Tag Info

New answers tagged

0

It's a rocker, not a toggle, but this switch has a reset coil. It has an 11\$\Omega\$ 5V coil that requires that you limit the on time to 50~100ms maximum on and at least 5 seconds off (2% duty cycle).


0

I've seen something that does work a bit like that, but not exactly. It's a no-volt dropout or zero-volt dropout switch. It's like a normal switch (a rocker switch in the type that I've seen) in that it has two positions, on and off, and will stay put in either of them. But, it only stays in the ON position if there's a power supply (from the mains). If the ...


0

You have the solenoid wired in series with the LED, so they are both drawing the same current - which is enough to light the LED but not to operate the solenoid. You need to connect the solenoid in parallel with the LED and 3.9k resistor, ie. directly from battery positive to the transistor's collector. Also you need to increase the base current to keep the ...


5

No, this transistor cannot be expected to do this job for long, if it can do it at all. From the datasheet, look at the "On characteristics" on page 2. First, its free air power rating is 0.625W, which means Vce had better be 1.25V or less at 500 ma. Then, gain (hFE) is shown at different Vce voltages and currents. But significantly, not shown at Vce=1V ...


2

To select a switching transistor, you'll need to know how much current will flow through your 12V LED. Choose an NPN transistor that can pass that current, allowing a good margin of safety. You're looking for the Ic(max) parameter in a data sheet.


2

If you follow BJT modelling, then you can conclude these facts Emitter is doped highest but has considerable bigger area than base and lesser than collector Base is doped considerable more than collector but less than emitter and is having least area Collector is doped least and has maximum area among all three. Now this will lead to Band diagram as ...


1

Why are you not driving the motors directly by L293D? Its a reliable working H-Bridge and also the it can make your circuit less troublesome.. There are other ICs avaliable for H-Bridge based on persons requirements..


0

What is guaranteed about the strength of the x3 input, and what are the output drive requirements? The most straightforward implementation using transmission gates would use two transistors for the inverter, two for inverters on x1 and x2, and four for each mux, for a total of 18. It would, however, pass through the x3 input directly without buffering in ...


0

The above problem can be solved by using TG We have 3 inputs now for sake of less number of MOS to be used lets get \$x_1 x_2 x_3\$ in true logic as well as inverted logic So MOS needed for 3 inverters is \$ 3 \times 2 = 6\$ now all we need are 3 TG to be implemented NOTE: since we already have inverted logic form of \$x_1 x_2 x_3\$ we don't need ...


0

Usually to operate BJT in active region Base-Emitter Junction must Forward Bias and Base-Collector Junction must be Reverse Bias In above circuit if you assume it as PNP BJT. then Base is at lesser voltage level compared to Emitter and higher compared to collector


1

I eventually found the solution to this problem, and I forgot to update this post with the fix I needed. First it might be worth me explaining how this circuit actually works: The capacitors charge in parallel through the respective resistors and once they are fully charged (or near enough) a voltage can be pulsed to the base of Q2 (as well as the ...


2

I don't see anything dynamic here. No positive feedback, no bistable circuit. There is no AC input (except maybe ripple at R1, where it does no good). There is no transistor switching. The transistors Q1 and Q2 are assembled as a two stage amplifier, rather than as a bistable oscillator - why? Perhaps R6 should be tied only to R7 and C2... There are no ...


2

The problem here is that you're misunderstanding (or not having the basic understanding) of how MOSFETs work, in particular the \$V_{GS}\$. On an N-channel MOSFET the gate voltage has to rise a certain amount above the source pin. On a P-channel it has to fall a certain amount below the source pin. In your schematic the N-channel's source is connected to ...


1

I've solved problems like that in the past with a Sziklai pair. It's like a Darlington, but with two complementary transistors. In your case, use a PNP transistor with its emitter connected to the alarm chip's VCC, and the base, through a resistor, to the open drain alarm output. The collector goes to the base of an NPN transistor (again, through a bias ...


1

You could devise a scheme where you put the burden of bias for the switching circuit onto the 9V supply instead of having to introduce the coin cell. The circuit design idea below would support switching your 9V supply to a load. As shown the load could be up to ~400mA. The bias load on the 9V supply when the alarm output of the RTC is off is about 2mA with ...


0

No. OR requires 6 transistors. NOR can be implemented with 4. You can't put NMOS on top in a simple digital circuit because there is no voltage available to turn it on. You can put NMOS on top on a linear analog circuit, but you will not be able to drive to the upper rail, unless there is some higher voltage available to drive the gate. If you need to ...


0

What you have to consider is the gate-source voltage needed to turn on the FET. For example, when an input is high, you want the corresponding N-FET to pull the output to VDD. But if you factor in the gate-source on-voltage requirement, it does not work as needed.


