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0

Usually you want them 1/10" or more for easy soldering (and soldering). More height also helps to put some thermal resistance between the transistor and the soldering iron. I once made an assembly house put a transistor 1/2" (about as high as it would go) because I needed thermal resistance between the transistor and the board. The board had some sensitive ...


2

If it's a hobbyist project, just follow what you see on existing commercial hardware. You want it to sit far enough above the board so that the leads are not stressed, either because they're too short and everything is jammed together in assembly, or because they're too long and the part is moving from shock and vibration. Ideally TO-92 transistors should ...


0

It looks like I found a solution highly dependent on the device parameters. Low leakage parts are the critical factor. Take careful selection of Ids @ Vt and Vgs Two inversions are done: The 1st must have a gate threshold of << 1/3 to 1/2 of Vbat and not 2 to 4V. This is what causes decay on the 1st time constant. hi-side mechanical switch, low-...


3

While this is an old post, I recently had this same problem when I was making an EEPROM burner for retrocomputing (where I needed to write some 27C512-style EEPROMs). The currently accepted answer has the correct idea (which I used), but it is missing the logic level (5V) control and protection, and pull-down on the output side. The following circuit adds ...


0

I think, this question needs - at first - some definitions. (1) Common mode is clearly defined: Vb1=Vb2 (2) Unsymmetrical diff. mode: Vb1 finite and Vb2=0. (3) Symmetrical diff. mode: Vb2=-Vb1 (4) General symm. mode: Vb1 finite and Vb2 finite (with Vb1 not equal to Vb2). Input resstances: For the first three cases, it is a realtively simple task to find ...


1

the BJT will be in saturation, it might over-heat and burn in practice. Also this circuit will definitely not be amplifying anything as its in saturation and proper coupling doesnt exist b/w microphone and BJT. The reason should be obvious, the Vbe on = .7V and Ic=Is*e^(-Vbe/Vt)$ where Vt=26mV. Thus Ic rises exponentially with increase in Vbe and results in ...


0

I'm wondering if everyone is overthinking this. The question states (paraphrasing) the base current is zero amps (open circuit) By definition \$I_{CEO}\$ means "collector-to-emitter current with the base open" at some defined junction temperature and some defined collector-to-emitter test voltage \$V_{CE}\$. For a PNP transistor, if the emitter voltage ...


0

This circuit may suffice simulate this circuit – Schematic created using CircuitLab


0

You're OK with using multiple transistors per bit, right? In that case, you should study the following topics: Logic gates (AND, OR, NAND, NOR): what they mean How to create logic gates using transistors (CMOS) Latches and flip-flops: how to create them using logic gates This is a pretty standard introduction to creating a 1-bit storage element. It's not ...


0

Your device has a minimum high input voltage of just 2.0 V, so you can trigger it directly with your 5 V signal:


9

Assuming you're asking about the die, then parts with the same part number from the same manufacturer will typically have identical dice inside regardless of what package they use. There's no rule that says they have to, though; they could use specially-designed dice to fit more conveniently in different leadframes if they had reason to. Economically it ...


-1

The equivalent DC source (and its effective series resistance) applied to the base is what they are calculating so, in the example in the first reference they calculate the voltage at the base under the assumption that the base takes no current (3 volts) and, the effective series resistance of that 3 volts. That effective series resistance is the parallel ...


0

You treat Vcc as ground during resistance calculations, and determining the resistance of the circuit. (Short independent voltage sources and open independent current sources).


0

The equation for RB comes from finding the Thevenin Equivalent for \$V_{IN}\$. If you assume that \$V_{CC}\$ is deactivated (set to zero volts) then R1 and R2 are effectively in parallel. So, the Thevenin Equivalent resistance from \$V_{IN}\$ to the base is equal to R1||R2.


0

And for what it's worth there are multiple \$\beta\$ factors to consider when designing BJT circuits: \$\beta\$, \$\beta_{DC}\$, \$h_{FE}\$ - These refer to the BJT's DC forward current gain when the BJT is operating in forward-active mode (small signal amplification). Used when performing the DC design/analysis. (n.b. The 'FE' refers to Forward current ...


3

Two things for you to check. 1) Ensure that the power supply rise time is FAST. Generally don't switch the power supply on with the inverter connected, the inverter can get into a linear mode that prevents fast switching of the MOSFETS and leads to destruction. Connect the inverter to an already powered and enabled supply. 2) Check your inductors are not ...


