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1

Its not really practical to design any circuit without calculations and Alfred's answer gives a comprehensive approach (so +1 from me). However, there are other 'rule of thumb' type" approaches that might suit a 'practical' designer to get started. The thinking behind this approach is very much based in the theory developed by scientists and engineers over ...


2

Since there are 4 resistor values to choose, one needs 4 (realizable) constraints to uniquely choose the 4 values. Some examples of specifications that impose constraints are (1) input impedance (2) output impedance (3) open-circuit voltage gain (4) operating point (5) positive and negative clipping levels (6) idle power dissipation You've only ...


0

For audio purposes I often use the IF3602. Its over $40 each and has a huge lead time, but is the lowest voltage noise FET I've ever seen: < 1 nV/rHz at audio frequencies and has a large input impedance. A good BJT PNP is the SSM2210. With a few of these in parallel you can get an input referred noise of ~0.5 nV/sqrt(Hz) from 0.1-10000 Hz.


2

You didn't state BJT or MOSFET, low or high current, low or high voltage, through-hole or surface mount package, digital switching or audio use, ... SS8050DBU / SS8550DBU : BJT, TO92, 1.5A. 2N4401 / 2N4403 : BJT, TO92. 2N5401 / 2N5551 : BJT, TO92, Vceo = 150V. MPSA42 / MPSA92 : BJT, TO92, Vceo = 300V. TIP2955/ TIP3055 : BJT, TO247, 15A. 2N7000 or ...


3

None of those transistors (nor the TL431 regulator) are particularly suitable. Look for an NPN silicon transistor with an ft of 600-800MHz. The best fit is the high-voltage 2N5401 with an ft of 300MHz, but at 100MHz+ FM it won't be that great.


0

4 LEDs in series, each dropping 3 V isn't going to work well with a 12 V source. The LEDs are using up all the voltage, so there is nothing left for a switch to drop or some mechanism to make sure the current is reasonably regulated. The "12V" battery will vary a bit depending on temperature and state of charge. LEDs have a steep current curve as a ...


1

You need to get some headroom, so split the serially connected 4 LED array into an array of two parallel strings, as shown below. R1 and R2 will dissipate about a half-watt each, so it wouldn't hurt to use two standard 150 ohm 5% carbon film 1/2 watt resistors in parallel for each string, each parallel pair in series with its string of LEDs.


2

It appears that you only need a switch to switch the LEDs on and off using an Arduino pin, setup as a digital output. The comment about a transistor being an amplifier seems to be a distraction. You could do this in a variety of ways. You could use a MOSFET, BJT, Darlington transistor, or an IC. The simplest approach might be to use an N-Channel MOSFET: ...


0

To answer in simple terms: the collector behaves like a current sink, and the collector voltage settles to whatever value lets that amount of current flow (though it can't go any lower than approximately Ve+0.2V). In your example circuit, the collector-emitter junction can be thought of as a variable resistance whose value depends on the electronical ...


2

You've got a couple of good answers. I'll try to add some intuitive insight. When the transistor is biased with so that it is not saturated, it behaves like a current sink (recall that a perfect current sink has infinite impedance), so the collector-load junction looks like a voltage source with a Thevenin equivalent source impedance equal to the load ...


5

The BJT collector current equation is $$i_C = I_S\cdot e^{\frac{v_{BE}}{V_T}}\left(1 + \frac{v_{CB}}{V_A}\right)$$ where \$V_A\$ is the Early voltage. But, this formula is often written as $$i_C = I_S\cdot e^{\frac{v_{BE}}{V_T}}\left(1 + \frac{v_{CE}}{V_A}\right)$$ Thus $$\frac{\partial i_C}{\partial v_{CE}} = \frac{I_S\cdot ...


2

If the supply voltage and the load resistance remain constant, then as the base current varies, the collector voltage and current will vary. Such being the case, then for any collector current there must be a resistance between the collector and emitter such that: EDIT: $$R2 = \frac{E2R1}{E1 - E2}$$ Where R2 is the transistor's collector-to-emitter ...


4

I'm inclined to agree with Spehro here - you are better off with an LDO or a switching regulator that come with an ENable pin. However, having just done this myself I'll add a picture of what my high side P Mosfet switch looks like (cropped quickly from a larger schematic): In this case the power input is the line coming into the P mosfet on the left. The ...


4

If you pull the ADJ pin to ground with a transistor you'll reduce the output voltage to 1.25V. If that's good enough, all you need is a transistor like a 2N2222 with a base resistor (note that the base will draw current to turn the regulator off). It will still draw current, but it should be reduced quite a bit (the regulator with resistors will draw at ...


0

From my experience, most of the time a transformer makes noise, is due to a loose lamination or a loose mounting. A mechanical chopper makes noise because the reed that "chops" the current moves/vibrates. Obviously anything that moves, makes a sound. A transformer usually produces a 60 Hz hum, while a chopper depends on the frequency for which it was ...


