# Tag Info

5

You need a PNP transistor ideally controlled from an NPN transistor. simulate this circuit – Schematic created using CircuitLab The reason why a single NPN emitter follower doesn't work is because the emitter output HAS to be lower than the base (3V) to get any current passing from collector to emitter. So the emitter might be 2.5 V when base ...

0

Here is another answer that will work if your microprocessor PWM frequency is much higher than the oscillator frequency and/or if your desired frequency range is small in percentage terms Use the pwm to switch two resisters via a analog CMOS switch like a 4053 or a 4066 which has worked for me the two resistors correspond to Fmax and Fmin your ...

2

Something like this should work well, at least up to a couple kHz. It uses a Fairchild FOD817 optoisolator. simulate this circuit – Schematic created using CircuitLab

3

You cannot use the NPN in this fashion. When the base of Q1 is greater than the collector voltage by 1 diode, you will forward bias the base-collector junction. This is not a normal operating point for the transistor and depending upon the impedance of the 5V node could cause excessive current flow through this path and damage Q1. If all you want to do is ...

3

You could use a DDS which is easy (the hard work has been done for you). Assuming you don't want to do that, read on. For low frequencies (through the audio range) the easiest way is to generate triangle waves with a VCO which can then be shaped into sine waves if necessary. This can be done with a DAC to generate the control voltage and an oscillator made ...

2

Golaž answer is the correct procedure to follow- and you should accept his answer. I'll just comment on the power dissipation factor. For the purposes of steady-state analysis consider the relay to have a fixed resistance of R= 5V/0.09A = 55.6 ohms. The transistor is thus an element in series with 55.6 ohms with a total of 5V across it. Power ...

3

Since you will be using transistor as a switch the rule of thumb for a saturated BJT is to use beta (current gain) of 10 or 20. Meaning you can directly drive a general purpose BJT (with at least 100mA of collector current) from your Arduino pin. This is how I would do it: simulate this circuit – Schematic created using CircuitLab NPN always ...

0

You are doing it wrong. You need to more powerful transistor, but for now yours will work. Connect the motor between collector and + of the battery. Connect 3v through 500-1000 Ohm resistor to the base and emiiter.

0

Bipolar transistors act as current amplifiers. Inject a small current into the Base, and you can extract a proportionally larger current from the Collector. But how much current does your motor need, and how much Base current is required to get it? The behavior of your motor indicates that it is not getting enough current, most likely because you aren't ...

1

First: Without the proper tools you're doomed to fail, so at the very least do yourself a favor and get yourself a digital multimeter and learn how to use it. Second: When you ask a question you should provide at least enough information (Like a schematic diagram of your circuit) to show how you connected all of the circuit elements together. If you ...

3

Short answer: no. Long answer: That power supply plus a transistor can act as a switch to let a bigger current pass through (that came from another power supply). So you're better off using that bigger supply in the first place. You cannot magically increase a power supply's current and/or voltage, unless you open it up and mess with the internals, which ...

2

R2 sets the current through the LEDs. It must soak up the voltage difference between the supply and the LEDs. A typical 5mm super-bright white LED drops about 3.3V at 20mA. A 3.6V AA NiMH battery should deliver between 4.2V and 3.3V depending on state of charge, with an average of about 3.8V. The transistor will drop about 0.1V in saturation, leaving ~0.4V ...

1

The curves appear to correspond to the min-typ-max of the BC639-16ZLT1 (third entry in the table at 150mA). Gain for individual BJTs varies wildly. If you need a better number than 100-ish, someone's going to have to test individual components.

4

As they've not labelled the different curves it looks pretty dubious to me. Normally the graph would apply to typical values so you would not necessarily find them in the tabulated values (this one does not show typical values). I might guess it's showing you either temperature (highest trace would be highest temperature, lowest trace lowest temperature) ...

3

It's a badly written data sheet. Sometimes you must just give up on a transistor if you can't trust the data sheet. For instance, there are three curves in the graph and they are unknown i.e. On-Semi have failed to give relevant information about what those curves mean or represent. The graph next to it has several curves and referred to is a transistor ...

2

I would use a MOSFET and the simplest arrangement is an N channel load switcher: - Ignoring the requirement for switching it on and off at high speed (i.e. kHz upwards), when operating at 200V, certain precautions may be advisable. The first relates to the common earth/0V/ground point. You may want isolation between control signal and the load ground so ...

0

The spikes are coming from C4. Everytime the input switches, C4 passes a short spike to the base of the transistor. C4 and R5 and R11 form a sort of high pass filter that lets the edges of the signal from the sensor through. Remove C4 and the spikes should go away.

0

I think what you are looking for is cascoding two transistors as transistors can be CASCADED and CASCODED. that totally depends on the application here is a cascade example, as the output of the first transistor connected to the base of the second transistor "The cascade of a Common Emitter amplifier stage followed by a Common Collector (emitter-follower) ...

