Tag Info

New answers tagged

0

You're best bet is to use the following: Connect your bulb to your 12 volt source in series with your resistor, which is in series with your transistor (Collector side on an NPN BJT). Then connect an arduino pin, in series with a resistor, to the base junction. Finally, connect your Emitter to ground. The problem with your calculation is that it assumes ...


6

The input is digital at 3.3V and should output 5V TTL levels. For this situation you very likely do not need any conversion circuit at all. Both 3.3 V and 5 V TTL logic switch with a threshold of about 0.8 V. Therefore no conversion circuit is needed to drive a 5 V TTL input with a 3.3 V logic signal. To be absolutely sure, check the minimum Vih of ...


11

This circuit tends to be used quite often: -


13

Absolutely. This circuit takes advantages of the properties of a MOSFET to bidirectionally switch a signal between two different voltage levels.


0

Max collector voltage swing will be twice the smallest difference between Vc bias point and nearest rail, i.e. V+ & 0V. More than this swing will appear to clip.


0

I assume you are talking about a common emitter circuit, with perhaps a feedback resistor in the emitter line. To calculate the maximum allowable voltage swing, you must consider three situations: The situation without signal, where the base is only fed the bias current. The situation where the transistor is cut off, i.e. no current flows through the ...


0

The circuit does not work because you have the bases of the MJ15003's strapped to +5V. Also, the TIP50 can not supply enough current. Also, you confuse the device's MAXIMUM ratings with the dynamic characteristics. MAXIMUM rates tell you when the device will release the magic smoke. You do not specify the maximum current you want to draw, I am assuming the ...


0

1 - As shown, with no load each MJ15003 will get a continuous base drive of ~ 40 amps. That's assuming they don't blow out and you have a really hefty 5v power supply. Assume 1 volt Vbe at these current levels. Then the base current is (5 - 1) / .1 . Assuming a 3.5 ohm load, to give you 5 amps load current per transistor, your base current will drop to 35 ...


1

Generally, yes. You can use a Triac with a heavier rating in an application that was intended for a lower rating. Devices with heavier ratings are generally more expensive, that is why devices with lower ratings exist. There may be some complications: Heavier devices may have larger parasitic capacitances. Heavier devices may have a different resistance. ...


1

\$V_{gs}\$ defines the thickness of the channel under the Gate (in a 3D mosfet model). Think of this channel as a hallway whose width you can increase as you increase \$V_{gs}\$. Think of people walking through the hallway as electrons ( \$I_{d}\$). Think of yourself as a god who can force people to have to run through this hallway (force = applied ...


0

There are many 'High-side switch' IC's available for this purpose. Check the application notes for details on how to use them. But as mentioned by smashtastic, check that the vehicle is negative grounded. The wetting current can be provided by bridging the two contacts with a capacitor (from http://en.wikipedia.org/wiki/Wetting_current).


1

Whether the 2N2222 will be able to do the job depends on the current drawn by the relay coils. I would guess it would be no problem. There is no fundamental benefit of BJT's over MOSFET's (or vice-versa) in this application: you just need to switch the current; any transistor capable of handling the current & voltage needed to energize the coil will do. ...


0

The instant that transistor, TR1 switches "ON", plate "A" of the capacitor immediately falls to 0.6 volts. In an NPN BJT, the collector (c) plate will normally never drop below the base (b) plate. This is because a BJT consists of two diodes, one from the base to the emitter, the other from the base to the collector. The BJT diagram actually shows the ...


0

Looking at the circuit, I would think that by swapping the + and - inputs of the opamp you get a current source. Haven't tried the circuit though! To analyse stability, use a thought experiment: what happens when I raise the voltage on the plus input. As a result, the output of the opamp will be raised as well. If this raise leads to a raise on the - input, ...


3

The voltage that must be greater than \$V_{TN}\$ is the voltage from the gate to the source, \$V_{GS}\$, not just the voltage at the gate. Once the source voltage rises to \$V_{DD} -V_{TN}\$ the voltage from the gate to the source will just equal \$V_{TN}\$. Any increase in the source voltage will cause \$V_{GS}\$ to be less than \$V_{TN}\$ so current flow ...


1

Yes, connecting the body to the source eliminates the body effect. However, this may or may not be possible in your intended fab process. If you want to connect an NMOS source to its body, and you expect the bodies of different NMOS transistors to be at different voltages, you need to have a process with P-Wells. If instead the process uses a P-type wafer ...


1

A triac (and snubber circuit, if present) will leak a certain amount of current, so it's not realistic to measure the output without a load. The best way to test the output is with the motor connected as a load. If the voltage measured across the motor is not more than zero, measured on the AC VOLTS range of any multimeter, it's not going to turn. It's ...


