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2

When R2 is connected to the + of the battery the transistor will be turned on. When R2 is disconnected the transistor won't have a clue what to do since you aren't telling it what to do. An output of LOW on the Arduino doesn't mean "disconnected", it means "connected to ground". To turn OFF the transistor your should connect the base to the - of the ...


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With your original idea the LCD circuit is most likely finding a ground path along the other Arduino pins that are connected, and strange things are happening. Trying to enable a circuit by pulling down the ground point is usually not recommended except for simple circuits with components like relays, bulbs, or LEDs. It is possible that the LCD system ...


1

Here is the basic idea. When VCC goes high, C1 will start to charge up. After a few ms, M1 will turn on. D1 is intended to make sure the cap discharges promptly when VCC goes low. It may not really be needed. Larger cap or larger resistor will increase the delay. This is an imprecise delay. There will be variation from unit to unit and over temperature. ...


2

My recommendation: Forget the link which gaves you the above information, which is false resp. misleading. (By the way: This link leads you to other "explanations" which also are wrong). Hence, you should not blindly trust any information available in the internet. The text says that the "25 mV value being the internal voltage drop across the depletion ...


1

This is what I consider to be a differential cascode amplifer: - Note the important difference - the output is taken from the drains of the upper transistors. What you appear to have is Vout taken from the drain of the 1st pair of transistors but this voltage will tend to be clamped by the sources of the 2nd pair.


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Yes, except for the location of the output connection (the first circuit looks wrong), the two circuits are topologically equivalent.


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The BuckPuck is a step-down ("buck") converter. You must provide it with a voltage at least 2.5V above the forward voltage of the LEDs. The LED has a typical Vf of 3.2V, so you want your input voltage to be at least 5.7V. Call it 6V or higher. The BuckPuck will take an input voltage all the way up to 32VDC, although the efficiency drops a bit. It is ...


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The collector current with an 83mA/12V relay (T9AS1D22-12, if I got that right) and an LED is in the 100mA range. An appropriate base current is in the 5-10mA range to keep the transistor well saturated. Note that the saturation voltage is specified at Ic/Ib = 10 (from your datasheet): The 2N3904 is not particularly well suited for this application- ...


0

The driver you have is a constant current driver, and checking its datasheet it should work OK for driving individual LEDs provided that you meet the requirement that the input voltage is 2.5V higher than the nominal forward voltage of the LEDs. You don't need a resistor in series with the LEDs in this case. The datasheet doesn't make it clear whether LED- ...


2

Assuming a 0.7 drop for Vbe and a 5v Arduino system the base current should be about: (5v-0.7v) / 1000 ohm = 4.3ma. That should be well within the transistor's normal operating range for base current. (If you have a 3.3v Arduino system the base current is about 2.6ma, with the same assumptions.) There doesn't seem to be any problem. (If you needed ...


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I was initially very annoyed by this 'resistance looking into' concept.But now I can see how simple it is.I will try to explain it in lay man's language. How do you calculate resistance between two points in any circuit ?You apply a voltage find the current and do V/I. Here is something that beginners overlook as they are used to working with 2 terminal ...


1

Assuming moderate collector current, the region where about 0.2V < Vce can be analyzed as anywhere else in the active region, with suitably adjusted hfe. So you could use small signal analysis about the point where Vce = 0.7V without much risk, for suitably small signal amplitude (such that the deviation from linearity is 'small'). Consider the curves ...


1

Once a junction gets saturated, it takes awhile for it to react and come out of saturation. I believe this is where the "Charge Control" model, or Gummel-Poon model, comes into play. The time delay for a saturated junction to react is modelled by waiting for a capacitor to discharge. From Wikipedia, the basic model is: Courtney of Berkely, you can look ...


2

SCRs turn on and then stay on until the current through them drops to zero (perhaps because they're switching AC). They're typically used at mains frequency up to a few kHz and not above. Here are some switching characteristics from an inexpensive 8A SCR (the BT151-500R) The dVd/dt is a maximum rating tgt and tq are typical turn-on and turn-off times ...


1

I've come up with something that might work. You will need to breadboard it and see. simulate this circuit – Schematic created using CircuitLab Q1 & Q2 need to be identical transistors, preferably from the same batch. They should also be touching each other if possible so that they are at exactly the same temperature. Cover D1 and adjust ...


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If you really want to use an LED (not good but cheap) you should probably use it as a current device and make a one op-amp current-to-voltage converter. Very simple and the output would be linear. Use a TL061 for its low cost and its low bias current which might be important depending on the amount of current the LED puts out. Do not use a TI brand TLO61 ...