0

This circuit is a current source. The transistor provides an inversion as does the opamp. The net change is positive or no inversion. However, the point of feedback is on the collector of the transistor which means the controlled voltage is at that point. This means whatever voltage is dialed into the potentiometer will be duplicated at the bottom of R3. If ...

2

The FETs can hog current upto 500 mA. For higher load, it is very easy to find FETs with low Vgs threshold. Before pressing the switch SW1, M1 will be off since Vgs is zero. Once switch is pressed, M3 turns on followed by M1 followed by M2. M2 holds the ground to gate of M1 there onwards. When SW2 is pressed, M2 releases the Gate of M1, making the Vgs of M1 ...

3

Imagine you (using a curve tracer) increase the collector voltage (with zero base current) until the collector current is 1mA. That voltage will be a minimum of 45V. Similarly for Vces, collector current of 0.1mA and emitter current of zero. And again, increase the negative voltage on the base until you see 10uA. That won't occur until a minimum of 5V (as ...

5

Usually the columns min/typical/max on datasheet refer to the spread of the mentioned parameter among a manufacturing batch. This means that, for example, Vceo is guaranteed to be at least 45V for any specimen in the batch (at the given conditions). In other words, if you buy 100 such BJTs and determine their Vceo you'll discover that each one has a ...

0

To answer your question: Yes, it is normal for (power) transistors under load to become very hot while operating. Most are rated for temperatures well above 100 deg C. Even 60 deg C is too hot to touch, at least for exposed metal tabs and such. Note that it is only normal for transistors to become hot when operated with substantial current. The transistors ...

5

It's saying that the transistor can be operated up to 45V. This has to be specified as a minimum value because if it was specified as a maximum value then you wouldn't know the lower limit that might cause it to fail. It's a guarantee on performance - if you bought a car that was guaranteed to do 150mph you'd want the MINIMUM guarantee to be 150mph. It ...

1

The Vceo and Vces [and Emitter Base Voltage] are marked down as minimum values ... Yes, and if the voltages go above those then the transistor is likely to be destroyed. Those are therefore the minimum dangerous voltages. If not, how am I to know the minimum current gain if I'm not passing in 100mA? Look at the charts later in the datasheet. One ...

0

To calculate the current in Q2 we need to first determine the bias for Q1. If we make an assumption that the voltage across Q1 is 0.7V. The current through the 180 ohm resistor in Q1 emitter is (20-0.7)/(20000+180) = 0.95mA. The same current will flow through Q2 because of symmetry. The effective emitter resistance of Q2 will be 1000/(40*0.95)+180 = 206 ...

0

In short: B-E junction is forward biased and causes electrons (better: negative charges) to move from the emitter (it "emittes") to the base region (forward biased pn junction) which is very thin. Therefore, the majority of the carriers has enough energy to cross this depletion area and will be "collected" by the collector potential which is larger than ...

1

Very short (and incomplete) explanation: There are many things to understand first but I think the most important thing is to know that currents can be differentiated by their cause: Drift current is caused by a gradient of the electrical field (voltage). Diffusion current is caused by a gradient of concentration of carriers (i.e. electrons or holes). ...

8

The way they are connected they'll act as a bipolar 8~9V "zener". They would be for over-voltage protection, is my guess. Edit: This is because of the reverse breakdown of the emitter-base junction, more at this Wiki reference (thanks to @Leo). The ~7V breakdown of the E-B junction is in series with the forward voltage of the other transistors E-B ...

1

A single transistor will not provide an interface with your FET AND do it without inversion. You can do the job with two transistors, though. simulate this circuit – Schematic created using CircuitLab This will provide a decent drive for your FET with about a 1 usec added to on-time. That is, if you put in a 10 usec pulse, the FET will be on ...

8

If you are using the transistor as a saturated switch then you decide what Ic/Ib should be. The saturated characteristics of the transistor are guaranteed at Ic/Ib = 10. Most often you'll use a bit less drive, maybe Ic/Ib = 20 unless you are very close to the limits of the transistor. hFE is gain in an unsaturated condition (as an amplifier) with a ...

4

In the table, Hfe is specified at a collector-emitter voltage of 2V. In figure 3 the transistor is being forced into heavy saturation i.e. Vce is mainly below 0.1 volts. You usually find that driving a BJT into saturation causes Hfe to dramatically reduce. Bear also in mind that the graph is "typical" and it could be slightly worse.

4

$h_{FE}$ and $\beta$ are two symbols for the same thing. The reason the saturation plot shows $I_C=10I_b$ is because saturation is a behavior where the collector current drops below the level that would be predicted in the forward active region ($I_C < \beta{}I_b$). If the situation was $I_C=\beta{}I_b$, the part wouldn't be in saturation. In ...

0

I am actually working on a project with an RGB LED strip that requires 12V. I am using a 2N2222A (NPN BJT). A MOSFET will not draw current at the GATE, but a BJT will draw a small current through it's base. They are super cheap and super simple. I took this off the internet, but it is a similar setup: Since LED's are current driven, this is a good example ...