1

Read the datasheet of your MOSFET. Maximum Vds will be specified. In the case of IRF540, the limit is 100 V In this circuit you should also consider that the motor itself will produce a voltage when you try to turn it off. You can search for "flyback protection" circuits on this site or elsewhere to see examples of how to avoid this damaging your MOSFET. ...


13

Some of the biggest drawbacks of RTL were: much lower noise immunity ( 1x Vbe bias vs 2x) excessive static power dissipation for "0" output state due to low pull-up R for low impedance needed to reduce rise time for "0 to 1" output state to drive pF loads. Such as R2= 50 Ohms. Noise immunity and slew rates were very asymmetric if R2 was high ( e.g. 1k) ...


7

Using resistor pullups causes the output rise time to be slow, so TTL uses a "totem-pole" output that drives the output high. This results in much faster operation. RTL logic tends to use more power because large currents flow directly to ground when you need an output to be low. Believe it or not, resistors with reasonable values often consume much more ...


5

RTL came about before TTL, it was just harder to stuff as much stuff onto a chip, and people didn't know as much about analyzing the circuits. In many cases, doing the job of one resistor with two or three transistors may actually take less space on the die. TTL needed to be higher performance and easier to use.


16

Amplifier System Specifications The first step is to organize the specifications you provided in an earlier version of your question. end-to-end system specifications Two BJT voltage-gain stages with some independent specifications (see later.) Bandwidth: \$20\:\text{Hz} - 20\:\text{kHz}\$. Input impedance: \$R_{_\text{IN}}\ge 5\:\text{k}\Omega\$. Output ...


0

A typical N-Channel MOSFET behaves as if a capacitor exists between the gate and source terminals. Usually, when referring to "gate voltage", one is usually talking about the gate-to-source voltage in the context of 3-pin MOSFETs. This "capacitor", when charged, turns on the transistor. Your confusion lies probably in the fact that MOSFETs are actually ...


1

To get the BJT to work properly you need enough current to flow through the base terminal, and in your case due to the high resistance you are only getting: $$ I= V/R $$ $$ I = 1.2/10^6 = 1.2 \mu A$$ Thus the base current is too low to function. While when you reduce the resistance to 1K ohms only, the base current will be 1.2mA which is sufficient enough.


0

Yes - it is entirely "doable". A transistor "works" regardless of external applied collector or drain voltage - but you can see that it is working externally only by applying voltage. Be CERTAIN that S1 and S2 are NEVER both on simultaneously :-) :-( !. You could use a bipolar or field effect transistor for both locations, but a FET is liable to be easier ...


2

You have mid-band gain of 820/100 * 700/180 = 8.2 * 4 = 32X. Thus 1vpp in should cause 32 VPP out, and you have only +-12 volts or 24vpp. I see Q2 biased to about zero volts, which allows +12v swings, but allows much less (-) swing, the polarity where your output stage clips. I'd drop R7 to be about 1/2 of R6. That allows +- 1/3 of 24v on Q2 collector; ...


2

Or they actually just add the resistances without caring about the dependent source for some other reason? The \$g_m\$ source produces current in the C-E branch, not in the B-E branch. So it doesn't contribute to determining the B-E equivalent resistance. If you wanted the equivalent input resistance of a common-emitter amplifier with emitter degeneration (...


3

The IRF530 FET has a high gate threshold, and will not reliably turn on with 3.3V gate to source. Use a logic-level FET that's specified for 3.3V gate-source, or use another EMH1T2R in place of M1.


3

Since you confirmed in comments that the problem was the signal level, I'll explain that a little bit. You are using an NPN transistor as what is usually called a "low side switch." Simply put, you are switching the connection between your load (the solenoid) and ground (usually the negative terminal of a powersupply or battery.) When used this way, an ...


1

The gm is a small signal parameter which is only calculated and available after a DC operating point simulation. Note that an AC analysis first does this DC operating point calculation as it needs that gm because an AC analysis uses the small signal model for the transistors. Your transient simulation uses the large signal behavior of the NMOS so gm isn't ...


2

These devices look like vertical bipolar transistors (as opposed to lateral) that are different sizes or multiple emitter/collector. They are probably all the same type of transistor. If I had to guess, the purple area is the emitter, the yellow area is the base, and the blurry light blue rectangle on the bottom is the collector. Devices with multiple purple ...


