filter. Perhaps choose R1 for <0.3V drop max @ 3mA or <100 Ohms and C to attenuate expected ripple f for mV(pp). The TSOP384xxx series has better AGC for CFL stray light pulses. The ...


0

If you're not too concerned about the effects of self heating, then a circuit like the one shown in fig. 1 might be useful. simulate this circuit – Schematic created using CircuitLab Figure 1. Resistor R1 is the current sense resistor. If you know R1's resistance value (by measuring its resistance with an ohmmeter), and if you measure with a ...


1

Use the application circuit diagram to connect to the Arduino. Since you are using 3.3V, you don't need R1 and C1 (short instead of R1, simply omit C1). The "Output" pin should be connected to a GPIO of the Arduino. The "Max Supply Current" is the maximum current that the part will draw, not what you need to limit it to. You rarely have to limit the ...


0

1) No. That specification means you need to give it 2.5V - 5.5V, and it will draw no more than 3mA. Putting a series resistor in there (beyond their suggested RC network) will only screw things up. 2) Correct. You can probably leave it off. Don't worry about it -- just connect Vs to +5V or +3.3V, whichever you're using. 3) No.


3

Basically, a LED is just a diode. If you take a diode and bias it in reverse, no current flows through, because there's no hole/electron recombination happening. Now, that "no" is a bit of a stretch: of course, due to thermal effects, pure luck (tunneling), a few electrons and holes will still recombine. Now, in an LED, a photon hitting the carrier-...


0

Have a look at I/O Expander. There you can control behaviour of individual pins or multiple pins in a bunch. It is produced in different packaging for any application.


0

Forget about using a decoder at all. Simply drive the enables directly from pins on your MCU. Then you can select any combination, including more than two at a time if necessary. Note that to select an arbitrary combination of 0, 1 or 2 outputs, you need a total of 8 output pins from your MCU anyway (3 select bits and an enable bit for each of two decoders),...


0

This design is going to deviate from expected performance due to insufficient GBW requirements. With a gain of 3.3 in each stage with a Q of ~14 in each stage at staggered frequencies to achieve a net BW of 10% fc= 10kHz, if one truly expects a Butterworth response 1 octave up, means the minimum GBW product is >> 50 MHz. A GBW of 10MHz reduces the output ...


0

1) There is an error showing V+=Vcc (U1-7) power connected to an input node. Since this circuit uses a 0V reference with R4 to 0V, it needs a bipolar supply. 2) V- = Vee (U1-4) is open on schematic and needs some negative supply. This is why you have no output signal.


2

Positive corona indicates the HV electrode is the anode. Negative corona indicates the HV electrode is the cathode. In Electronics IEEE has defined the flow of current from +ve to -ve for "logical" reasons to read +ve voltages, but it is opposite to the actual flow of -ve electrons. So electrons flow towards the positive electrode. A sharp electrode ...


2

Step back and look at the definitions of cathode and anode for a moment. From the Wikipedia cathode page: A cathode is the electrode from which a conventional current leaves a polarized electrical device. From the Wikipedia anode page: An anode is an electrode through which the conventional current enters into a polarized electrical device. ...


1

You have done an transient analysis, do an AC sweep ( I don't know what version of multisim you have but here are instructions for the latest version) An AC sweep will allow you to get an amplitude vs frequency plot. If you still want to find dB, you can do this for a transient plot, you need to know the amplitude of voltage in and voltage out, then ...


5

Barrel connector with switch -- 2 conductors, 3 contacts. Here are some similar parts Digi-Key has listed. Most will show a diagram in the data sheet. Terminal 1 is the sleeve connection, and terminal 2 is the tip. Terminal 4 is a switch contact that is connected to the tip when there is no plug inserted, and disconnected when the plug is inserted.


11

That signifies a jack with a connection that breaks when a plug is inserted. In the case of your schematic, that connection isn't connected to anything anyway, so it shouldn't matter.


1

Adding an MCU to a Ding-Dong chime doorbell The circuit is likely like this currently: simulate this circuit – Schematic created using CircuitLab The Reed relay can be almost any 5-12VDC low current unit such as LittleFuse HE3621A1200. This operates from 12V to 22V without problems, and the 8V supply when rectified would give about 9-10V DC, more ...


2

Just use an ordinary relay, whose coil is rated for the 8 or 12 VAC of the chime — but won't pull in on the amount of current the bulb uses — and connect its coil in place of the chime coil. Now you have an independent contact closure you can use any way you like.


