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7

Unless you are good at iterative math, you'll have to make some simplifying assumptions: The transistor is on, so its \$V_{BE}\$ is about 0.6V. The transistor is in saturation, so that its \$V_{CE}\$ is about 0.2V. Assuming the LED is white, its \$V_F\$ is about 3V. From these, you can calculate the base current ...


0

The dutch ( in Groningen) store okaphone sells the 2N6027 http://www.okaphone.nl/ for 95 euro cents piece. they also have a webstore order.


0

Individual transistors are amplifiers. They may handle very small voltage changes in analog circuits, or in digital circuits they are driven to turn fully on or off, in order to give the voltages that correspond to 0 and 1 (typically 0 is 0V and 1 is +5V). A flip-flop or "bistable" circuit needs two transistors. They are wired to stay in whichever of the ...


1

This question has probably already been answered 100's of times. But here is maybe the 201'st time. Use a circuit that looks like this:


1

You could also use the very common 2N3904 for this application.


3

The 2N2222 should be totally fine. Voltage, hFE, etc. all look equivalent for your purposes. There's nothing critical about the VU Meter design since it's merely switching the LEDs on and off.


0

The ULN2803 is an inverting open collector driver and the noise you are seeing is likely due to high frequency pick-up when the device is open circuiting between pulses. Try lowering (or connecting) a pull-up load to your local Vcc and see how the noise reduces in amplitude. Here's what a single channel looks like: - If you need to drive a digital signal ...


1

Real-world results: A green LED was partially lit by reverse-biased CB leakage current on a 2N3904 when the base was disconnected (or 3-stated during reset). Adding a path to ground shunts the CB leakage current out of the base region, and the LED was now completely dark. Not an issue with an LED, but had it been say a motor, there could be undesirable ...


0

Frequency response is important in any circuit which can be employed to process a signal. In particular, op-amp are often referred to as ideal amplifiers in basic textbooks, i.e. components that can multiply a signal by an arbitrary constant gain. This is just an oversimplification valid for very low frequencies. As any real system, an op-amp will react ...


0

Here are observations about the significance of the frequency response graph in your question. The Frequency scale is probably logarithmic and should show units of log(Hz). The Output scale should probably show units of decibels dB The 0 dB point is low to the right on the curve, and the frequency there is the UNITY gain bandwidth. The high plateau at the ...


4

The differences are numerous, but I'll attempt to summarise. An opamp is a differential amplifier. That is, it amplifies the difference between its two inputs. An ideal opamp has the following characteristics of interest: Infinite voltage gain Infinite input impedance Zero output impedance Real opamps deviate from this, obviously, but generally share ...


3

To answer your questions: Correct: it is ok to tie directly to tbe base pin, given the internal resistor. The reason one size fits all, is the combined gain of the darlington pair is so high, and that the output pins' current ratings mean any current at the output within specs will be sustained by the internal resistor at the base. Yes your calculation ...


9

The ULN2003A is a Darlington array, made up of Darlington pairs. The Darlington pair has higher current gain than an individual bipolar transistor. The side effects include a higher \$V_{be}\$ (in common emitter configuration, it will appear as about two diode drops instead of one) and \$V_{ce}\$ (\$V_{be}\$ of one transistor, plus \$V_{ce}\$ of the other).


-1

In both cases, the output voltage is measured between 'ground' and whichever point has been defined as the output for that circuit configuration. So in a common-emitter circuit (1), the output is defined as being at the collector. But in a common-collector circuit (2), the output is defined as being at the emitter. In the first circuit, R is not the load. ...


0

You're right: in the CE circuit, if he were to measure the output voltage across the resistor, the voltage would increase with increasing input voltage. He is running with the assumption, however, that all signals are ground-referenced: from this perspective, the output voltage must be taken across the transistor instead. However, the output voltages are ...


1

No, you are mistaken in your analysis. When the base is driven harder in a common collector NPN circuit, the transistor passes more current from collector to emitter and this raises the emitter voltage with respect to 0V. In a common emitter circuit, with a higher base drive, the collector voltage falls towards 0V potential.


3

First, you mix voltage and current. Voltage is the "pressure" on electrical charges, so they want to move. Current is the flow of electrical charges, if the find a way. Case 1: To answer the question, have a look into this datasheet of a common BC337 The table on page 2 states a Collector Cutoff Current of 100nA for V_CE=45V and V_BE=0V. So yes, there ...


1

Yes, this will work, but don't forget to use different resistors for different colors, because each color has different voltage. But my opinion is to connect LEDs serially and use higher voltage (but you need to use other LEDs with NO common wire) then you get much better efficiency like example: simulate this circuit – Schematic created using ...


