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2

You say: I use TL082 as comparator with negative supply (-Vcc) connected to ground directly. In the datasheet for the TL082, under APPLICATION HINTS, it clearly states: However, neither of the input voltages should be allowed to exceed the negative supply as this will cause large currents to flow which can result in a destroyed unit. Your ...


1

In most basic terms, you need some kind of amplification to make the circuit oscillate. That's the purpose of the transistor--to provide the amplification. The other requirement for oscillation is some kind of feedback from the output to the input. The actual power to the circuit comes from the battery to the tap on that coil. Then, the topmost ...


1

Basically this circuit transforms the variations of voltage v(t) of the photoresistor due to the light intensity in variations of current i(t) in the oscillator circuit. This transformation is done by the bipolar transistor, it is its function to amplify the v(t) signal. The variations of i(t) change the frequency of the circuit oscillator circuit, it ...


0

Jfet's are becoming rare. You will only need those current source jfets if you want to get to zero volts. (If you can live with 1.25 V as V min, then...) So what you want to do is match the zero gate (Vgs=0) current drain. Here's a list from digikey. Your job is to find out the value for the jfets in your circuit. (I would think that they could be ...


7

Q1 sucks enough of the base current (coming through R1) from Q2 to keep the emitter at around 0.6V higher than ground, because when the transistor is in the active region, that's Vbe. More voltage means the transistor diverts more current from the base of Q2, less voltage and it diverts less. There are limits- if the value of R1 is chosen too high, then Q2 ...


5

Both Q1 and Q2 are BJT transistors. That is indicated entirely by the shape of the symbol. Furthermore, they're NPN type BJT transistors. BJT transistors look like this: MOSFETs look like this: This circuit limits to a specific current only if the value of R2 is correctly chosen to correspond to that current. You want the voltage across R2 to equal ...


2

Put the switch on the output and you won't need the diodes. A CD4053 Triple 2-channel analog multiplexor will do the job (you could use it to switch both the input and output if you wish, though this shouldn't be necessary). Alternatively you could just switch the amplifier gain down to +1, and then it will pass the signal through unamplified.


8

Change your load-driving PNP for a P-channel MOSFET. Also, you want a pull-up resistor on the gate of that MOSFET to ensure it switches itself off when the NPN switches off. The NPN should also really have a pull-down on its base to keep it off when its input is HiZ or disconnected. This is a circuit I use all over the place:


0

You don't need those transistors in your circuit. The DIG outputs of the MAX7219 are designed to sink current directly from the common point of your LEDs. If for some reason you really think you need transistors in there, then use PNPs instead of the NPNs you have.


3

A MOSFET may be a better choice for the following reasons: Today's power MOSFETs are amazing devices with options for extremely low ON resistance and quite high current capacity. The MOSFET uses a voltage control input which can be easier to drive in a high speed manner to reduce switching time loss in the device. There are even devices commonly referred ...


0

I assume you not trying to learn the digital logic concepts and transistor circuits at the same time. Once you have learned each separately, it is most helpful to know that a digital output of '0' or '1' is achieved by two transistors acting in a coordinated manner such as when one is "on" the other is "off". This allows the output to be "driven by" the 5V ...


0

For RGB strip you indeed need 3 pcs transistors or MOSFET's. Most commonly RGB LED strips have "common anode", so they take in 12V from one wire, and 3 other wires need to be connected to ground make each color glow. So connect the power to 12V and take 3 Logic level N-channel MOSFET's (http://www.bgmicro.com/fet.aspx), put the MOSFET's between the R, G and ...


1

There is another clever technique to quickly switch off a transistor by reverse biasing its base emitter junction. This scheme is demonstrated in this high voltage switching circuit whereby the upper transistor is forced off by having the load current pass through a diode when the lower transistor turns on. The load current through the diode applies a ...


6

When transistors saturate there is stored charge in the base that must be removed before it will turn off. One way of doing that is to use a speed-up capacitor, which helps to suck the charge out of the base by increasing the drive current during switching. More information in this on-line source (source of the below image). It's usually better to ...


2

Okay, I see a few things. First, the LM2917 has an internal (nominally 7.56V) Zener diode across the power supply rails. If it's really an LM2917 (not an LM2907) and you've really got it connected that way, more than about 7.5V on the power rail will destroy the chip. Now, on to the design details. You've set the threshold for comparison at 0.5 of the ...


2

Back-EMF. Remember the motor is also a generator. Except when it's stalled. Why does the generated voltage have the same sign? The magnets are the same way round, the motor is rotating the same way, so the generated voltage will have the same sign. Understand one important thing about an electric motor's operation : that back-EMF is always there when the ...


1

It's not an officially recognized term by any standards body, but in the south-eastern United States (and possibly elsewhere) "cut-in" (cut-on) is the colloquial opposite to "cut-off" (cut-out). It is used by many in everyday life in reference to light switches -- "cut-on the lights" = "turn on the lights". Ergo, this is the informal name for the threshold ...