2

The ceramic part is a Balun. It's used to convert the differential-drive RF output from the IC to a single-ended 50Ω output that is suitable for feeding into an antenna (I have used similar parts when designing similar 2.4 Ghz transciever PCBs). The balun may be specially designed to reject frequencies out of it's pass-band, but it's fundamentally still a ...


3

Most Likely a Multilayer Ceramic Band Pass Filter. For the Antenna Stage. (The Schematic below uses RFBPB 2012 in a 0805 sized package, but any similar Filter would be used.) Most Likely a Voltage Regulator. The IC uses multiple voltages (3.3V VDD, 1.8V, VBAT [Likely 5v]). Alternatively, it could be an opamp, but you need to trace the pins out to know where ...


3

In the picture, what is the component of above red box? This looks like 2.402 GHz chip band pass filter like the one shown in the datasheet here. What is the component of below red box? Top of this component, It's written "DH=M6C". But I couldn't find any datasheets about "DH=M6C" That may be 5 pin LDO for supplying the power to ...


6

You should be able to see immediately from inspection (without doing any calculations) that the transistor is saturated. Generally you figure the C-E voltage of a saturated transistor is 200 mV or less unless the current is unusually large, which in this case it's obviously not due to the size of R1 and R3. The answer for any real electrical engineering ...


2

This isn't an amplifier as such. The transistor is acting as a switch. Let's have a quick look at the numbers using rules of thumb. (I'm assuming those voltages are DC with no AC component) 10V - 0.6V (Vbe) = 9.4V at the base. Divide by 25k = 0.37mA roughly base current. If we multiply that by (hand wave) a hfe of 100, we get 37mA into the collector before ...


2

Moving membranes or piezoelectric materials obviously produce sound waves, but how can "purely" electrical circuits such as transformers or DC/DC choppers (and others) often have an audible noise? Is the material microscopically expanding and shrinking with the current? While others have explained the part about the material moving nicely, one key point ...


1

There's been a lot of theory here. In practice, usually loose wires of inductors are involved. Tapping around on the coils (not!!!! with anything magnetic like a screw driver: trying that on coils in CRT flyback circuitry is something you don't do more than once) may help locating the culprit, and suitable warm glue or nail polish may help getting it under ...


0

Since your input only has two states (+ and -) you can use a simple H-bridge to do this. The H-bridge must tolerate minus inputs and interpret it as a '0'. For example:- Driven by this: simulate this circuit – Schematic created using CircuitLab


2

Here is one more Sound by changing properties of surrounding plasma or gas due to exposure of an electrical field and/or electrical discharge Based upon the "Singing Arc" which was discovered around 1900 by William Duddell, the Ionophone or as it is mostly called plasma speaker/tweeter (is actually used in speakers) produces sound waves by charging plasma ...


55

What you are really asking is how can electrical circuits cause small motions. After all, sound is motion of the air. The answer is that there are various ways electric fields or electric currents can cause forces or motions. These effects are harnessed in the design of various tranducers, which exist to deliberately cause or sense small motions. ...


1

Another effect not yet touched upon is wire straightening under load - wires do tend to straighten when current is passed through them, whether microscopically or visibly. The wire within a power transformer's windings tries to straighten very slightly 100 to 120 times per second (depending upon the frequency of municipal power). This phenomenon can be very ...


12

An ideal inductor or transformer might be a purely electronic component, but a real inductor or transformer produces a (rapidly changing) magnetic field. It is a design aim of such a component to keep that magnetic field within the component (for instance inside the ferromagnetic core), but that won't be achieved for 100%. The 'leaking' magnetic field will ...


5

It isn't expanding or contracting the material, that emits the sound in transformer or inductor-based circuits. However the parts are moving. Transformers are subject to significant mechanical forces caused by the alternatine electromagnetc fields. That causes wires and laminations to move, and hence emit sound. DC-DC converters often have wound inductors, ...


0

The diodes aren't protecting the motor; they are protecting the transistors. So put them across the transistors. The current-both-ways problem isn't present at the transistors.


0

One option would be to not use any protective diodes, but to parallel the winding for each motor with an MOV and a smallish capacitor, while making sure that your transtors' VCE is higher than the MOVs' clamping voltage. The capacitor will help level the inductive flyback, and the MOV will prevent that spike from getting very tall.


0

Firstly I must point out that your circuit will NOT work as you have wired the transistors but assuming this is just a picture rather than an actual schematic read on... You connect them across all four transistors in the H bridge. For instance Q1 emitter has a diode anode and Q1's collector has the diode's cathode. For Q2 the emitter has the anode and the ...


4

From your other question you need to be able to switch 3A and you need to switch 3A so that when the MOSFET is on it isn't going to get too warm so maybe say "it shouldn't dissipate more than 100 mW at 3A. This means its on resistance (\$R_{DS(on)}\$) is a maximum of: \$\dfrac{power}{current^2}\$ = 11 milli ohms. Next you need to be able to turn that fet ...


1

Much of the time Complementary MOS (CMOS), i.e. circuits built using P-Channel and N-Channel MOSFETs, is not conducting. The power consumption is mainly during switching from one state to the other. N-Channel MOSFETs will either use less power to switch, or will switch faster than P-Channel MOSFETs for similar geometry. (PMOS is 'slower' than NMOS, because ...