0

If the base is kept fixed and the emitter voltage is reduced, the transistor sees a higher voltage between the base and emitter and it is turned ON harder. If the voltage on the emitter increases, the transistor turns OFF as the difference between the two is reduced. This is exactly what happens in this circuit. The 5pF capacitor between the collector and ...


0

For a circuit to oscillate it is necessary to have a LOOP GAIN of unity (Barkhausen criterion). Hence, the amount of output voltage that is fed back to the input of the amplifying unit depends on the gain (magnitude and phase!). In the present case, only a part of the output voltage is created across the impedance seen at the emitter node because of the ...


3

It's not a short circuit because the resistors restrict the current flow. You can't use just one pole of the battery, because there would not be a "circuit" if the current couldn't flow around. It isn't a transistor, that is a programmable unijunction transistor, and it has an anode, a cathode, and a gate (no base). By the way, contrary to what one ...


2

You should be able to run the relay from either a BJT or a MOSFET. For this application the important thing to remember are; the voltage / current requirement of the relay coil, the operating voltage of your circuit and the voltage of the control signal. When choosing components with so many parameters i usually use a suppliers parametric search feature, ...


1

simulate this circuit – Schematic created using CircuitLab The IRFZ44N would likely work, but it may be slower in switching and definitely overkill for your project. The 2n2222 transistor should work fine. I would take 200mA divided by the minimum gain of the transistor (hfe=35) and you'd get the current that you need to send into the base of ...


5

You will notice the different hFE values are for different conditions of operation. Ask yourself which of those is closest to the way you want the transistor to work when on. Looks like you want Vce to be very low (less wasted power) at Ic=200mA - so the closest match would be the value for Vce=1V, Ic=150mA which gives hFE>= 50 and therefore Ib=4ma so you ...


0

Can you cut out the 1st stage and drive R3 with R7 pullup and use C1 positive feedback from output R6 to base of Q2? Or do you need inverted output too?


0

You're using an RC delay so the effect of the supply voltage is hard to mitigate. You could charge the capacitor to the value of a reference (zener diode is a cheap and not very good ref, but you get the idea) rather than to the rail.


3

Treat this as an energy/power problem. Capacitor C1 holds a certain amount of energy when fully charged: 100nF @ 36V (worst-case) is 65 µJ. This energy will be dissipated in Q1 whenever you discharge C1. Clearly, any single operation will not endanger Q1. The question then becomes how frequently can you do this before the power dissipation (energy ...


1

To protect Q1 from surge currents, we identify that the capacitor is potentially the problem. At a maximum there can be 36 volts across the capacitor. BC556 can handle 200mA maximum current. 36/0.2=180 ohms. If you place a 180 ohm resistor in series with C1, you will guarantee protection from C1 to Q1. RC time constant would be 180*100nF = 18usec. That's ...


4

With a distinct lack of detail, I shall prescribe this component: http://www.digikey.com.au/product-detail/en/MMBT123S-7-F/MMBT123S-FDICT-ND/815722 This is a surface mount NPN transistor. It has a beta value of 140-150, at 400ma load. Your 3ma max will drive 400ma through it. It will dissipate ~100mW through it during this time, but it is rated to 300mW. ...


0

It's possible that it's just a simulation singularity/bug. I have done a lot of SPICE modelling and simulations and they are quite frustrating indeed. What happens if you put a simple 10uF capacitor from VDev to ground? the original poster followed this, and fixed their issue.


1

I think that N-channel MOSFET is a good choice here but you have to take care for Threshold voltage (Vgs). This is minimum gate-source voltage necessary to open transistor. In your case you may search for MOSFETs with Vgs=2.5V or lower.


1

You are right that you need a transistor to switch lots of current from a GPIO pin. It is often better to put the transistor at the negative side of your load. It means that the GPIO pin does not have to see the 3.7 volts, or have to go that high to turn your load on and off. It looks like your research was around BJTs, (bipolar junciton transistors, ...


0

For just enabling and disabling you're best off with an N-FET type transistor. It is most commonly used between GND and the load. Connections go like so: Load +: supply + rail Load -: FET Drain ATMega output pin: FET Gate FET Drain: supply GND In essence the N type FET is a voltage controlled switch, where voltage between Gate and Source controls the ...


3

Try this: - A GPIO pin from your MCU going high activates the BC547 and this draws current through the 1 kohm resistor and applies about 15V on the P channel FET's gate-source. This turns on the FET. When the GPIO pin goes low the BJT turns off and the P channel FET's gate-source voltage discharges in a few microseconds to zero thus turning it off. Pick ...


