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1

Just a short answer: According to my experience the following may be helpful. 1.) At first, connect a voltage source Vin to the respective node and convince yourself how many different pathes are available which allow a current to exist. During this step you consider all dc sources as signal ground. 2.) Based on Ohm`s law (and/or other laws, if applicable) ...


2

For output impedance start by understanding the right hand picture below: - It's the right hand picture I'm interested in - it shows several curves for a BJT and these curves are based on a certain base current (omA to 4mA) and for each current (held constant), the voltage across collector-emitter is increased from 0V to 20V. As this voltage is increased, ...


4

These diodes are for temperature compensation. As there is a diode connection between the Base and Emitter terminals on the transistors, to compensate for any characteristic changes with changes in temperature the diodes are put in place to match these characteristic changes. For example: Say that the voltage drop over the diodes increase due to a change ...


2

The circuit is built from two identical halves that can be considered independent. Without context it is hard to say what the function of the diodes is, but it seems an attempt to compensate for the base to emitter voltage across in the related transistor in such a way that the voltage across R4 and R1 is identical. However with the large difference in ...


1

What if you dispense with the buffer transistors and add some capacitance to your 12V rail? The continuous current rating will remain the same, but a few uF should prop it up for the brief surges required to charge the gates. There's also the output current rating of your logic to consider, mostly the peak rating.


0

Neither A nor B are needed - the PNP transistor will pull down the gate to about 0.7 volts when the input to the BJT push-pull circuit is at 0V and this is neither aided nor hindered by either resistor. What I will say is that the MOSFET shown in your circuit may not get adequately turned off and this could be a problem. This is because the PNP is an emitter ...


2

You take a TG based XNOR: That is 4 transistors. You add two Invertors to generate the complements, /A & /B so that is an additional 2 X 2 = 4 transistors. Since this is a XNOR you need to invert the output and that provides a consistent load to the TG's. That is another 2 transistors. Voila! 10 transistors


2

Homework? What are they trying to teach you? Look at the various versions and see how they work. Going from eg 12 to 8 they replace 4 with passives in some manner. Why did they replace 4? Could they have just replaced 2? Could any 2 of 4 have been been replaced ... or just 2 ........or 'this 2' or 'that two' but not just any 2 ...? These are probably ...


1

This paper references a 10T XOR gate: http://eprints.uthm.edu.my/1716/1/Nabihah_Ahmad_FKEE_(ICEDSA).pdf The 10T XOR in this paper probably isn't the most straightforward design, though. It's optimized for low power, and likely makes several tradeoffs.


0

The way I like to visualize it is to consider the transistor as a variable resistor which the opamp adjusts automatically in order to keep the voltage at the opamp's - input equal to the voltage on its + input. That way, since the current in a series circuit is everywhere the same, the current in the load, the transistor CE junction, and Rset must be the ...


1

I thought the collector-base junction was reverse biased which means that the electric field created by the external bias adds to the potential barrier? How is it that holes spontaneously cross this huge potential barrier? The built-in potential stops the diffusion current due to majority carriers diffusing from one side to the other. For ...


0

Another approach is to model the op amp as a large finite gain and the take limits. This gives the op amp output as \$K(v_\text{set}-I_\text{load} R_\text{set})\$ from which we have \$K(V_\text{set}-I_\text{load} R_\text{set}) = I_\text{load} R_\text{set} + 0.7\$. Dividing across by \$K\$ and letting \$K \to \infty\$ gives the desired result, ...


5

The opamp is acting as a unity-gain buffer, though it may not be obvious: The rule for opamps is that the output does whatever it has to to keep the two inputs equal, provided that it doesn't clip of course (run into its own supply and stop there). The transistor is used as an emitter-follower, in which the emitter voltage follows the base voltage minus a ...


9

The circuit employs negative feedback and utilizes the very high gain of the op amp. The op amp will try to keep its non-inverting and inverting inputs at the same voltage \$V_{set}\$ due to its very high gain. Then by Ohm's law $$I_{set} = \frac{V_{set}}{R_{set}}$$ Negative feedback causes the op amp to adjust the transistor base voltage so that ...


4

My first reaction is to use a opto-coupler. That isolates your microcontroller from the PC ground, which is a good idea. This solution still requires that the device be installed with the correct polarity, which personally I don't think is a big deal. I also don't think that worrying about the voltage being high makes any sense. The front panel button is ...