3

First off, your transistor is backwards. Emitter should connect to ground. Switching to an N-channel MOSFET instead of a BJT would make it simpler, since it is a voltage driven device, not current, You will need to bias the transistor so it can switch on when the voltage across the water sensor rises past a certain point. Using a simple resistive divider ...


2

The circuit you show has the transistor inverted which is .. bad. BJTs still work (sort-of) when inverted but the gain will be very low (probably less than 20), and the breakdown voltage will be less than 10V most likely. This should more-or-less work, if the water is actually 200K. R1 represents the water, R2 and R3 are real resistors. I still don't ...


0

RS-485 is good for 100kbps at 1.2km, using two wires for data and two for power. You could probably stretch it to 2km at 1200bps. Here is a TI application note for a 2-wire power + data solution. It couples the data transceivers through large capacitors, and connects the power source and sinks through normal-mode transformer filters. 1000BASE-LX gigabit ...


0

Alas, you are trying to get a DC signal out of an audio jack. i.e., Your audio jack is AC, with an average DC value of zero volts. (You might even find that your audio driver feeds the signal to the audio jack through a capacitor...) So, you need to choose a way to convey your information on an audio signal. This can be accomplished by using some means of ...


0

One bit of advice. You'll probably find that transmitting unmodulated low speed data difficult to get the impedance right to avoid reflections and possible corruptions. Transmitting 100 Mbps is quite achievable on most coax cables over 100m these days without much hassle so please consider upping your data rate to push your transmit spectrum into the low MHz ...


0

Notice how at \$Io \gtrapprox 0\$, Vo drops sharply from Vcc down to 3.5V. This is the output voltage with just enough load to bias Q1+Q2 so \$ V_o = V_{cc} - R_1 \cdot I(R_1) - V_{be}(Q_1) - V_{be}(Q_2)\$ with \$I(R_1) = I_b(Q_1) \approx 0\$. That's a bit less than 5v - 0.7v - 0.7v = 3.6v.


2

Your basic architecture is flawed. The circuit producing the audio output will high pass filter the signal somewhere in the path. Even in "HiFi" audio, 20 Hz can be attenuated 3 dB with gain dropping off arbitrarily below that. The net result is that there is no DC information in the signal. There is therefore no positive and negavite PWM. The DC ...


1

As the part count in this circuit continues to increase, you might consider using a microcontroller instead. There are plenty of options available in whatever your preferred architecture (e.g. PIC, AVR/Arduino, STM8, ARM Cortex-M0). They have built-in ADCs and oscillators, and with an extremely simple program you can quickly get whatever LED behavior you ...


1

The question is confusing, because while it asks explicitly about "clipping" distortion, the rest of the data seems to imply that the limit is a thermal one, which does not produce clipping per se. But let's follow the reasoning regarding the thermal limit. First of all, with a bias level of 0.01% of the peak current (this turns out to be about 1.4 mA), ...


2

This can be easily done with some logic gates (2 IC's): I am using NAND's and AND's instead of OR's because the outputs of the LM3914 are active low. I found from another answer that the outputs are open-drain, so I added 100K resistors to all the inputs. All inputs on the left come from the outputs of the correspondingly named LM3914 outputs. If all ...


2

If the logic is "LED2-10 On, Transistor On" aka "LED1 On, Transistor Off", a simple PNP transistor would work. See Figure 19 for a similar setup. The LM3914 outputs are actually Open-Collector inputs. If On, the internal NPN is on, connecting the output to ground. If Off, the internal NPN is off, leaving the output floating. With a PNP and a weak pull-up, ...


5

If you power LEDs 2 to 10 via a small resistor, you can look at the volt drop across this resistor to see that at least one of those LEDs is activated. You might need to use a comparator. Alternatively you could diode-or outputs LED2 to LED10 to "acquire" the same information. Any one of those outputs going low would be seen on the common point of the ...


0

For a DC operating point solution: 1) Download LTspice, for free, here. 2)Download this file (your LTspice schematic), left-click on it, and then run it by clicking on the running man icon on the menu bar. The simulation is a DC operating point simulation, and you'll see a table displaying all the DC data for the circuit. "X" it out and the schematic ...


1

I need to find it by running a simulation So, enter the circuit into your preferred simulation tool (like CircuitLab, for example), and run an operating point analysis. In CircuitLab you just need to add a 15 V source to your circuit, then go to simulation mode: 1 From here, just press the button that says "Run DC Solver" and wait about 0.05 s. Then ...


0

Your schematic shows a typical common emitter single transistor amplifier. To understand why the input signal is inverted at the output consider the following: The input signal voltage appears across the emitter resistor. As it increases in voltage, it increases the emitter current. This also increases the collector current since, in an active transistor ...


3

This seems to be an academic type of a question. So there are some undefined prerequisites. Are the parameters of the NPN exactly known? If yes, you only have to measure the voltage on Rc or Re. You then know the collector or emitter current and can calculate the base current. As there is no small signal source in your circuit you have already gained the DC ...