4

The use of an external transistor is to allow for a linear regulator IC to deliver more current than usual. A 1.5A 7805 5V regulator can be made to drive loads of 3A+ using an external transistor like shown. It can also be used to make the external transistor take the heat during dissipation rather than the regulator IC. Assuming full charge on the 4-cell ...


16

The 7805 regulator is setup as an adjustable regulator as adjusted by the 10K pot. The transistor is connected as an emitter follower to increase the load current available to the LED load. Supposedly the 7805 is not able to supply the load current needed for the LEDs and so the transistor acts as a current booster. This circuit has a fundamental problem ...


2

Try this: simulate this circuit – Schematic created using CircuitLab It's a simple voltage-follower circuit. The base of Q1 is held at 5.6V by D1, and the emitter is about 0.65V lower, or very close to 5.0V. It'll deliver almost 2A at 5V, assuming you put a heatsink on Q1 that can handle the 15W of heat produced.


0

The voltage at node Y is not what you say it is: $$V_Y\ne V_{DD}-V_{THM1}-V_{THM2}$$ Rather, $$V_Y=V_{DD}-max\left(V_{THM1},V_{THM2}\right)$$ So, for \$V_{DD}=5\rm{V}\$, \$V_{THM1}=0.6\rm{V}\$, and \$V_{THM2}=0.5\rm{V}\$, then \$V_Y=4.4\rm{V}\$. M1 is in saturation with \$V_{DSM1}=0.6\rm{V}\$, while M2 is in triode with \$V_{DSM2}=0\rm{V}\$. If you ...


0

A small DPDT relay will do you the mechanical part of separating the two halves of the buss, including adding termination resistors to both halves if necessary. What you can't control is the amount of contact bounce. This is not going to be a problem if the buss is idle - the termination resistors generally hold the buss in a specific idle state and ...


6

I'll try to discuss the circuit without going too much into all the details, cutting a few corners here and there to make it easier to understand rather than being 100% accurate. Important to realize is that amplification actually consists of two things: voltage gain current gain (often forgotten) I think the black magic is really in the DC bias of the ...


0

The value of drain current from the source can be controlled by the potential applied to the gate i.e. the electric field b/w gate and source. That is why such transistor is known as F.E.T.


6

None on them are "simple resistors". If they were then you'd find a resistor there instead of a transistor. Every transistor in that circuit is in some way involved in amplifying the signal (although there could be some argument about Q2, since that one is in the feedback loop). No sane circuit designer is going to use a transistor to do the job of a simple ...


3

First question: No, the circuit isn't perfectly symmetrical. The current mirror performs a differential- to single-ended conversion. If you wanted a perfectly symmetrical circuit, you would make M3 and M4 current sources, and then use some sort of common-mode feedback to set the appropriate current (so that \$V_{O1}\$ and \$V_{O2}\$ stay in a usable range). ...


14

The diode is in place to protect the transistor from reverse \$V_{BE}\$ breakdown. If you reverse-bias the base-emitter junction by taking the transistor's emitter to ~\$6\rm{V}\$ more positive than the base, it will breakdown and begin conducting. This reverse breakdown damages the base-emitter junction, causing a degradation in \$h_{FE}\$. By placing a ...


2

For the sake of completeness: If you wanted to provide more current than your MCU can deliver, you can still use two transistors, but this time one NPN and one PNP. NPN in this case works as a non-inverting switch, PNP is inverting. Note: This only works if your Drive voltage (MCU pin) goes all the way from 0 to 5V. Not a problem with the ATMega. ...


2

I would use a relay for this application. The Arduino would operate the relay, probably with the aid of an NPN transistor, and the relay contacts would be connected in parallel with the switch. The relay provides galvanic isolation, and the contacts won't care whether they are switching AC or DC, or what polarity DC, or what voltages the contacts are ...


1

A good way to deal with this is to stop putting the LEDs in to the emitter part of the circuit. Instead put the LED's in the collector part of the circuit in series with the current limit resistors. Another thing to consider is that if the MCU pin has adequate drive capability and a 5V voltage swing you may consider simply connecting up the LEDs as ...


2

Two words - source, sink. The output pin of an arduino is capable of sourcing (supplying an output current) as well as sinking (taking a current into the output and down to ground.) see http://arduino.cc/en/Tutorial/DigitalPins You don't need any transistors to switch the LEDs. In this circuit when the output is LOW (near 0V) the green LED will be ON, ...


1

Looks like you have voltage leaking from the 5V through the second transistor base to the emittor, which is giving you a faint glow (Similar to the problem I faced in this post). As the current from the Arduino is enough to power 1 LED at a time (with a ~300-470Ohm resistor), you could just use the circuit from my post to make this all work. Saves you a ...


