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1

Just a few thoughts. Place a generous value (220uF or more) low ESR capacitor right at your circuit in case your USB supply is not stiff enough. Place a heat sink onto your transistor so that it will not fail while you are testing. The frequency of this design is set by the coil geometry (and capacitors) and not totally predictable, it may not always ...


1

If the current through the resistors is approximately the same \$(I_C \approx I_E)\$, and you want the voltage drop across them to be the same \$(\frac {V_{CC}} 3)\$, then by Ohm's Law the resistors should be the same value. $$V_E = V_{CC} - V_C$$ $$I_E \approx I_C$$ $$R_E = \frac {V_E} {I_E} \approx \frac {V_{CC} - V_C} {I_C} = R_C$$


0

It looks like something was erased in the area containing the short. Perhaps it was another resistor to form a DC biasing circuit with R1. Are you sure your secondary coil isn't picking up 50 Hz or 60 Hz interference from the mains?


0

LvW covered most of it. As far as why the collector can't swing below 4V, consider the bias voltages on the transistor. The emitter is at 4V, which means the base is at about 4.7V. If the collector goes below about 4.7V the base-collector junction will be forward biased and the transistor won't be in forward-active mode any more.


1

Of course, you are allowed to make Rc=Re. However, the question remains if this make much sense! Hence, I think the first rule you have mentioned (Vc=Vcc*2/3 and Ve=Vcc/3) is a good trade-off between allowable swing and good stabilization of the operational point. As you probably know, the resistor Re provides negative feedback for DC and stabilizes the bias ...


0

I was reading some thermodynamics when I saw something that can answer my question (?!) It holds $$ \left. \frac{d I}{d T} \right|_V = - \left. \frac{d V}{d T} \right|_I \left. \frac{d I}{d V} \right|_T $$ The second factor can be easily calculated from the Ebers-Moll equation.


4

There is a big problem with this circuit - the \$V_{CE(sat)}\$ of the TIP120 is 0.7V at ~100mA (near our operating point). That means that there will only be \$12-0.7 = 11.3V \$ across each LED string. Since \$3\times3.8=11.4V\$, this means you won't quite have enough voltage available to power the 3-LED strings. Bummer. Never mind, we can fix this. ...


0

I'm not sure this will work as designed - it'll be very LED dependent. I'd recommend a couple of things: 1. Use a MOSFET instead of a BJT. Edit: tweaked this section due to the TIP-120's higher Vce(sat). Switch your BJT out for an N-channel MOSFET. BJTs have a saturation voltage across the collector and emitter, and it varies with current. The TIP-120's ...


2

R7, R10, R11 are for thermal stability. They add a little emitter degeneration, so gain is dependent on characteristics of a resistor (stable over temperature) rather than a BE junction voltage (changes a lot with temperature). I would expect R8 to perform a similar function, linearizing Q7. Except that it is bypassed by C1, which removes the effect at ...


1

Just for your understanding: The base emitter voltage does NOT decrease if the temperature (and with it the collector current) are increasing. The sequence is as follows: For a constant VBE voltage the collector current Ic increases for higher temperatures of the transistor body (increased carrier mobility). And the bias voltage VBE must be externally ...


2

Due to C2 the amplification of the circuit is very high. Russell's comment about the gain being 4.7 is correctly based on the ratio of collector resistor to emitter resistor BUT the emitter resistor is shunted with 20 uF and that has an AC impedance of 8 ohms at 1kHz hence your gain is through the roof. Remove C2 and check that the gain is about 4.7. What ...


1

Your choices of components look good. Ideally Vc_DC shouild Vbattery/2 and the 4.635V you see is entirely close enough to Vbattery/2 = 9V/2 = 4.5V/2 for this purpose. (1) As shown the gain is very high. It is ABOUT Rc/Reinternal where Reinternial is the effective emitter resistamce of the transistor and is approximately equal to 26/Ie_ma. The reason for ...


0

Your two circuits have a vague resemblence to a Hartley oscillator, which should look something like this: simulate this circuit – Schematic created using CircuitLab EDIT: OK, here's the schematic linked to from the video: It is indeed a Hartley oscillator, and it looks a lot more like mine than yours!


1

In these formulas it assumed that the voltage across the PN junction is set externally and that it also decreases with -2.1 mV/K. If you would do that indeed the current would change significantly. In practice there would need to be some mechanism in place to control the current. A series resistor will do the trick, it will limit the current. But then you ...


1

The motors/drivers you've specified are an excellent match, as long as you realize that you'll need to provide a fairly high-speed PWM (Pulse Width Modulation - say, 10 kHz) signal for your speed control. The drivers are rated for 1.2 amps average, and the motors are claimed to run at ~100 mA, although I suspect that's with no load and they may well draw ...


2

The latching current is associated with turning on the thyristor, the holding current to turning off. If the current through the load is not large enough the thyristor will not turin on correctly and will turn off as soon as the load is reduced. The holding current is the current required to maintain conduction once it is established with 0 gate current. ...