2

I assume you use the base as an input. The base potential is set by the voltage divider \$R_{b1}\$ and \$R_{b2}\$. \$V_b=V_t\cdot\frac{R_{b2}}{R_{b1}+R_{b2}}=15\cdot\frac{56k}{166k}\approx5.1V\$. The base-emitter junction is like a forward-biased diode, dropping about 0.7 volts, therefore the emitter potential is \$V_e=V_b-0.7=4.4V\$. Now the emitter bias ...


1

The CD4050 is a digital logic buffer that can be used to shift signal levels from high to low voltage. It is not a voltage regulator - however you don't need one anyway because the EK-LM4F120XL already has an on-board 3.3V regulator. But perhaps what you are really worried about is the signal levels between your 3.3V MCU and 5V peripherals. The ...


1

You can use the flashing bar display circuit shown in the LM3914 datasheet (figure 20). To have multiple flashing LEDs in dot mode you just have to add a diode between the timing capacitor and each output that you want to flash.


0

My answer is likely more than you bargained for but if you’re curious, you’ll appreciate the effort I put into it. A typical OP AMP has an open-loop gain of at least 100,000 (very high). Its output takes the difference of its inputs [V(+) – V(-)] and multiplies them by its gain [Av]. Vo = Av x [V(+)-V(-)]. Here, V(+) = non-inverting input and V(-) = ...


1

I have read several forums regarding this topic but with my components I used different approach and created overall schematic of Arduino that measure the battery voltage and turn it off when it reaches the low voltage limit. A0 measures the battery voltage via voltage divider 1:1 D5 is high and when the battery voltage reaches the lover limit, Arduino pin ...


1

The problem in your design isn't the output transistors and neither will this be solved by using darlingtons - in fact the problem will become worse with darlingtons. Look at the spec of the NE5534 and read the section about maximum output voltage swing. On page 5 it tells you that the max swing might be as low as 24V on a +/-15 volts supply i.e. a 30V ...


1

Common Asian parts for this application are the 8550 (PNP) and 8050 (NPN) available in both TO-92 (various pinouts) and SMT. If you buy the highest gain bin, the hfe will be the range of 160-300 at 100mA (costs slightly more than the low gain bins).


9

In your design, assuming you're using a common small signal NPN BJT (2n2222, BC547, 2n3904…), you're wasting about 20-30mA of current when the switch is closed. That is, about 10-15mA through the upper 220Ω resistor and 10-15 through the lower one. In addition, driving the base of the transistor with 10mA or more in this situation makes no sense: using a ...


4

Short answer, yes. Long answer, yes and use a circuit simulator! Green trace is the switch, blue is the debounced output.


3

So when you press the switch You turn On the transistor and when you release the transistor is cut-off. So why use transistor in first place can't you just connect your switch instead of transistor.? I mean Like this: simulate this circuit – Schematic created using CircuitLab (Don't care about component Values)


1

Here's an overview of the design process to get you started. I'll let you work out the exact calculations. I would replace \$R_{load}\$ with an independent current source \$I_{load}\$ for your simulation (you can use your CircuitLab schematic for simulation once you add resistor values). Set \$I_{load} = 25\$mA since that is your worst case. Pick a ...


1

Let me say this as a answer so we don't get yelled at for too many comments. OK let's first pick R3. The purpose of R3 is just to keep the transistor happy when there is no load. you want to have a little current going through it. A nice number might be 1mA of current so at 5V take R3 to be 5kohm. R3 will then always have 1mA flowing (and we can now ...


1

It might be easier to see this with an example. Let's take a look at the CD4001/CD4011 gates which are NOR or NAND gates respectively. Here is the internal schematic of a single NOR gate and a single NAND gate from the datasheet: You can ignore the four transistors that form the buffer for the output: Now for the NOR gate you can see that there are ...


2

It's actually quite simple, and it's based on two requirements: Every node in the signal path must be connected to either Vcc or GND, to be at logical level 1 or 0 respectively; There should never exist a conductive path between Vcc and GND, else it will short the supply and start draining a lot of current and possibly burning some parts here and there. ...


0

Why you would want to use a darlington transistor like the TIP 3055 on a small 250 mA load, but yes. Use the lowest hfe you see to calculate the base current needed. Any value resistor that would work on the Ice of 1 A will work just as well for 250 mA, putting the transistor in saturation. It will be inefficient on the base side though, so make sure your ...


2

Microchips are made using a very wide variety of process steps. There are basically two main components to each step - masking off areas to operate on, and then performing some operation on those areas. The masking step can be done with several different techniques. The most common is called photolithography. In this process, the wafer is coated with a ...


1

It's a photographic process, similar in some ways to a film camera with separate exposure and development steps. They don't have to print the features in actual size; they can print them in a size they can handle and use lenses to focus that image onto the silicon.



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