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0

what is the difference between transistor and zener diode?


1

There is perhaps no need for a flyback diode at all if the ULN2003A output is clamped to a reasonable voltage through the COM line (as it would normally be). When the output voltage exceeds the supply as the magnetic field is collapsing, the current will flow through R2, R4 and the clamp diode in the ULN2003A causing Q1/Q2 to turn off slowly enough that Q1 ...


4

Normally a diode would go directly across the relay. Since one of those points is not availale to you, you have to find another way to limit the voltage across Q1 to safe levels. There are two common solutions for this: A snubber. This is a resistor and capacitor in series between the collector of Q1 and ground. When Q1 turns off, the instantaneous ...


1

Since the Sonifex can be driven by 5 volt logic, you should be able to connect your logic output directly to the input lines, without the transistor. You can add the transistor if you need a logic inversion, and/or if it makes you "feel safer". In any case, your logic ground must be connected to the Sonifex ground. If there is any problem connecting the ...


3

Use an N-channel MOSFET with low \$R_{DS-ON}\$ at \$V_{GS}=5V\$ on each ground connection. simulate this circuit – Schematic created using CircuitLab Also, you will want to buffer the signal through a voltage follower with a very high input impedance since the 1MΩ resistance range will give incorrect results due to it being too high an ...


1

simulate this circuit – Schematic created using CircuitLab The red led has 1.4V forward voltage drop while the green one has 2.2V. When the red one is on, the green one will not have enough voltage drop and will be very dim. simulate this circuit If you want to turn off the green led completely off, you can use a MOSFET switch like this. ...


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When S1 is made, current flows through K1's coil, pulling the armature down and connecting the Common contact to the Normally Open contact. That will allow current to flow through the green LED and its ballast resistor, lighting the LED. When S1 is opened, current will no longer flow through K1's coil, the armature's spring will force the Common contact ...


1

It appears that you are using the B-E junctions of Q1 and Q3 to provide the bias for Q2 and Q4, respectively. This will only work if the four transistors are coupled thermally so that they are all at the same temperature. Also, the value of 39 for RES1 and RES2 is ridiculously low. What are the operating voltages of this amplifier? Overall, this is a very ...


2

There is an old technique that I saw many years ago that looks like it is applicable to your situation. By the looks of things, you are using a linear CMOS inverter as an amplifier. I say "linear" as opposed to Schmitt Trigger. Note that these should be un-buffered inverters such as CD4069UB. The "UB" suffix indicates un-buffered. The issue is that you ...


1

There are a couple of ways I can think of approaching that problem: 1) Is the input signal always the same amplitude and always present? Does it matter if the output of the inverter oscillates at a low frequency with no-signal present? If so a very easy way is to use a Schmidt trigger inverter (e.g. 74HC14) couple the input signal with a capacitor and ...


1

You could try a simple blocking capacitor with a variable resistor to 'add back' a DC level. The capacitor C1 removes the DC level of the signal and passes only the AC (sinusoid) to the wiper of VR1 forming a high pass filter. By varying the wiper position you can set the new DC level between O and +V. Values should be chosen to be appropriate to signal ...


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One possibility is 60hz being coupled in by you acting as an antenna. try using the AC range of your meter and see whats across the LED.


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The circuit you got from the video is incomplete - it doesn't show how to drive the transistors. Your assumption that they can be turned on by simply applying a voltage is wrong. Bipolar transistors are designed to amplify current (a small current flowing from Base to Emitter controls a much larger current flowing from Collector to Emitter). The ...


1

Like the BJT high-side switch shown in your question, this does the same thing using an N-channel and P-channel MOSFET: If your microcontroller can tolerate 5 volts on its output pins, then you could hook up the output directly to the gate of the P-channel MOSFET in an open-drain configuration, and not need the N-channel device. R1 is there to insure the ...


1

As far as the second circuit goes in the original question, using a NPN transistor as a "high-side" switch is generally not a good idea because you would have to bring the base voltage all the way up to Vdd to turn the transistor on, which would generally be a problem, for example, if Vdd is +12v and you are trying to drive it from a microcontroller output ...


1

In my toolbox I don't have a negative voltage rail, so I'm planning on using an op-amp to invert my positive rail into a negative and use it to rail other devices. No, no, no this won't work. An op-amp produces an output voltage that is limited to values within the range of it's own power rails. If you wanted to generate -5V by using an inverting ...