1

Take NMOS as example, it has p-type body, and n-type drain and source. If you apply a negative voltage on the gate, holes are drawn to the semiconductor-insulator interface. A conducting surface extends from the bulk all the way to the interface. But because both source and drain are n-type, in accumulation mode, it's like a npn transistor (source and ...


6

BJTs don't work how you apparently think they work. Remember that the B-E junction looks like a diode to the outside circuit. You are seeing a 650 mV drop (5.00 V - 4.35 V) from B to E, which is squarely in the expected range. You are using the transistor in emitter follower configuration. You will therefore get significant current gain, but no voltage ...


1

The photo-transistor collector is connected via 330R to the local supply. Try shorting the 330R out and decouple the transistor's collector with 100nF to ground (bottom of R3) to reduce miller capacitance slowing the response of the transistor down.


1

Very simple problem with an equally simple fix. Your entire circuit is floating. You need to use this for GND: Edit: Node 0 is the internal reference in SPICE for all voltages, and the ground symbol with the graphic '0' is connected to it. Every node in your circuit must have a DC path to node 0 (even if that path is a 100G ohm resistor). The ground ...


0

As a rule, it's common for Chinese, Taiwanese and even original Japanese transistors to drop the "2S" (or similar) prefix from the marking on the case. For example, I have a Chinese-made power supply that has (on its primary side) transistors marked with "C4161". These are clones of Sanyo 2SC4161, no doubt. As for your "C8550", it's hard to tell who made ...


2

The arduino interface won't work as-is, but the rest looks good. When the switch is on, the arduino's transistor will probably not like the emitter and collector both connected to +24v while the base is at +5v or whatever the arduino gives you. That's -19v, which is probably outside of the spec. When the switch is off, the arduino will only be able to ...


0

You did not provide a datasheet link or a part number for the opto, but I can make some educated guesses. First, the 470% CTR is a suspiciously high number as a guaranteed minimum at 5mA LED current for a phototransistor output isolator. If it's actually a Darlington output, then you'll need to modify the circuit to get to work reliably as the saturation ...


0

My guess is that you have the Base and Emitter swapped over. Here's an example that is similar to yours. Note how inductor L3 is connected (through R5) to the Emitter, and the erase head is connected from ground (through R3) to the Base. The circuit is based on a Common Collector Colpitts oscillator. The erase head in combination with C6 and C7 form a ...


2

Assuming your schematic is really how you've hooked it up, I'd hazard a guess that all your problems (except the pot burning up) are caused by the fact that you have the power and ground reversed. +5 goes to pin 8 and ground to pin 4, not the other way around. Once you fix this, the question of offsets may or may not become relevant, but replacing Rsense ...


0

The three main factors that determined the maximum switching freq are 1) on/off delay. If a MOSFET took say... 1second before it even started switching you would immediately have a factor with regards to switching freq 2) rise/fall time. Similar to #1 3) thermal The higher the switching freq the more losses and at some point you will not be able to ...


1

The first thing that I notice from the datasheet is that the opamp has 3 mV input offset voltage, which is clearly biasing your circuit. Try to use larger signals, since you have enough output voltage headroom. (100 mA * 22 Ω = 2.2 V). I'd start by using something in the order of 0.1 V as input signal, then maybe reduce it until you get too much error due ...


0

This might be overly simplistic, but I think that the main factor usually is that a MOSFET is generating a lot of heat during the transition, as there's a voltage drop across it when it's not fully conducting or in cut-off. For example if you have PWM frequency of 1 KHz (2 transitions per millisecond) and there's a 100 microsecond lag in the MOSFET, you ...


0

About the bipolar transistor symbol. First transistors were made with needles tucked into a germanium crystal, and were derived from point contact germanium diodes. Galena diodes (for radio) used also nails set over the crystal surface, not soldered connections. The symbol is litterally that. A specific junction transistor symbol once existed, but the ...


2

In short, the arrows show the current direction of a PN junction when forward biased. In BJTs the PN junction is the base-emitter one. In JFETs it is the one between gate and channel. In MOSFETs it is the one which exists between the channel and the substrate (the terminal where the arrow is placed in the symbols you posted), which is not available ...


0

Two suggestions.... Place a small valued capacitor from one of the transistor bases to GND. This will imbalance the circuit so that it will start to oscillate at startup. Simulations are often too prefect in behavior and the two sides of the vibrator will come up balanced. In a real world this would be much less likely becasue the two transistors, two ...


8

For BJT's there is a PN junction between the base and emitter. The arrow indicates the order of the junction (base to emitter or emitter to base). An NPN has stacked N, P, and N doped channels. The PN junction (between base and emitter) goes from the center out. PNP likewise is the opposite. Observations, not necessarily fact: In a MOSFET, the body is ...


1

The arrows in the bipolar transistors (emitter) show the direction of flow of conventional current. That is, current flow from positive to negative. Diode and LED symbols are the same. EDIT: clarified, the emitter arrows.