0

This won't work as you want. For a start the BS170 is an N channel mosfet and does not have a collector or emitter. Secondly you are attempting to describe it being wired as a source follower - this will not amplify the gate drive voltage at all. In fact it's likely you'll get a smaller gate drive voltage onto the IRFZ44. What you could do is have the BS170 ...

-1

Not an EE myself, I would try some BAC's. "Big Ass Capacitors". That would give you the "boost" you need when starting the fans. A lot of peeps use them with sound systems when needing instant power. In addition, there are no indications of how the coils on these relays are wired. Never mind I got it, had to follow the wires! DUH Here's another thought, ...

2

The circuit cannot work-220K is too high for that transistor due to internal resistors. D2,3 do no good- they should be across coils of K3,5 (reverse biased, or you'll fry the transistors). You could try replacing the transistors with power n-channel MOSFETs or greatly reduce the resistors and increase the capacitors proportionally to compensate. Something ...

1

Although the Pi can supply 16mA (ish - drive strength is a poorly controlled parameter) at 5V, energy is conserved. Since Power P = I*V, where I is current and V is voltage, stepping up the voltage means the current supplied to the LED will be lower than the current from the GPIO - even with 100% converter efficiency, you'll get 4.4mA max from a 3.3v GPIO. ...

2

In general, you need a boost converter on the 5V supply input, to 12V. You can typically see 85% efficiency. You still want to use the transistor on the RPI output, since it is 3.3v and low current. Any typical step up or boost converter will work, considering you keep in mind the minimum regulating load. Alternatively you can find a boost converter ...

2

If you have a 12 volt LED and only a 5 volt supply you need a boost regulator and a means of activating the LED from a gpio pin. So, look for a suitable boost converter (TI or LT) and then choose a BJT that can be activated by a gpio pin via a resistor. Emitter connects to 0 volts and collector to the LED cathode. Anode to 12 volts and MAKE SURE that the LED ...

0

A PMOS can be used as a pull-down device, but it isn't because of its poor performance. In books like Rabaey Digital Integrated Circuits they refer to this phenomena as the PMOS passing a strong 1 but a weak 0. The reason behind this is the regions of operation during pull-up and pull-down. To synthesize: Pull-up The PMOS is mostly in linear region ...

2

If you look at the Figures 10, 11 and 12 you can see a set of graphs for Vce vs collector current at various base currents. Choosing the 4.8mA base current curve as being closest to what you want, you can read the Vce value at around 150mV - 200mV for all three of the transistors at 150mA of current. This is quite typical for a BJT device in saturation. ...

0

This answer assumes that the P channel device is connected like this: - Source connected to the "high" voltage Drain connected to ground/0V Gate used to control the device. It can be used to pull a "high" (rating dependent) voltage to ground/0V but to do so requires that the gate voltage be BOTH taken below 0V AND constrained to be within about 15V of ...

2

The things you are overlooking and how, in a nicely ordered list are as follows: The LED takes 3.3A. 100W at 6V is 16.7A. That's assuming 100% efficiency, which is much more likely to be 80%, which would make 20.8A at 6VDC input. But even in "perfect" conditions you are trying to "melt" your transistor that has a maximum current capability of 5A. ...

1

The TIP122 has a Hfe (current gain) of 1000, so to get 3300mA (3.3A) out of it, will require 3.3mA into the base. The Zener diode (which is drawn backwards by the way) will "remove" 5.6v of the 6.3v supply, leaving 0.7v. Problem is, Vbe(on) in the datasheet can be as much as 2.5v... So correct the zener orientation, and try a value more like 5v or even ...

2

I would recommend using a GPIO expander for this. I have used the Microchip MCP23017 in the past for this very setup. This device is a 16 channel programmable GPIO available in either SPI, or I2C (the part number is slightly different for SPI). You can connect the gpio outputs from this device to N-Channel mosfets with a pull-down resistor to achieve your ...

1

If you don't need a true ground on the load and the load is not remote then the first circuit is preferable (an example would be an integrator where the load is a capacitor). It would not be great if the load was a temperature sensor with 20m leads. The transistor circuit has the load grounded, will work even if the load is below the V- of the op-amp. It ...

1

You need a current limiting resistor between the base and the microcontroller. Otherwise you are driving the micro controller pin to a low impedance input which drives it low, your micro controller pin is currently switching between 0V and 0.7V and wasting a lot of current. I recommend trying a 1k resistor between the micro and the base.

3

Pick a transistor that will be able to easily handle the current (Ic(max) >> relay coil current), and check the SOA (Safe Operating Area) to ensure that it can withstand full coil current at 3*Vcc for a millisecond or so. The voltage rating should be at least 3-5x Vcc. Chances are you can use a good SOT-23 transistor but you might need an SOT-89 or ...

3

What do you think about used SMD instead of THR? I think you should use whatever fits your budget, BOM, capabilities, and requirements. Will transistor be able to dissipate power? Your application is not high-power in the first place, so any TO-220 or TO-263 will more than suffice, and possibly be overkill. MCU, would it deliver ...

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