2

The current through a resistor is proportional to the voltage across it: $$I = \frac{V}{R}$$ This can also be written in terms of the change of current and the change of voltage: $$\Delta I = \frac{\Delta V}{R}$$ If the collecter end of R1 was connected directly to a fixed voltage source VC, the voltage across it would be Vin – VC, and any changes ...


0

It takes a year to a year and a half of college-level classes to lay down the prerequisite groundwork: calculus through differential equations, chemistry, physics, basic electric circuit analysis. It takes about half a semester of semiconductor device theory to understand the processes in bulk silicon and PN junctions, leading up to the bipolar transistor. ...


1

Here I don't get why VC > VB is necessary Actually, it is not : If the transistor is saturated, you can have (for example) Vbe=0.6V and Vce =0.3V. When NOT saturated, you will have Vc>Vb. It looks to be if VC is bigger than VB, then the current will flow thorough the base. No, it won't. In NPN transistors, you have a diode junction between base ...


1

You could get an inverted vocal signal by rotating a microphone 180 degrees - does this sound bad? Given all the reflections going on when you record live music AND all the reflections and acoustic reverb you get when you listen to music, it should be no surprise that doing a single electrical inversion has no audible effect whatsoever.


-2

No, it would NOT sound fine. A single BJT common-emitter amplifier only provides an active sink or source for one quadrant. You need a two-quadrant driver for a speaker; one that can sink and source current to the speaker. And inverting the signal isn't important if all you are doing is driving a speaker.


3

In general, the absolute phase of audio signals doesn't matter. But if it's a problem, just reverse the leads to the speaker!


0

As Andy mentioned it is the gate capacitance that affects the simulation. For simulation purposes you could add a resistor of 1k between the gate of M2 and R3. Or a cleaner approach, modifiy your mAEn to look something like this: PULSE(0 3.3 10m 1m). (Initial 0V, Final 3.3V, rise at 10m seconds, risetime 1m second)


2

Due to parasitic capacitance within the MOSFET (from gate to drain), when you drive the gate low (in order to turn it on), this transient is coupled through to the output and you see a short spike as per your simulation. I'd say this should be expected. If you want to get rid of (or reduce) this effect try slowing down the activation signal mAEn or uAEn. ...


3

From the first 2N2222A datasheet on Google, in the TO-18 mechanical data: Note the lack of an isolation ring around pin 2 (collector).


0

As I understand it, the intrisic gain is the gain of the amplifier itself (or sometimes called the open loop gain, meaning there is no feedback loop). And the overall gain is the gain of the whole circuit with the loop (called closed loop gain)


2

R19 is picked to set the base current of the transistor when the output of the SIM340C is "high". The design equation would be: R19 = \$V_{OUT} - V_{BE} - V_{F} \over I_B + (V_{BE}/R20)\$ Where \$V_F\$ is the forward voltage of the 1N4148 at the design output current and \$V_{BE}\$ is the base-emitter voltage of the transistor at the design base current, ...


3

R20 can have two purposes. Without analysis of the full circuit and its purpose I can't say whether either is applicable here. without R20 ALL current through R19 will flow through the transistors BE and will thus be amplified. Especially when the pull-up resistor has a high value (in this case it is only an interal weak pull-up??) this can cause a ...


3

Given that R19 is so much smaller than R20 I'd say the main purpose of R19 is to limit the current into the base of the transistor. This of course needs to be done in all grounded-emitter configurations. The forward volt-drop of B-E is about 0.7V and it's quite easy to destroy a BJT without this current limiter. R20 is less obvious - some engineers will ...


5

All BJTs have what is known as "miller capacitance". It can be "altered" by a greater or lesser reverse bias voltage across the collector-base region. Note that C1 shunts this small capacitance to ground and that C3 returns this capacitance to the positive rail thus ensuring miller capacitance is effectively in parallel with the inductor. This change in ...


0

Use a 8 A 12 V or 16 A 6 V relay. The coil resistance will provide the wetting current required for the switch. The relay will switch the load to your lamps. Typically this will be high side switching and the low side from the lamps to the car's chassis for the 0 V return. Unless your car is positive ground. Then it will be low side switching on the 0 ...


0

I'm going to go out on a limb and assume that when you plugged in the base of the transistor to the 1.5V power source, you actually plugged it in across the motor. This would explain why the behavior was as you described. Assuming that your implementation of the circuit was correct, and the transistor isn't backwards (it gets me all the time; it's worth ...


1

In addition to Ignacio's answer, I would like to recommand this web-site to learn "How to use a transistor as a switch": Using Transistor as a Switch myself learnt that how to use a transistor as a switch by this web-page.



Top 50 recent answers are included