2

The analog-focused answers are correct, but there's also a digital way to look at it. The circuit you linked to is basically two inverters connected in a loop, not unlike an SRAM cell. Here's a simplified version to make this more clear: simulate this circuit – Schematic created using CircuitLab Homemade Circuit Designs added some resistors for ...


1

In order for a circuit to behave as a DC-stable latch, it is necessary that there be a feedback loop with a positive gain at DC which is greater than one. While bipolar junction transistors and field effect transistors may be combined with resistors in a variety of arrangements, all of them have a gain which is either negative or is less than one; ...


0

Mosfets with heatsinks can handle hundreds of watts. 10 watt zener diode cost about 9£ while 100 watt mosfet is about £0.7. Also if you produce many different power supplies and other products, you only need couple different mosfets in large quantities. If you used zener diodes worst case is that you need a different zener for each product.


1

A typical DC SSR has only two output connections, so the current for the transistor base must come from the load current. So, think of configurations that would do that- for example, a triple darlington with the phototransistor at the 'top', or a hybrid MOSFET/BJT output stage with the MOSFET as a driver. Edit: Here's a somewhat similar idea with less ...


2

It is worth pointing out that fundamentally the transconductance of a BJT is much higher than for a MOSFET. i.e. the current varies with the exponential of the applied voltage in the case of a BJT, whereas it only varies with the square of the voltage for a MOSFET. Ideally all systems would be a mix of BJT and MOS, but that is not how the world works. So ...


1

The circuit you've shown uses the transistor as an emitter follower, not the common emitter amplifier you referred to in your subject line. Wire it up common emitter, simulate it, and you'll answer your own question(s). Caution: If you choose to build it in the real world, common emitter, the load will dissipate about 550 watts, the transistor will ...


1

Is this a homework problem? I'll start with a hint: Imagine that the transistor is turned on all the way (negligible Vce). What is the voltage at the emitter? EDIT: Since EM spilled the beans, I might as well clarify. :-) Imagine that the transistor is somehow fully on. All of the voltage is dropped across the load, so Ve = 24V. The base voltage Vb can be ...


3

You have an equation for \$v_o\$ of the form \$v_o = kv_{gs}\$ but you need \$v_{o}/v_i\$. Use the hint (which is just KVL) to substitute for \$v_{gs}\$: $$v_o = kv_{gs} = k(v_i - v_o)$$ and solve for \$v_o/v_i\$. For \$R_o\$ you have forgotten the contribution of the dependent current source. To calculate \$R_o\$ set \$v_i = 0\$, apply a test voltage ...


1

You can make it a bit more sensitive by increasing the 10K resistor, but there is not enough current gain in one BJT to make it work in very dim light. If you replace the BC547 with an n-channel MOSFET you can increase the 10K to 100K or 1M and make it very sensitive. Alternately, connect a second BC547 to the first in a Darlington configuration (and ...


3

That's the body of the MOSFET. It's also sometimes called the substrate. The VN2222 datasheet from Microchip shows that the body is tied to the source internally, which is standard practice for discrete MOSFETs. Your simulation should do the same. I'm not sure why PSPICE thinks there's a fourth terminal on the package, unless it's using a generic model. In ...


8

The middle terminal is the substrate (edit: also body or bulk as Null says). In almost all discrete MOSFETs (including the VN2222) it is effectively connected internally to the source. It behaves a bit like a second gate (though not an insulated one) if you had access to it. You should connect it to the source if you're trying to simulate a real VN2222. ...


1

Umm, this question can be generalized. Here Vin is 0V, but if Vin is other values, you can think as below. No matter which region the transistor working, the charging current to the capacitor, that is \$i_{D}\$, should tend to 0. Then we can do some analysis on the \$i_{D}\$'s equation on these two regions. At nonsaturation region $$ v_{DS} < ...


0

In steady state, due to the capacitor, the current will asymptotically tend to 0, and therefore Vo will become equal to Vi. But as \$phi(t)\$ is applied to the gate of the transistor, you know that the source can only go up to the gate voltage minus \$V_T\$.


3

Your calculation of the collector current is correct. So let's focus on calculating the base current. Since we are dealing with a saturation mode, vce=0.2 making the drop across the resistor 9.8 v. The current going through the collector is Ic=9.8/10kΩ=9.8mA and assuming β=100 then Ib=0.098mA. β is a characteristic of the transistor in forward-active ...


0

VC does NOT equal 9.8V. It is 10 - 9.8 = 0.2 as in your assumption. The 0.2V is only an approximation - depending upon the transistor it may be as low as a handful of millivolts.


3

You cannot make both assumptions about \$V_{ce} = 0.2V\$ and \$\beta = 100\$ at the same time; they are for two different modes of operation for the transistor (there's also a third mode where the transistor is in cutoff, and a fourth less common one in reverse active). You start out assuming one of the possible states. For example, suppose I assume the ...



Top 50 recent answers are included