1

If you put a diode in series with transistor collector it is likely that this would work. If someone wired this up in reverse, placing a high value leakage resistor across the transistor (maybe 1Mohm) would protect the transistor. Duplicate this circuit so that one of the transistors would always work no-matter what the polarity and you have a bipolar ...


2

The voltage from V1 is divided between R1 and the B-E junction of the transistor. Since that junction is a forward-biased diode, the voltage drop across it is only about 0.65V; the rest of the voltage appears across R1, setting the magnitude of the base current. The voltage from V2 is divided among R2, the C-B junction of the transistor, and the B-E ...


0

For the right battery to reverse bias the collector diode( or collector base junction as it is called), the potential at the collector must exceed that at the base. Otherwise, the transistor is no longer in active mode. Note that it is the collector base junction which is reverse biased in active mode.


1

Let's say the base is at approximately 0.5 v potential above ground and causing a collector current of (say) 1mA. If V2 is at 10 volts with respect to ground (emitter) and R2 is 1kohm (for instance) then there will be 9V at the collector. With 9V at the collector and 0.5 volts at the base, the CB junction is reverse biased by 8.5 volts.


1

Both batteries share a terminal. So that the collector-base diode is reverse biased, \$V_C\$ must be greater than \$V_B\$, as $$ V_{CB} = V_C - V_B $$ Viewed another way, since the emitter terminal is grounded directly, must \$V_C > V_{BE}\$


0

Because the gate source and drain connected so MOSFET is in saturation so we can write:(Be aware that gate current is zero and current of drain is current that pass resistor so we can write I=v/R that I is current of drain and V is Voltage of gate) SO we can obtain V from this equation and then obtain current of drain. (Sorry for my bad English)


3

Use a comparator. Just be sure to get one with an input common-mode range that extends down to the negative rail. TL331, for example, might work for you.


0

Building on @Peter Bennett's answer, the voltage level at the source matters very much. The transistor, along with many other electrical components, could care less about the voltage level of an individual terminal. What matters is the relationship between the different terminals. For example, if the source was at 10 volts, the gate at 15 volts, and the ...


1

If you disconnected the speaker you'd find that the output transistor probably doesn't get warm - the speaker is like a resistor to ground of just a few ohms and this will be taking maybe 500mA thru Q2 without any input signal or sound being present. Like @jippie says in his comment, try adding an electrolytic capacitor in series with the speaker to prevent ...


1

A MOSFET is a metal–oxide–semiconductor field-effect transistor used for switching and ampflification. An LDMOS FET is a laterally diffused metal oxide semiconductor. They are typically used in power amplifiers for radio and microwave frequencies. From the Wikipedia page: Silicon-based LDMOS FETs are widely used in RF power amplifiers for base-stations ...


0

I guess we can actually identify those from the markings, as you can see here for 1ft and here for 3bw. The 2b seems more tricky, but take a shot here.


1

As mentioned already - the keyword is "negative feedback". Feedback can be made efficient for dc and/or ac signals. DC feedback is important for stability of the selected operational point (Q point) against temperature and tolerance influences. In both of the circuits shown we have dc feedback only because all ac feedback signals (above a certain corner ...


2

Simple amplification of audio is the most trivial demand for a BJT and virtually any transistor (including germanium) will do the job you want. Maybe look for BJTs that are referenced as good audio amplifiers but, given that the marketing boys would see some folk being put off buying a particular transistor because it is pigeon-holed by the label. The level ...


1

The difference is in the biasing of transistor and feedback used in both circuits. In the first circuit you are using voltage feedback and in the second circuit you are using current feedback. Self bias circuit , which is the first one, is recommended for its stability


0

Identifying SMD components is usually not possible. Markings, if present, rarely refer to what's inside the package. That is done because there is no room for useless information, and because printing them rises costs. Only "big" resistors, as 1206, are usually marked. That said your best shot is to desolder one of the working one, find out if it's a fet or ...


1

Normally, to amplify an audio signal, you should know what is the excursion signal you want to achieve over the load. A good method is to use the load line. This will let you know the maximum collector current to develop and the maximum voltage that must support the collector-emitter transistor. These are the basic parameters to select the transistor.


1

First, the two schemes presented, showing common emitter configurations. The name of this configuration, because the emitter terminal is shared by the input circuit and the output circuit. The main difference between the two, is polarization. While the first uses a resistor divider to set the base current at rest, the second uses the resistive circuit, ...


1

The transistor only knows about the voltage between its terminals. It doesn't know or care what the voltage is between any of its terminals and what you consider "zero volts" or circuit ground.


0

The substrate (body) is (almost always) connected to the source on discrete MOSFETs. See this diagram.