3

You want a P channel MOSFET. You need one that has nicely low Rdson with only 5 V gate drive, but that shouldn't be too hard since your voltage requirements are low. You can probably find one in the small 10s of mΩ range. For example, let's say you find one with 35 mΩ at 5 V gate drive. At 2 A, that will drop only 75 mV, so your output will ...


2

As you've noted $$V_{GS} = V_G - I_D R_S = 3.33 - (3000\Omega)I_D$$ Rearranging: $$I_D = \frac{3.33\text{V} - V_{GS}}{3000\Omega} \tag1$$ Your two equations for \$I_D\$ correspond to the MOSFET in the linear and saturation regions, respectively. Only one of the two applies to this circuit, depending on whether the MOSFET is in the linear or saturation ...


1

The first equation you wrote is correct. $$I_E = I_C+I_B \tag1 $$ But he other two equations are written by neglecting the reverse saturation currents \$I_{CBO}\$ and \$I_{CEO}\$. The original equations are: $$\begin{align} I_C &= \alpha I_E + I_{CBO} \tag{2}\\ I_C &= \beta I_B + I_{CEO}\\ &=\beta I_B + (\beta +1)I_{CBO} \tag{3}\end{align}$$ ...


0

Connect circled 1to an arduino output pin and the ground symbol to arduino GND and it will work fine. Maybe R3 is a little low. 1000 Ohms is more conventional.


2

You must connect the grounds together, since that is the only way to relate the signal coming into the base with the voltage on the emitter.


4

Something like this may work for you (2 x 2 matrix shown). Part numbers and values are placeholders, but you'd normally want the drivers without series resistors to have lower base resistors because they're going to be handling much more current than the other transistors (but for much less time). The drive voltage for the high side drivers (PNP transistors) ...


0

A simple transistor and gate. Both must be driven high to conduct, turning on the led. simulate this circuit – Schematic created using CircuitLab The transistors need to be on either side of the LED because the LED is part of a small LED grid display and one transistor controls the vertical line (which the anode is connected to) and the ...


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Your Va is too low. It should be at least a volt higher than Vcc, and maybe 2 or 3, depending on your base resistor and LED current. This is the classical problem with making a high side driver.


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First - don't power a pump from USB unless it will consume less than 200mA. Keep that in mind for future applications/projects. DC motors will draw very large start-up currents, and this can blow your USB hubs or their fuses, and other nasty things. Of course, a 1.5A USB phone charger plugged into the wall would be fine for most things! Always be careful ...


0

For this situation (not knowing what the load will be) it's better to use a P Channel FET as a "high side" power switch. The PFET should be pulled up to 12V to turn it off, with a resistor, and the 5VDC USB signal can be used with a simple NPN transistor to pull the PFET gate to 0V and thus turn it on. simulate this circuit – Schematic created ...


0

I'd use a relay where the ignition key either ON or in the ACCESSORIES position would energize the relay, and its normally open contacts would be wired in series with the line going from the lighter socket's fuse to the socket.


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With a bipolar transistor, a small current flowing though the base-emitter junction due to a high impedance element (your finger) allows a great current to flow through the collector-emitter pair. With an NPN, the small current through the base needs to come from the +, not the -, so that current can flow from + through the base then the emitter to -. ...


1

Okay, here's how the voltages add up:


0

If you look at the Truth Table for a S-R Latch, found here wiki S-R Latch you will see that the condition you are describing is "Not Allowed". If you wish to have a circuit that toggles the outputs when both input are high you may want to consider a J-K Flip Flop; which is described further down on that same page.


1

it is not actually a real one you need two add (-VEE equal two VCC) to the emitter or use large capacitance in the output and input nodes otherwise pnp transistor has not appropriate bias


0

Re in pic 2 provides a negative feedback against temperature induced current drift. The more current the junction starts to conduct the higher the drop and therefore less bias on the emitter junction. Circuit 4 is also a feedback, don't have pencil at hand but it probably makes the amplification a factor of the resistor ratio rather than the beta which ...


5

As Olin said, circuit shown in #1 and #3 are completely open loop. So the bias stability of the circuit is less and it can even lead to thermal runaway. The bias stability can be improved by including a negative feedback mechanism in these circuits. Circuits #2 and #4 does that. Negative feedback in circuit #2: Assume that the collector current increases. ...