1

The width differences arise from the difference in mobility of the electrons and holes with holes being a factor of ~2.2 slower than electrons. \$ \frac{W}{L} \mu_e = \frac{W}{L} \mu_h \$ Having balanced \$ g_m\$ on teh PMOS and NMOS will give symmetrical rising and falling edges times. Which is not a necessary condition for low power as you state. Low ...


0

Check any electrolytic capacitors in the audio amp section. Could be they have dried up and are killing your audio.


1

The simplest calculation is to assume the base current is 0, then Vb = \$Vcc\frac{R2}{R1+R2}\$ and Ve = Vb - 0.7V. Once you know Ve you know the emitter current Ie= Ve/Re, and since the base current is assumed to be zero, you can find Vc = Vcc - Ie * Rc. That is valid if \$\frac{R1R2}{R1+R2}\$ << \$\beta Re\$ If you want more accuracy you ...


0

You would really need to know the output resistance of the input voltage source to Vb. Without it you can't really know for sure. If an op-amp or a similar amplifier stage drives Vb, then the output impedance should be quite low. If it's low, then it's as if you're applying a true voltage source to Vb. In that case, all of the bias resistors really don't ...


1

You must analize the Thevenin's equivalent circuit: where $$ R_{th} = \dfrac{R_1\cdot R_2}{R_1 + R_2} $$ and $$ V_{th} = \dfrac{V_{CC}}{R_1+R_2}\cdot R_2 $$ then, the equivalent circuit where \$V_b\$ is \$V_{th}\$ and \$R_b\$ is \$R_{th}\$ Then, you can write KVL, for the input circuit: $$ V_b = i_b\cdot R_b + V_{BE} + i_b\cdot (\beta + 1) \cdot ...


5

instead it offers a resistance called ac resistance. why it is so ? why dont it rectify ? Let the voltage across the junction be of the following form $$v_{BE} = V_{BE} + v_{be}$$ where \$V_{BE}\$ is positive and constant while \$v_{be}\$ is AC and is, in some sense, small (I'll later clarify what that means). Now, recall the equation for the base ...


2

You will never be able to get accurate amplification of a 1mV full scale DC signal with a simple circuit using discrete components (I do know a way, but it's not simple). Even a crappy LM324 will do better (because the transistors are matched and on one chip). Tell us what you have and we can, perhaps, help. An instrumentation amplifier can be made with ...


3

The full operation of the circuit is analyzed in two parts: the small-signal model and the bias circuit. In the bias circuit, one considers the nature of the rectifier base-emitter base-collector junctions to establish the operating point (point Q), together with external components needed. In the small signal model, considering all the factors that alter ...


2

It is rather a property of the small signal model than a property of the diode you are applying the model to. It is the very purpose of the small signal model to look at any non linear device as if it was a linear device, i.e. a resistor. Of course it can only be applied with restrictions, e.g. that the function is differentiable at the operation point. ...


3

Because if you look closely enough at a small enough portion of a nonlinear curve it looks pretty much like a straight line. There's a detailed mathematical treatment here, but I'll also reproduce the Wiki graph to illustrate intuitively (I hope) what is going on: In the small signal model we are looking at very small changes from the bias (operating ...


0

A circuit using a phot-interrupter would look something like this: simulate this circuit – Schematic created using CircuitLab D1 and Q1 are inside the photo-interrupter. There will be a gap in the package between them so something can interrupt the light path between the LED and photo-transistor. If light can pass from the LED (D1) to to Q1, ...


0

The easiest approach is to use logic-level output photointerrupter like this one. There is no need for two power supplies, connect the transmitter side through a dropper resistor to set IR LED current (usually 50mA at 1.2V). Output of this particular series is open collector, so if your load draws less than 50 mA,you can connect it directly between output ...


3

The variable g is used for what are called the inverse hybrid network parameters. It is indeed defined as \$g_{22}=\dfrac{v_2}{i_2}|_{v_1=0}\$ The inverse hybrid parameters are very rarely used in practice, and they don't have conventional names except "the inverse hybrid parameter \$g_{22}\$". The z parameters are conventionally called the impedance ...


1

Your first answer may be what they are looking for, if one assumes Vbe and \$\beta\$ are constant. The second is slightly off, as you failed to take into account the base current. The third is way off. It's approximately 100K/\$\beta\$ but I'll let you work out the exact value.


1

No, the output impedance is not zero. To figure it out, try increasing or decreasing the value of the current source by some small increment, and then calculating the resulting change in output voltage. ΔV / ΔI gives you the effective impedance of the resistor+transistor circuit, which is also the output impedance, since the impedance of the ...


1

When we specify output impedance of a two port network, does it imply open circuit output impedance or Thévenin equivalent impedance(short circuit output impedance)? If an output produces a 1V signal when open circuit and can drive a 10mA current signal when short circuited, the output impedance is 1V/10mA = 100 ohms. It's the same as have a ...


1

It's the equivalent of sticking a resistor on the output of a zero impedance output



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