# Tag Info

0

I'm not sure what your second diagram is trying to show, but it together with all the stuff following (I tuned out after the first diagram) suggests that you are making this complicated. It is good to see that you have thought about this problem and put some effort into solving it, but it's easier to just solve it than to try to figure out exactly what you ...

1

I came across this site, All About Circuits, and as someone with virtually no knowledge of electronics, I found it helpful in explaining why electronics work. I am still at the beginning of Chapter One, but my overall impression is very positive.

5

This isn't practical. Since modern Ethernet has a code, there are transitions even when there are no packets. The LEDs you see on networking equipment are driven by the PHY, not straight off the cable. Even given that, the transitions are way too fast to see it go off an on with tranisions. The light would be apparently always on, and not blinking. If you ...

0

Apart from all those books mentioned one very good approach, is that of Rizzoni, Principles and Applications of Electrical Enginnering. It's a book that start from point zero, develops in a rather simple and not demanding way the methods used in analysis and gives a great variety of solved problems let alone the numerous examples of circuits used in ...

3

Microelectronic Circuits by Sedra and Smith; I've used this textbook along my electronics course in Computer Engineering at "Sapienza" University of Rome, and it is one of the best textbooks I have bought these years. It doesn't talk much about the really basics (resistors, capacitors, inductors), but it extensively talks about everything else, especially ...

0

Electronics: A Systems Approach by Neil Storey, I've found particularly useful while studying. I'm in my second year of Mechatronics, and anything to help me understand analog I've considered gold. I agree with the top two answers Malvino has been hailed by my professors as the definitive textbook for Electrical Engineering, but I felt it made too many ...

12

BUT I need to know why the components are placed where they are placed for the circuit to work. As a consumer electronics technician, I felt the same need. That's why I left the field and enrolled at the School of Electrical Engineering at the Georgia Institute of Technology. You won't find the answers you seek in any single book; learning "why" is ...

5

The problem which you have correctly identified (good for you!) is that the incoming signal has small voltage swings around 0V, but the transistor requires around 0.6V to begin to turn on. The solution is to bias the transistor to a given operating point: providing a base voltage which causes a degree of turn-on that exists when there is no signal. Then the ...

0

You are thinking of the transistor as a digital switch. You think that .2 volts isn't enough to turn on a transistor that needs .6 volts to do so. But a transistor was also meant to operate as an analog device. Such circuits "bias" the transistor by setting the base voltage up into the "turning on" region, and they do this through a high resistance. Then ...

7

If you want to know answers to why-questions in regard to an electronic circuit which appears in any book or schematic that you can find online, stop looking for another book and try this website. But as far as books go, take a look at The Art of Electronics by Horowitz and Hill;and also Electronic Principles by A. Malvino.

0

Looks like Kurt sees it the same as me. AC / audio trigger circuit: - V2 is the audio input. It can be made to run off a smaller supply but I think about 3V3 will be the limit and also it'll have to be a normal LED that drops about 1.8V. Higher supply voltage (no more than 12V) means you can trigger a higher power LED.

0

here i post the working circuit(constructed myself using MSP430G2553 with internal oscillator & used 2XAA EVEREADY cell.when i decreased the voltage of the R8 below 0.8 and again increased to 0.9 or Vcc,controller freezes,then i reset the controller even you can use watchdog timer to avoid this problem) ...

0

Look at my circuit it may be suitable for you: https://www.circuitlab.com/circuit/dsyqc8/uc-switching-by-bjt/ BJTs Q1 and Q2 form an electronic switch. When SW1 is pressed the microcontroller will get +Vcc and in order to keep the uC powered the HOLD uC output should be made high. This works even at very low input voltage.

0

Why does C1 need to be directly connected to the speaker? Don't forget that C1 is is connected to the collector of Q2 too and this is actually critical to the circuit design. The fact the the speaker is connected to that node to is simply to convert the Q2 collector voltage to sound. How does the capacitor discharge? When Q2 is "off", the ...

0

Why bother with a bipolar and relay. How about using a Pch connected from +5V input to the Vdd supply pin (arduino)? Use a switch to drive the gate from 5V (off) to gnd (on). The gate signal can be latched or toggled with a 5V logic device (eg. D type flip flop). Choose the PFET such that its Vt is -1 to -1.5V and an Rds-on (100mohms or less) that doesnt ...

0

There is a lot wrong with your circuit that other people have mentioned. I'm not going to go over those issues, but instead present you with an alternative. When you say this: I'm designing a circuit that will use a combination of switches and resistors to allow a microcontroller to identify which switch was pressed based on the voltage read and ...

3

CircuitLab allows you to simulate this: simulate this circuit – Schematic created using CircuitLab Open the circuit, click on "simulate," and figure out how it works yourself! Note that I had to simulate the inductance of the speaker (important in this circuit) by inserting a series inductor, as CircuitLab didn't include this in the speaker ...