3

The big difference between #1 and #2 is that #1 is completely open loop. Transistor gain varies widely, so it's nearly impossible to come up with values for RB1 and RB2 so that Vce is near the middle of its range. The emitter resistor in #2 provides some feedback so that the operating point is less a function of the transistor gain. The downside is that ...


4

Key principles: An N Channel FET such as a 2N7000 has drain more positive than source and needs its gate driven positively relative to its source to turn on. A P Channel FET such has drain more negative than source and needs its gate driven negatively relative to its source to turn on. You can use 2N7000 as the high side switch ONLY if you are ...


1

You'll want to drive the diode (or diode connected transistor) with a current source. 10uA is a nice value. An easy(accurate) current source can be made with a voltage reference, opamp and resistor. Edit.. I should add that there are a lot of power supply connections and bypass caps missing from the above schematic. (for R1 try two 1 Meg's in ...


2

The easiest is to get a off the shelf temperature sensor IC. These are cheap and plentiful, and do a much better job than you can cobble together on your own. These devices are available with a voltage output you can run into the A/D of your micro, or with a digital interface like IIC or SPI. Trying to make your own just doesn't make sense, especially if ...


1

To test what the input impedance actually is, 1) Put a variable resistor in series with the input to the amplifier, 2) Send in a signal with known peak to peak voltage, 3) Measure the voltage across the variable resistor, 4) Turn the resistor so that the peak to peak voltage is exactly half the peak to peak voltage of the input signal. 5) Measure the ...


3

\$V_{CC}\$ is small signal ground so \$R_3\$ is connected to small signal ground along with \$R_4\$ (and therefore they are in parallel with one another). This resistance is in parallel with the impedance looking into the base of the transistor, which is \$r_{\pi} + (\beta + 1)Z_E\$, where \$Z_E = R_6||C_{19}\$. This total impedance is in series with ...


1

It is perfectly valid to run AC through a BJT between collector and emitter. What happens is the BJT will (should) act as a rectifier, conducting for only half of the period. The B-C junction of a BJT is a P-N junction, hence it can become forward biased if the collector voltage goes lower(NPN)/higher(PNP) than the base voltage. You do not want this to ...


2

An NPN BJT will conduct significantly from collector to base (it is a junction) if the collector is brought below the base voltage by more than a few hundred mV. That could cause problems. Either block the voltage (series diode) or conduct it to ground. In one case the compound switch is 'open' for voltages << 0V, in the other case it conducts. ...


3

The simplest solution applies to frequecies only above the cut-off frequencies of the high-pass effects which are caused by C8 as well as C19. In this case, it is simply Zin=R3||R4||rbe. The dynamic resistance rbe is identical to the small-signal parameter hie (or h11) and we have rbe=hfe/gm=beta/gm (beta: current gain; gm: transconductance at the selected ...


0

There's nothing wrong about the circuit at all. It's a class B push pull amplifier and will work fine if your ears are able to tolerate cross over distortion.


2

This is a very simplified pic of the way that most audio power amp stages work. It is just trying to show that one of the two output transistors delivers current to the load at any time. In practice lots of other stuff would be needed to get this working, it's not a practical circuit as it stands.


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Another side of the "so many transistors" story is that these transistors are not individually designed-in by a human. A modern CPU core has on the order of 0.1 billion transistors, and no human designs every one of those transistors directly. It wouldn't be possible. A 75 year lifetime is only 2.3 billion seconds. So, to make such huge designs feasible, ...


10

It appears most people drive relays through a transistor rather than directly from a digital output pin on an Arduino. I had wired the output pin directly to my relay before realizing this and it worked fine. So far, you have been lucky. What's the transistor for? The GPIO ports tend to have high output impedances. This roughly translates to ...


14

If the relay coil requires a small enough current, then you can drive it directly of the microcontroller (μC) pin. I have done that myself too. Importantly, you still have to have a flyback diode. If a coil requires larger current than an output pin can provide (sink or source), then a transistor (or MOSFET) is used. It can provide more current. If ...


2

Yes, it could be said that it protects the output pin from supplying excessive current. Also some (most) relays require more throw current than an MCU GPIO pin can supply. Did you include a freewheeling diode to protect the relay driver from the backlash when you release the relay? Failing to do so will kill your driver pretty quickly.


1

How much current do you need? There are two easy ways to do this with simple components. One method is good for only a few mA current - enough to feed the input of something active like a VFD. The other method is good for as much current as you want to make it for. I'll show the higher-current option first. simulate this circuit – Schematic ...



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