0

Figure 1. A very simple polarity reversal circuit using two GPIOs. GPIO tricks. GPIO1 GPIO2 LED ----- ----- ----- 0 0 Off 1 0 Green 0 1 Red 1 1 Off


1

After looking into this a bit more, I think I have a better answer than my comment above. Since the LM148 is (as the datasheet says) "a true quad 741", I had a look at the actual schematic of the famous 741 opamp. Which you shouldn't be using, read here why. Of course for educational purposes, studying the 741 (or LM148) is fine. This site has an excellent ...


0

The problem in the second half of your question comes about because you're trying to solve a differential equation as if the left and right halves do not depend on each other. You write: $$C_1 \frac{d\,V_o}{dt} = g_{m_1} V_{ekg}+g_{m_2} \frac{V_o - V_{ref}}{100} - g_{m_2} V_{HPA}$$ Then you integrate both sides. You cannot do that, because it is a ...


0

GM1 and GM2 are transconductance amplifiers, per the data sheet. This means that they generate currents that are equal to \$g_m \left ( V_+ - V_- \right)\$, where \$g_m\$ is the transconductance of the amplifier, \$V_+\$ is the non-inverting input, and \$V_-\$ is the inverting input. The output of GM2 feeds back to the op-amp. So what is integrated is the ...


0

As mentioned above, an H-bridge driver with four external power MOSFETs will give you an all-electronic motor reverser that needs only one control line from the Arduino (two if you want a third state - off). The larger the MOSFETs, the lower the heat. International Rectifier is big in h-bridge control chips. Here is the general idea:


1

You have a temperature sensor with analog voltage or current output. Different operating modes, but one is the basic process controls 4-20mA two-wire current loop meant to be fed into a remote <600Ω load resistor. 3.5mA or 23mA (NAMUR NE43) are used to indicate errors. And 0mA indicates an open circuit or unconnected sensor. Not the same, but relevant ...


0

I believe what they're telling you is that although the supply voltage is nominally 24V, you'll never see more than 18V at the output terminals, even if you leave them open-circuit.


2

It will work until something overheats or otherwise fails. The speed of a DC motor is proportional to voltage. If a 3.7-volt DC motor is connected to 12 volts, it will try to run at 12/3.7 = 3.2 times the normal speed. The mechanical power required to drive a fan is approximately proportional to speed cubed. That means the fan will try to load the motor to ...


2

You'll need to take the transistor off the board and follow these steps (for an NPN transistor): Step 1: (Base to Emitter) Hook the positive lead from the multimeter to the to the BASE (B) of the transistor. Hook the negative meter lead to the EMITTER (E) of the transistor. For an good NPN transistor, the meter should show a voltage drop ...


0

Following @DaveTweed 's comment, a low-power cross clipper has become obvious to me. There are different issues with the circuit now: cross-clipping is heavily dependent upon input amplitude, so if the signal is loud the relative distortion is low, a low-amplitude signal will be dominated by the distortion, if it passes at all. It would be nice to figure out ...


0

From the 1976 ARRL Handbook. My apologies for the poor focus. Note that the inductor and cap values are for 455kHz-ish, not 14MHz. A search on "diode switched crystal oscillators" may prove helpful.


-1

How about this simulate this circuit – Schematic created using CircuitLab


1

One option that is quite robust is to have two oscillators and feed square wave clock to the chip. Another what I have seen is to use an analog multiplexer to switch one pin from the chip to one of the crystals.


2

A J-FET junction is like a diode when the junction is forwards biased. that will reduce the amplitude of your signal generator output.


4

We call this a "netlist", but generally don't edit it directly as text It is possible to represent a circuit as a serialized text representation, versus the graphical schematics we normally use. The traditional name for this is a "netlist", as it is a list of components and the nets (nodes) that connect them together. While there are a wide variety of ...


1

This is my understanding of the circuit. We need to know the minimum and maximum pulse widths on the incoming serial data. I have just assumed a square wave to get started. We also need to know more about the load on IC9, I have just assumed two load states. You need more signal conditioning, the transistor that was removed was probably useful. I added a ...


3

LTSPICE schematics save as SPICE netlsts which are text annother other way to do circuits in text is ASCII art eg: to get 50% duty cycle the easiest way is the 1 resistor circuit, ----+--- vcc | +-[R1]--------|--------+ | ...


1

A 1st order approximation using : For a 1kHz square wave with T/2 sag or tilt by 10% \$\tau_1=5ms =0.5ms/ 0.1(sag)= 0.5ms/0.1 = 1/\omega_1\$ = HPF Sine tr=0.35/f_{-3dB} BW for rise time 10 to 90% \$\tau_2=4.55us=1/\omega_2 =t_r/(2\pi *0.35)=10us/(2\pi *0.35)\$ = LPF assuming 10~90% for rise time Then the transfer function can be made with gain constant ...