0

Actually, I have been wondering about something like this myself (in the past) after seeing 900V-rated transistors packed in TO-220. It turns out that the topic is rather complex. There are various safety and "just functional" levels of distance, or rather of creepage to use the technical term. There is plenty of literature on this topic. A good source ...


8

Yes, you would typically apply a compound to seal the pins after mounting. Even for much larger spacing this is typically done since the leads often have sharp corners (more prone to corona and breakdown). We routinely add something like Corona Dope to even rather large components (HV relays, etc) when the voltage gets up and over 1kV. This provides ...


0

High voltage primary considerations are clearance and creepage at the physical layer. Clearance is the shortest path between the points of interest and the standard usually used is IPC-2221A. Creepage is the shortest electrical path on the PCB If either of these distances is less than found in the above reference then as you surmise, a compound with better ...


7

It is not clear what the actual minimum distance between the collector and the other pins is but it seems to be a little bit more than 1 mm. Probably in a sealed housing with dry air that would just be sufficient (assuming anyone would use it near the maximum rating !). Another possibility is to apply a conformal coating. BUT, the fact that the transistor ...


15

Hmm, it does seem tight. The pin pitch is 2.3 mm, and the maximum pin width is .85 mm, leaving 1.45 mm minimum space between pins. The transistor is specified for 1.4 kV C-E, which are on adjacent pins, so that's just about 1 kV/mm. As I said, that seems tight, and you'd have to be careful in designing the PCB footprint to not make this worse. Usually I ...


3

Your idea 'current follows the path of least resistance' is an approximation, and it holds only when there is a limited amount of current. In your case you use a lab power supply, which is a constant voltage source, which will supply all the current you ask from it. Hence both the transistor and the red LED parts get current. A way to switch the other LED ...


0

Look at the datasheet of the opa2341: rail-to-rail input and output range supply 2.5 - 5.5 V That should would work just fine in my opinion. If you want to use a 12 V supply, feed it to the power jack on the Arduino board, this will take the 12V to an on-board regulator which will make 5 V from that 12 V. I would just connect the suppy pins of the ...


0

Perhaps I am way off base here, but interpreting the OPs questions, there are a few, I think he is trying to build a "demo" of all of the logic working off 2 switches. See below. Top Level Diagram: simulate this circuit – Schematic created using CircuitLab After adding the remaining gates to the circuit, the OP would need to realize the gates ...


1

If you just want to make your own logic gates to see them work, then RTL (resistor-transistor logic) is probably the easiest for you to implement. The basic gate output is a pullup resistor to the supply with a NPN transistor connected between it and ground. When the transistor is on, the output is actively driven low. When the transistor is off, the ...


1

The gates you show can be used only under very limnited circumstances. Except for the NOR and the inverter (which lacks a base resistor!), they are based on emitter followers. A property of an emitter follower is that the output is ~ 0.6V lower than the input. This is OK for one or two stages in series, but after a few more the logic signal is lost (will be ...


2

Your basic problem is PNP1/NPN2. With the bases tied together, you are getting maximum current flow through the two emitter-base junctions from +12 to ground. Any real transistors in this configuration will die instantly. If you must use BJTs and want to keep your circuit ideas as much as possible, replicate NPN1/PNP2, and use a fifth transistor to invert ...


0

In a shameless bid for rep points, I'll expand on my comment. The circuit is very simple, but not very good. Its biggest problem is that its output impedance is fairly high. In other words, its output voltage will vary with load resistance, which is what allows your fine adjustment to produce results. Ideally, of course, a voltage source will not vary with ...


0

Your circuit won't work (simulate) because the BJT Transistors (you have shown) are current controlled devices and when SW1 is open there is no current flowing. You can have a look at the tutorial here it uses BJTs too.


2

I hope you have series resistors in your circuit that you left out in your picture? Otherwise you are pushing an undetermined amount of current through the transistors and LEDs. To 'block' a lower LED you could use this basic circuit (untested): simulate this circuit – Schematic created using CircuitLab As others have stated in the comments, ...


1

Your original schematic is somewhat confusing, as it does make it look like you are trying to short-circuit the USB power rail. I have redrawn it below with a circuit that should do what you want. It uses the switch to pull down the gate of a P-channel enhancement mode MOSFET. simulate this circuit – Schematic created using CircuitLab The main ...


0

First off your switch in the second diagram will short out the 5V supply when pressed. Secondly put a resistor in series with the switch and move the lower connection of the switch to connect between R2 and the base of the transistor.


1

That looks like a decent stab at an emitter follower based regulator, with internal feedback (transistor beta). Assuming it is operating correctly, the output voltage should be around 0.7v or so less than the base voltage. I'm not sure what your RV2 is supposed to be doing. If the regulator is working correctly, it should have no effect on the output ...