8

P-channel MOSFETs tend to have a higher Rds(on), making them less efficient (for the same price). For a small H-bridge, the simplicity of using them makes them practical However, for high power applications, the extra complexity of driving N channel FETs can be justified by improved efficiency from lower Rds(on).


0

There is another major difference between a transistor and an SSR. SSR is an on/off device, current either flows or it doesn't. A transistor CAN be used as a switch, but in many cases, it operates as an amplifier, ie an extremely small current through the base allows a much larger current through the collector (assuming bipolar transistor). There are ...


8

One essential feature of a relay, solid state or not, is that the input and output are isolated. In practise this means optical isolation in the SSR (solid state relay) case. In contrast, ye olde phashioned mechanical klunkety-klunk relays are magnetically isolated. One could conceivably make a solid state relay using magnetic isolation in various forms ...


1

The short answer is no. An SSR is typically optoisolated. This means that the input is, for all practical purposes, an LED. It may have a current-limiting resistor included. The light from the LED then drives the switching element. This may be a MOSFET with an optically active gate, or a triac, or it may be a photosensitive driver circuit which then ...


1

You can use a IGBT for the high side switch. But you need a supply with a voltage higher than 100V to drive your IGBT. I usually do it with a separated and isolated supply just for the gate driver (a small switched supply), there are some on chip solutions for that (Power Convertibles HPR 105 for example). The supply's GND is tied to the IGBT emitter. Then ...


1

Obviously you didn't build this, right? This is a simulation? Connecting the virtual ground of the opamp to GND is a no-no. Your op amp will have no swing. You either need to run both positive and negative supplies and define a ground in the middle or use voltage dividers all over the place to run the circuit off a single supply. If you try to build ...


2

2N363 is the part number for a TO-5 germanium PNP BJT. The closest equivalent you're likely to find in any quantity is NTE102.


0

use 2n2222 transistor.i had made this 8x8x8 led cube see the transistor at left end.use 2 transistor to control one cathode array.if you need more help i can give you full details.


1

There is another way to implement a current sink when you have a regulated voltage available. It often works well enough. You can consider using it since you have 10V available. The base current will be IC/beta. Choose R2 and R3 so that the current through them is 10x the base current or more. It is a good idea to keep Vemitter at around 0.5V or more. ...


0

The circuit you have drawn will output a constant ~600μA. It's sometimes referred to as a "\$V_{BE}\$ over \$R\$" current source, because Q3 and Q4 act in a feedback loop to maintain the voltage across R5 to be a \$V_{BE}\$, which of course appears across R5, and the resulting current flows through Q3. Q4's \$V_{BE}\$ is fairly invariant to the value of I1 ...


1

I think this is more-or-less what you have in mind. If you change RL from 0 ohms to about 2.5K, the current will stay pretty constant at about 3mA. The resistor value R1 sets the current- it's the Vbe of Q1 divided by the desired current. Vbe is typically around 0.6 to 0.7V, so R1 should be 200-233\$\Omega\$, so I used a standard E24 value (5% resistor). ...


0

Bob Pease addressed this question in one of his inimitable "What's All This ..." articles http://electronicdesign.com/analog/whats-all-logarithmic-stuff-anyhow. See particularly figure 7. Basically, you set up two transistors to draw the same current, and match for Vbe. While you don't say exactly what dynamic range you're looking for, I'd guess that ...


3

Let's have a look at the datasheet: At room temperature the maximum leakage is 15nA at 30V, implying a 150mV voltage. It will be less at 3V, but not necessarily proportionally less. The maximum leakage at a Tj of 150°C is 5uA which would be excessive. So, how much of that is because they don't want to make time consuming low-current tests on an ...


4

Interesting question; the data sheet doesn't actually say. Note that this transistor has very high gain, especially at low collector current values — the "C" variation has an \$h_{FE}\$ of over 500! This means that at a collector current of 0.3 µA, it only takes a base current of about 1 nA to saturate it, and this is much smaller than the ...


6

The circuit IS is a comparator and can be remarkably useful more or less as shown. I have used a circuit essentially the same as that in production equipment to meet a requirement which was difficult to meet easily and cheaply by other means. There are several ways to look at the circuit. The circuit does not care which way you look at it - but one or ...


0

The first reason for the bad performance you are experiencing is what other's have already said: You are saturating the transistor. Then other reason is, you are using a very high collector resistor. Read the data sheet of your transistor. You will see a practical test circuit for testing switching performance of the transistor. You will probably see a very ...


3

Germanium alone makes for some pretty poor devices. It was the original material used for transistors primarily because it is easier to work with than Silicon. Mixing Silicon and Germanium in a heterostructure CAN be helpful for high-speed devices because it increases strain in the device. In some cases strain adding in specific ways can speed up ...



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