4

It's common-base. You can recognize it because the base is bypassed with large value capacitor to a supply rail or ground (it's at AC ground). Large is in relation to the operation frequency, which in this case appears to be about 100MHz. C1 keeps the base grounded for RF, R1 provides bias when the input is open (or capacitively coupled). L1 and C2 + C3 ...


0

The formulation is correct. You must be careful with the \$h_{fe}\$ value you use. Search the device data sheet curve \$h_{fe}\$ vs. collector current, to the value of \$h_{fe}\$ in saturation state. EDIT: from the datasheet You must use the lower curve to size the output current and by the ratio $$ \dfrac{I_C}{I_B}=1000 $$ obtain the required value of ...


3

You have two basic problems. First, the top FETs should be P channel, not N channel. That means a gate voltage below the source will turn them on. This is necessary with your drive arrangement since the gate voltages will always be between ground and the supply. This prevents the N channel FETs at top from turning on. Second, you have to consider break ...


2

Lots of problems here. First, to turn the high side transistors on, the gate voltage would have to be higher than 12V. How high depends on your choice of transistor. Second, what FETs are you using? Depending on their gate threshold spec, they may not be turning on all the way anyway. A FET H bridge is best done with an H bridge driver IC. They have the ...


2

You're not turning Q5 on properly. It needs a certain Vgs to turn on fully. Measure Vgs on Q8 which works and arrange for the same Vgs on Q5. (Start by working out the source voltage on Q5 when it is fully on). This usually means arranging for Q5 to be driven from a different signal (generated by a level shifter). Which means a supply voltage greater than ...


1

No, you shouldn't do that. The pin will rise to about 4V and then more than 3mA will flow continuously. This is outside the normal operating mode of the micro. There's effectively a diode from the input to Vdd. It could cause the 3.3V line to rise as well, possibly damaging something.


0

Remember the V(CE) vs I(C) curve and consider an NPN transistor. Thus forward biasing simply meaning there is positive voltage on the p-type junction relative to an n-type junction. Thus, in our case forward biasing the BC junction means having a voltage V(BC) > 0. Now, when collector voltage V(C) drops below the base voltage V(B) and forward biases the ...


0

Electrons have no trouble entering the collector, since even at saturation, there still a small positive bias on the collector. If this bias disappears, then all of the emitter current will flow to the base.


1

Let's clear up some terminology here for the benefit of the reader. A positive photo resist is one in which the exposed areas are removed in subsequent processing steps. i.e. where ever the light shines, the photoresist (PR) is removed. EBL means Electron Beam Lithography. In lift off, you are depositing a layer of material on top of a PR layer, there ...


3

Pretty close. You should give the PNP a lot more base current to make sure it saturates. Use a forced beta of 10-20, which means that the resistor R2 should be more like 470R. The value of R3 is not critical, 10K would be fine, so would 20K. You don't need a Darlington to drive the PNP transistor, a regular NPN will be fine. Use a base resistor of more ...


0

I suspect that the emitter of the transistor is not connected to the Arduino ground. If you see 5 volts between the base and emitter, the transistor is definitely dead! You also need resistors to limit the current through the LED, and to limit the base current into the transistor.


3

That seems like a lot of complexity. If you want to do it as an exercise, or a homework, that is fine. I'd do something simpler. Firstly, P-Channel MOSFETs would be an easy way to drive the common Anodes. The P-MOSFETs will need their gates to pull-up to the chosen LED supply voltage 6.3V to turn off. AFAICT the MSP430G2335 MCU doesn't have 6.3V (or ...


0

Since your problem goes away when the fan is disconnected, it sounds like your problem is due to the inductive spike caused from switching off the fan. If it were a DC circuit, the solution would be a flyback diode, however since this is an AC fan, you need a Snubber &/or a high-voltage clamp such as a bidirectional TVS diode (transient voltage ...


1

The answer I personally use in my circuits: Use a sawtooth voltage (i.e. from simple BJT sawtooth generator) and a ADC with LED driver (LM3914 comes to mind, can drive 10 LEDs, no resistors needed) - you only require 1 IC and LEDs to do the visual part, but you can easily change the behaviour of the LEDs by changing the voltage ramping (sine, tri) and ...


5

I would take a 555 square wave generator output from here 555 Oscillator Tutorial and drive shift register clock input from here The Shift Register. Shift registers input should high all the time.


4

It is possible using 10 bit Johnson counter with 2 Hz clock frequency, but it repeats the cycle after the last LED turns on (10th bit).



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