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Yes, you can do this with just NPN transistors. 60mV is small, but probably enough. Resistor values will depend on your voltage source, but try this: simulate this circuit – Schematic created using CircuitLab In normal conditions, R1 and R2 set Base voltage about 0.57V - not quite enough to conduct. Adding 30mV gets you 0.6V. With no Emitter ...


2

In linear circuits with both AC and DC components, one can split the circuit, using the Superposition Theorem, into a DC equivalent circuit to model the effect of the DC components and an AC equivalent circuit to model the effect of AC components. By "components" I mean voltage sources and current sources. For AC analysis, you are only concerned with ...


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A perfect voltage source has zero output impedance. A perfect current source has infinite output impedance. Don't replace a DC voltage source with a ground connection, replace it with a short circuit. There is often a difference.


2

You can do what you want if you add some amplification to the output signal from the LED. A single op-amp can do this or you can use a pair of transistors configured as a differential amplifier (the "long-tailed pair" that Brian Drummond mentions above. I'm going to show you an op-amp version because it is easy. I like using the LM358 or LM324 op-amps for ...


1

What is "preferred" makes no sense since the two circuits are different. The first, assuming well-match transistors, is merely a way to control two different current sources to the same value. The second is a current mirror as you say. The purpose of that is to have the current source produce the same current ("mirror") as the current drawn by a load. ...


3

In your first circuit, it's uncertain what output current you'll get for a given input voltage. First, because if you build this circuit on several chips, the PMOS behavior will not be identical from chip to chip. Also because as the temperature chagnes the PMOS device behavior will change. In the second circuit, you know the output current will be ...


1

Sorry folks for disturbing you i found out the answer myself what i needed to do was just make the ac equivalent circuit for the given circuit and further proceed with problem . For ac equivalent the capacitor behaves as a short so resistors such as R3 becomes parallel with Rc and so on.......


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strong textHere is another - straightforward - way to solve the task. You only need Ohms law, which gives you a system of 4 equations with 4 unknowns, which is easy to solve. Assuming that the current through the 91k resistor is I1 (and the current through the 150k resistor is I1+IB) you can apply Ohms law to the three resistors (base resistors as well as ...


2

Another way to do this is to determine the Re equivalent (Ree) as seen by the Base. With B=100 the Ree seen by the Base is (100 x Re), or 330k. Replace Re with Ree (330k), add the 0.7v for the emitter diode, calculate the actual Vb and Ib. Now go back to the original circuit and calculate the rest of the currents and voltages. (Another ...


1

Iterate. Start by offsetting the circuit so that the -3V line becomes 0V and the +3V line becomes +6V. You can return to the plus minus 3V when you are done. Calculate the Thevenin equivalent voltage at the Q1 base, assuming no base current. Use that to set the emitter voltage, calculate the current through the emitter resistor. Use the current through ...


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This doesn't satisfy everything that Dwayne speaks of, but for a simple tester it should serve your purposes. simulate this circuit – Schematic created using CircuitLab Q1 can be either an NPN or a PNP. This is really just testing the polarity of the base terminal by assuming that current will flow out of the base in a pnp and into the base for ...


0

What you have drawn will work but it may not give you the actual result that you are looking for. Yes: one or other of the LEDs will light if the correct transistor that is working properly is installed into the correct socket. However, the LED will also light up if a transistor is inserted not correctly. For example, if you swap the B & C lead on ...


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Your output resistance is mostly controlled by Rc, which is 4.7k, so you are going to get a lot of loss driving such a small load resistance. To maintain gain and drive the low resistance you need a common collector buffer, as Photon said.


2

I have recently characterized several signal and even power mosfet devices (e.g. FDP6030BL in TO220 case), at room temperature, using an Agilent B1500. To my extreme surprise, the OFF-state (Vgs=0) drain current in most of the cases (even in power devices!) was between 0.1 pA and 10 pA (at about Vds=20V), despite it was 1uA in all the datasheets. The worst ...


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Your electronics class has probably taught you the hybrid-pi model and given you some complex (yet accurate) formulas for gain, input resistance, and output resistance of the various amplifier topologies. It might help your understanding to have some simpler, approximate formulas. These come from the always-helpful Art of Electronics by Horowitz and Hill. ...


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According to the above drawing, it looks like the fan wants to see a 0-10 volt swing on the PWM input (yellow wire) and will supply 10 volts DC on the Voltage output (red wire). simulate this circuit – Schematic created using CircuitLab If you simulate the above circuit, it looks like a 3.3 volt input to R1 should be enough to saturate the ...