1

Task 1 - find Ib. We know that VBE is 0.7V (given).{quick check - is this possible with the given values? - without the base connected R2 would produce a drop of about 0.75V - this is a good result because we can always drop this voltage by taking current into the base}. For VBE = 0.7 and R2 = 6.3k by Ohm's law I(R2) = 0.7/R2 = 0.11111 mA The 11.3 volt ...


2

Whenever the 5V device transmits, it pulls the bus down, and the diode conducts to 0.5~7V, when it releases or pulls the bus UP to 5V, the diode blocks the current from flowing, and a resistor on the 3.3V device should pull the bus on the 3.3V side back to 3.3V. Whenever the 3.3V device transmits, the "HIGH" (about 3.3V or a bit less) is supposedly inside ...


1

In addition to Nicolas answer, the LM386 is definitely not suitable for this circuit, and nor is the 'zener'. The circuit to implement the zener is not in regulation (as already noted) and is not actually doing anything for you. Either use a real zener at the voltage you need (which will be, by definition, below 9V) or come up with some other reference. ...


1

First of all, a 9V battery will not output 2Amp. This alkaline battery http://data.energizer.com/PDFs/522.pdf can give you 500mA but even then the rated capacity is halved. Next, your "Zener circuit" will start conducting above 17V (it starts conducting when Q2 base is 0.7V above ground, so when current through voltage divider reaches 0.7V/4.7kOhm=0.149mA ...


6

You can connect a MOSFET directly, even without a resistor if you like (though it's not nice to put such a heavy load on the output), but if the MOSFET fails with a Drain-Gate short your Arduino will be scrap. I strongly suggest using a higher voltage MOSFET than that one- it's only 12V and you will at least be subjecting it to at least 12.5-13V when the ...


0

A flip-flop cannot be implemented by a single BJT or transistor,there is no feedback hence your output only depends on the current input,its not a flip-flop. You have shorted the base and the emitter hence they are at same voltage,giving trigger will add same potential to both terminals,hence no current flows,when you add a capacitor between the base and ...


0

1) To current flow , \$ V_{BE} \$ have to be around .6~.7V based on the transistor type. When you shorten them out, it become zero and so no current flow. 2) When you add a resistor , as there is no voltage difference between base and emitter , it has no effect.But if you add a capacitor , capacitor create a voltage difference \$V_C\$ . But after some ...


1

The reason this did not work is because the base needs to be at a higher potential then the emitter, by about 0.6v. the reason the capacitor worked is that you stored the pulse energy in the capacitor above the potential of the emitter and then the capacitor slowly discharged through the base. When the potential of the capacitor fell bellow the threshold of ...


0

Ico is the reverse saturation current when in CB mode the emitter is assumed open hence the,there is a reverse bias between base and collector,and it is dependent on the value of applied voltage. Hence the BJT acts like a p-n juction and the equation seriously satisfies with the Shockley diode equation . So the current will decrease as Vc increase (Vc is ...


0

1) The polarity of the capacitor is backwards. 2) You need a discharge resistor for the capacitor. Place a high-value resistor across the capacitor. Adjust the resistor value for the desired delay time. 3) Check the value of R3 to make sure that it is the value shown in your schematic. 4) As others have already said: this is a poor circuit design. When ...


-1

I suggest that you consider a Silicon Controlled Rectifier (SCR) for this application. Remove C1, R3, R4, and both 2n2222s. Connect the upper end of R1 to the gate of the SCR. The Anode and Cathode of the SCR would go between Rs 2 and 5.


0

Your circuit is not working properly because it is not designed correctly. You haven't really given enough information, but based on what you have; I am thinking you want something similar to one of the circuit on this page Link. There are several there, if you are just interested in driving the LED then the two NAND Gate circuit (the first one) would be ...


0

This circuit is reminiscent of a boot strap. When Q1 is off the space between Q1 and R5 is at ground level. When you press the button you charge C1 (reversed polarity!). Once C1 is charged it will supply current to Q2 that will intern supply current to Q1. When Q1 begins to conduct the spot between Q1 and R5 begins climbing in voltage pushing the base of Q2 ...


1

A Darlington is not a very good way to drive a motor from 3.3V. The voltage drop is too high (At 3A the voltage drop could be as much as 2V, but at 2A it's typically a bit over 1V- even so it takes about 1/3 of the motor voltage typically). It's better to use a MOSFET. You can search distributors using parametric search. For 3.3V drive you should pick one ...


1

The TIP120 will work at 3.3V as well. You do not apply a voltage to a bipolar transistors base, but you apply a current to the base. The resistor between your micro controller and the TIP120 will convert the voltage to a current. You might need to reduce the resistance your base resistor a bit at 3.3V. Using a Darlington transistor can be effective for ...


1

You didn't provide a link to the TIP120 datasheet, so I'll take it on face value this is a darlington transistor, presumably NPN. The advantages of two transistors arranged in a darlington configuration is that the overall gain is the product of the two individual gains. However, one drawback of this approach is that the saturation voltage will be much ...



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