0

There's a huge problem with your circuit: You have the relay in series with the emitter, which means that the relay can only get whatever voltage is on the base, minus the 0.6V or so B-E drop of the transistor. For what you are trying to do, you need to put the relay in the transistor's collector circuit, and use a second transistor driven by the Arduino to ...

3

Yes; best case the hFe of a BJT will be 100 or so, which means your 0.017 mA will turn into 1.7 mA, which is not enough to power the relay coil. There's another problem, too: The current out of an Arduino will not be sufficient to drive the coil of a relay, as the typical spec is 25 mA per pin out, and typical relays use 35-100 mA of current for their ...

1

This circuit appears to be an oscillator. Likely it is designed to produce some form of siren. I think C1 is connected in this way to ensure that the base of Q1 can be stimulated by the oscillations that will occur. Correct. As the voltage across the speaker increases, the capacitor will release more energy through Q1. This in turn causes the current ...

1

MOSFET cascode: - Because Q1 has very little AC signal on its source (due to the gate being decoupled), miller effect on Q2 (drain-to-gate) is greatly reduced. Q1 provides the voltage amplification due to it being injected (via its source) with AC current from Q1's drain. Q1 doesn't suffer from miller effect because its gate is decoupled. Two MOSFETs in ...

3

I'm going to ignore the reference to tetrode, I have never understood why an exact analogy reveals a fundamental truth. The miller effect arises in situations from a connecting capacitance across two nodes that that have an inverting voltage gain/relationship between them. it doesn't have to be in transistors either, but in MOSFET's you have $C_{GD}$. ...

1

A dual-gate MOSFET can be used as a cascode amplifier — see Figure 1 in the Wikipedia article. By holding the "upper" gate at a fixed voltage, the effective drain voltage that the lower gate sees is also fixed, eliminating the effects of capacitive coupling to the lower gate.

3

Output impedance is specified by you as 3kΩ, this sets R4. The ratio R4/R3 defines your gain in this circuit, which you specified as 3. You can calculate R3 now. The parallel circuit $R_1//R_2 = \dfrac{R_1×R_2}{R_1+R_2}$ is specified by you as 15kΩ. The transistor's base will load the voltage divider, but this is easily less then 10% when the transistor's ...

0

It is not clear why the experimenter expects $I_B$ to be constant in the face of changing transistors, including substitutions of Darlingtons for regular NPN's. We can assume that the voltage source is ideal (valid: since these results are from a simulation), then the magnitude of $I_C$ does not disturb the voltage. (Indeed in the circuit on the right, ...

0

In older device technologies such as TTL (which was the dominant technology during much of the integrated-circuit boom), the circuitry that pulls outputs high is considerably weaker than the circuitry that pulls outputs low, so a "high" output would only be pulled up to a fraction of VDD (somewhat more than half). Using a higher supply rail would the ...

0

Your fundamental assumption is incorrect since less than 5 V, sometimes significantly less, is still often a logic high level. You are also apparently assuming that all digital logic works on 5 V, which is far from the case. Even when 0-5 V is used for digital logic signals, there still has to be some margin. If the power supply is exactly 5 V and the ...

0

A Digital 1 or 0 is rarely exactly a single voltage. Not only is there variations at a given level (ex. 5v might actually be 4.8v or 5.2v in a given circuit), but the crossover between 0 and 1 is not exact either. It depends on the technology (ex. CMOS vs TTL) to define the cutoffs. In most cases, a datasheet for your chosen part will tell you what the ...

1

A well behaved (ideal) current source keeps the current constant no matter what its voltage is. There are of course lots of different topologies for making current sources, just like there are lots of ways of making voltage sources. However, what this question is referring to is a special property of BJTs that can be exploited to make simple current ...

0

No, the two circuits are not the same, even after ignoring the dead short of the input in the second circuit. Both circuits will produce about the same output for a valid input voltage. The difference is the load presented to whatever is driving the input. In the top circuit, the transistor is used in what is called the emitter follower configuration. ...

0

An ideal current source has infinite output resistance. A reasonable one has very very high output resistance. What this menas is that it will drive, the designed for current, into the load regardless of what that load is. In a simple model if you have a 1A source and you drive a 1 Ohm load the current source has to be able to maintain that current at 1 ...

0

If I understand the data provided by Danny & Kaz correctly, the strict beta measurement by simple current/voltage measurement is neither possible nor sensible due to changes in operating conditions (thermal emissions etc), reverse current gain and other variables related to specific transistors. Ib itself is varied based on the Vbe/Vbc voltages and ...

-1

Electron mobility is always higher then hole mobility, and since mobility is closely related to conductivity, for the same semiconductor, NPN transistor is always faster, have better current gain and better current capabilities then his PNP counterpart. Now, it has perfect sense to turn NPN with it's current gain, speed and current capabilities into PNP ...