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While this does not seem to be a typical regulator boost circuit, I do see the basic idea. With some adjustments it should work fairly well, though stability could become an issue. First off, the PNP is not wired correctly, the emitter-collector connections should be reversed. With the pins corrected the PNP will turn on when R1 is passing about 375ma, ...


1

No, the maximum current will be less. Think about thermal considerations alone- the breakdown voltage will likely be in the 9V range for a modern Si NPN transistor so the power dissipation will be about 10x higher at the same current. A few mA should be okay. You should know that there is degradation of forward beta when transistors are operated in this ...


3

Not a nice design, alas. Will work OKish but easy to do far better. Vout will be ~~~= k x V_DAC - ~~~= 1.2 V Testing: Ensure polarity of all transistors is OK 1 x PNP (BC557) should test as if there was a conducting diode from collector to base and from emitter to base and high resistance in reverse direction. 3 x NPN (BC547, BD245B, BD137) tests as ...


2

Please investigate solid-state relays; they are opto-coupled "transistor-relays". Such as: https://www.sparkfun.com/products/10636


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Forward current is very different from the allowed reverse current. Many datasheets don't list reverse breakdown voltage and even fewer list the allowable reverse current. It sometimes helps to check datasheets from different vendors when they produce the same transistor, but this is basically the vendor's way of saying: We don't support this mode, you are ...


2

K., All the answers for your questions depends about somethings, like this: What Transistor would you use? You want to control the load current (limit the load current) or you want that transistors operates like a fully on switch? To dimensionate the resistors, you will need to read the transistors datasheet, and search for the hfe or β, the ...


0

Basically the first circuit (named common-emitter configuration or transistor switch) is more appropriate for some reasons (smaller dissipated power, arbitrary input voltage). You have only to add a base resistor. A disadvantage is the ungrounded load.


0

Another option would be to use an addressable RGB string. There are a number of variants depending on the chip that they use (LPD8806 and WS2812 are two common variants), and there are nice libraries for both for the arduino. You can also drive them with 5V, so you don't need two separate power supplies. Adafruit carries a nice selection of them.


0

this is probably not an issue, but you want to make sure that the power supply is able to supply all the current you need and maintain the voltage regulation when all the strings are lit.


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Looks fine to me. Be sure that the base resistor for each transistor is low enough that the transistor saturates. 1k is a good starting point. – Dwayne Reid My NFET suggestion reduces base/gate current from the MCU to "0". It looks good. There are ways to get more precise voltage. Really, LEDs are current driven devices, so current drivers are used ...


0

I'd suggest to you to use a CMOS gate instead of a transistor. These are extremely versatile, have a much higher input impedance than a bipolar transistor, and can easily drive an LED or provide an output suitable for the Beaglebone or almost anything else, provided you use the right voltage. You won't need that resistor the transistor requires, and the gate ...


0

Few simple options. Germanium Transistors would be better because of the lower VBE voltage drop of ~0.2 instead of Silicon Transistor drop of ~0.6. Resistor values would vary, you gotta play with them. Two NPN transistors, one as a simple switch, the other providing the current gain. Inverted, so led turns off as audio gets louder/sound plays. ...


2

Yes, it is obvious that the base current can exceed the collector current if you consider the limiting case where the collector is open-circuit. The collector current will be 0, but you can still drive from 0 to the maximum allowed base current thru the base. With the collector open, base to emitter just looks like a diode.


-4

base current can never exceed collecter current.you must have studied about doping of base,collecter,emitter there surface area .There is a physical restriction to have base current higher than collecter current in any mode


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It is possible. Consider the figure below. The collector current at saturation will be $$I_{Csat} = \frac{V_{CC}-V_{CEsat}}{R_C} \approx \frac{V_{CC}}{R_C}$$ The base current is given by, $$I_B = \frac{V_{CC} - V_{BE}}{R_B} \approx \frac{V_{CC}}{R_B}$$ So it is clear that the currents at saturation are entirely decided by the resistors and hence base ...


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If you're making an avalanche relaxation oscillator, most transistors are not really designed or characterized for this so it should not be an enormous surprise that one maker's part will behave differently from another, or even other parts with the same part number from the same source. simulate this circuit – Schematic created using ...


1

Here is a solution that minimizes the input threshold voltage as battery voltage changes: simulate this circuit – Schematic created using CircuitLab Adjust R2 to control sensitivity to low-level signals. Adjust R5 to control sensitivity to high-level signals.


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Edit: Actually, I found out the answer I posted is already given in the question you rejected as "too complicated". I ask you to reconsider that judgement, as anything involving operational amplifiers (suggestions also found in that question) will be even more complicated. While the commentors are right that your circuit is a really bad idea in practice, ...



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