0

A precise definition of RS latch behavior should define it in terms of R, S, and previous Q and /Q values, recognizing that Q outputs and inputs may be stable high, stable low, or metastable. If either or both inputs is low, the states of Q and /Q will be ignored. If both inputs are high and Q and /Q are in any configuration other than high-low or low-high,...


3

Since it's hard to reason about a circuit that's not given in schematic form, but as a wiring diagram, let's only talk about the shortcomings of that wiring diagram: That's a 1.2 MHz switch frequency converter. Hence, it produces very high frequencies (multiples of the switching frequency), and breadboard is almost always a bad idea in that case, because it ...


1

Take the Laplace transform of the input and output signals, and divide one by the other. In this particular case the signals are piecewise linear. For (simple!) example, if the input is a unit step, and the desired output is a unit ramp for 1 sec, followed by a constant value of unity, the input and output signals would be: \$\frac{1}{s}\$ and \$\frac{1-e^{-...


0

The simplest substitution would be to replace each relay with a MOSFET operating as a switch. I'm assuming that the potential difference across the input pins of your receiver is 24V, in which case you'll want to use a low side switch, using an N-channel MOSFET (or NMOS for short), rather than a high side switch as you drew in your original schematic. ...


1

Since it is a nice sine, you can also find the delay between V(in) and V(out) (assuming V(out) lags wrt V(in) by triggering on their zero-crossings, using: .meas TRAN delay_info TRIG V(vin)=0 RISE=1 TARG V(vout)=0 RISE=1 EDIT Or even better, is there a way to get the time (and use it a variable, of course) where the measure occurs on the .measure spice ...


0

This is a good circuit for switching resistances. The gotcha is the gate level of Q1 needs to be higher than the drain so this circuit is only useful if you have a rail that is higher than the one you are switching.


1

If you need to measure the Y-axis at a specific time point then, for your case, you would use this: .meas tving find V(vin) at <time_value> Similarly fot the delayed voltage. Don't forget that the measurement is directly related to the resolution of the data points. By default, it's 300 points, which can give inaccurate readings. To disable waveform ...


-1

Others have answered the basic question. This answer is just to show a more conventional schematic. simulate this circuit – Schematic created using CircuitLab Figure 1. OP's schematic redrawn in conventional manner. For ease of reading and showing the intent of the schema normal practice is to enable reading from left to right with higher voltages ...


0

For your purpose, at 500 mA or less of power draw for your leds through the IRFz44N, it would be fine. It has a Vgs Threshold of 2V min and 4V max, and the Vgs to Id graph shows it will pass up to 20 Amps at 5V Vgs.


0

the total voltage in a loop is the algebraic sum of the individual voltages. This is KVL, not KCL. It should be something like V1 = h11 I1 + V2 according to KCL. The voltage across the resistor is \$h_{11}I_1\$ and the voltage across the controlled voltage source is \$h_{12}V_2\$. Therefore the KVL equation for this subcircuit is $$V_1 = h_{11}I_1 + ...


0

It's more complicated than that. The resistors & capacitors more or less set the gain and filter characteristics simultaneously. The easiest thing to do would be to go back to the TI tool and enter in your new numbers. To solve the problem of your ADC range being from 0-3.3V, you can use a "pseudo-ground" that's right in the middle of your ADC range (...


1

s-domain analysis is unitless, so you are free to describe any transfer function. You may simply provide your instructor with a resistor with a resistance greater than \$ 1 \; \Omega\$, along with the current to voltage transfer function of your resistor. You could also employ an ideal transformer with secondary:primary turns ratio greater than 1.


0

What you want to do (a passive network with gain > 1 independent of frequency) is not possible if you use only passive components. If you use resistors, one of them will have to be negative. The last time I checked, negative resistors were not yet invented. It is possible to simulate negative resistances, using a device called a gyrator. A gyrator always ...


0

will it work with IRFz44N No, you are driving this MOSFET with a 5V Vgs, thus you need a MOSFET that has RdsON specified at 5V Vgs, otherwise it may work, or maybe not. For IRFZ44 it is only specified at 10V. You can use IRLZ44 for example, here's its spec: Your schematic is OK but: The LEDs are going to draw some pretty high pulsed current, so I'd add a ...


0

Amplifier A1 is setup as a non-inverting buffer. The voltage at (+) is just some voltage controlled by the DAC, say 1.0V. Since the opamp has negative feedback we can expect the voltage at (+) to be equal to the voltage at (-), as an ideal opamp would behave. So what happens is there is a voltage of 1.0V at +. The opamp will drive it's output until ...


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