3

You will need a somewhat different circuit configuration to achieve a hold on latch. A couple of transistors interconnected like shown here would do the trick. Note that with this configuration the maximum safe load current cannot exceed the maximum steady state current that the PNP transistor can sustain. For small signal transistors such as the 2N3904 ...

4

What you are trying to create is a thyristor and it is slightly different. simulate this circuit – Schematic created using CircuitLab To sustain a transistor on you need current through the gate, in this design the two transistors are feeding output through each other.

0

After you release the button, there is no path for current to enter the base of the top transistor. For the transistors to stay on, the base of the top transistor must be about 1.2V above Vout but your resistor will pull the base of the top transistor down to Vout.

0

Here's a link to a previous post that I answered and this was the circuit that I used many moons ago for an audio detector that was quite low quiescent current and turned a battery amplifier on when a signal was detected: - Although the previous question mentions a microphone it is a line-audio ac detector circuit and also ran from a 9V battery.

2

Here is an approach using a comparator and minimum additional components, that would serve the purpose. The key active part is an open collector comparator such as the Linear Technologies LT1011. Substitute this with any open collector / open drain comparator that operates with a single supply ranging beyond 9 Volts. simulate this circuit – ...

0

Here is a question: you say that the 5V power supply for the motor is not connected to the Arudino at all. In this case, I have my doubts that the circuit will work, unless it happens to be ground coupled somewhere else (for example, if your Arduino is powered by USB, which is coupled to the AC ground, and the motor power supply is coupled to the same ground ...

0

I wouldn't even use a NPN tranistor to begin with due to heat dissipation (Pd). Pd can be calculated using: NPN/PNP -> Pd = Vf*I = 0.7V*I As you can see here, if you're drawing 1A, then that transistor will heat up like there's no tomorrow. (0.7V)*(1A) = 700mW. MOSFET -> Pd = I^2*Ron = small number As you can see here, even at 1A, there's essential ...

1

There is an easy schematic for a class A audio amplifier with a darlington pair. A darlington pair is just two transistors that square the gain of the first transistor(if you are using identical transistors). So if the first transistor's gain was 20, then the gain of the darlington pair would be 400. This website uses a darlington pair in the form of one ...

3

BC337 is a silicon transistor. Silicon PN junction starts conducting between 0.6V and 0.7 volts applied to it with P terminal being more positive than N. In your circuit, when current in 2.7ohm resistor reaches 250ma, it causes 0.675 volts across it turning on BE junction of transistor T2.

1

Differential gain: If the base of Q1 moves down by $-\Delta V_{be}$, and the base of Q2 moves up by $\Delta V_{be}$, then the junction of $R_1$ and the two emitter resistors $R_e$ will remain fixed. Since no signal current flows through $R_1$, the signal current through Q2 will be simply \\$\frac{\Delta V_{be}\ - (-\Delta V_{be})}{2R_e}\ = ...

0

If you really only need 7 segments (and no decimal point or such), then you could have each PCF8574AN drive a pair of 7 segment displays, one common cathode, and one common anode (some manufacturers build both variants). You could then use e.g. pin 7 as the cathode/anode pin: If it’s 0, an 1 in the other bits will light the corresponding segment of the ...

0

You did not mention a few important factors: what is the current (and less important, the voltage) you want to switch, and is your load resistive or inductive? what is the gate voltage your FET needs to switch fully on, and does that match the voltage your 555 runs on? at what frequency do you want to switch A small current (let's say 100 mA) can often ...

3

Use a Comparator "Digital" NOT gates (inverters) are comparatively weak, since they are not designed to drive big loads. In the more general case, you want a circuit topology known as a comparator. A comparator compares the input voltage to a reference and outputs one of it's maximum extents (as close to its positive or negative supply as it can get) if ...

0

The diagrams mentioned by other answers on this page where a resistor is used can distort your signal significantly. The resistor will slow down the charging rate of the FET gates which will smooth out the leading edge of your inverted signal. Using an actual inverter or 555 inverter "squares up" the wave to keep nice sharp/hard edges. A solution worth ...

2

"logic chips are always hard ON or OFF" - This isn't exactly true. They do a good job forcing inputs to either logic 0 or 1, but if you have a slow, sloppy input transition, than you'll get a somewhat slow output transition as well. The transfer of an inverter looks kind of like an S curve: I doubt that this will give your downstream circuit any problems, ...

7

You're charging the capacitor directly from the battery. So the charging time is related to the product RC, where R is just the internal resistance of the battery. Try something like this: simulate this circuit – Schematic created using CircuitLab Here, I have split the base resistance so that the capacitor is charged through a large portion ...

1

If you use a MOSFET and place the resistor from gate to ground The MOSFET gate draws NO current at all (that you can detect) Voltage decay time constant is now entirely RC based. Turnoff occurs when Vcap falls to close to MOSFET Vgs_threshold. (More useful stuff to learn :-) ). Be sure that MOSFET Vgs_max is > 12V. Many are about 20V. Some are lower. ...

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