0

As mentioned above, an H-bridge driver with four external power MOSFETs will give you an all-electronic motor reverser that needs only one control line from the Arduino (two if you want a third state - off). The larger the MOSFETs, the lower the heat. International Rectifier is big in h-bridge control chips. Here is the general idea:


1

In addition to the answers you've already received, here is my suggestion. simulate this circuit – Schematic created using CircuitLab Resistor \$R_{LOAD}\$ is a "dummy" component; it is a placeholder for the actual circuit that you want to turn ON|OFF via Q1. To solve for the values of the "current limiting" resistors R1 and R2 proceed as follows. ...


3

It seems that when you were assembling your circuit, you forgot to add the transistors Q1-Q9 in your circuit. These are needed to allow the Arduino to control when the LEDs turn on/off without supplying them directly, by wiring them to the transistor base, hence you won't need to worry about how much current is available from the Arduino pin. If you have ...


0

It is unfortunate that the LED common side is ground, so you will need a high-side driver which requires more than one transistor per circuit. Or, you could use an IC that has multiple drivers. I have seen up to 8 per IC, this one has 4. Assuming that you have 5V available, three of these would be needed for your 9 LED circuits. This is a little overkill ...


0

It raises the gain of the overall circuit as often one MOSFET can't increase the gain dramatically. Also taking a quick look at your drawing did you forget the source resistors? It'll separate the FETs from ground and alter your resistances. For example the vgs will be nonzero so the dependent current sources will have to be taken into consideration. Google ...


0

The layout problems mentioned in earlier answers are becoming less relevant with manufacturers integrating the driver and the transistor in a single package, thus circumventing the problem of gate loop and common source inductance. So, to a large extent, the question should be: "By when are we using GaN everywhere?"


1

Enabling/disabling a strong pull-up with a microcontroller The NPN BJT you proposed will not work. You cannot force the BJT into saturation in the configuration you have. You could use a PNP BJT instead, but I'd suggest the best solution is to use a P-Chan FET as shown below. simulate this circuit – Schematic created using CircuitLab When the ...


0

The SMD Codebook is a useful resource for identifying these, of the parts you show, I find UMH11TN for the third one, and BAS21-03W for the fifth one, but the marking in the datasheet for the last one doesn't accord with the marking in the database.


3

Looks like NMOS transistor being used in depletion mode as an active resistive pull-up. See Wikipedia: Depletion-load NMOS logic.


1

Two transistors Q1,Q2 are identical Unfortunately you do not describe in which context this statement is made. I have seen such statements before in the context of analog IC design. On a chip we can make transistors that are "identical" in the sense that they have very similar transistor parameters like \$\beta\$ and what \$V_{BE}\$ you get under certain ...


1

There are a number of things wrong with the driver circuit. The high side drivers are all wrong, they will fry everything connected to the +5V supply (including the Arduino) tout suit. Please review other answers for how to drive LEDs and how to design high side drivers. If you need to use the 12V supply you can use another 2003 or some NPN transistors (in ...


2

Since you are using an NPN transistor, you should have it between the LED and the negative pole of the battery instead of the positive pole. Like this: simulate this circuit – Schematic created using CircuitLab An NPN transistor requires the base to be about 0.7 V above the emitter voltage so that it will conduct. To get that in your circuit, you ...


1

Your simple circuit has no voltage gain and instead it has tremendous voltage loss so an AM signal that is 3.4V peak will barely light the LED. The LED needs 2V if it is a red one. The emitter-follower transistor needs a base voltage of 2.7V for it to light the LED. The rectifier diode input needs to be 3.4V for the circuit to light the LED. An AM radio ...


1

Figuring out that "middle part" of the circuit is easier when re-drawn slightly differently. Especially the diodes and zeners are better drawn showing the voltage going down through the chain. The zener + diode in series are used when you don't want any current to flow when the current flows in the "non-zener" direction. That's what D1 and D4 do, they ...


0

In addition to the formulas as given by Warren Hill in his answer, I like to visualize the role of the emitter resistor RE and to show why is not too important to know the actual Vbe value (tolerances, uncertainties, temperature dependence). Instead, for calculation purposes we assume a fixed value of app 0.7 volts.


0

This is not a full answer but much too long for a comment and should give you some pointers We know \$ \beta = \dfrac{I_E}{I_B} \$, and \$ \alpha = \dfrac{I_C}{I_E} \$ so we can define \$ \alpha = \dfrac{\beta}{\beta+1} \$, we can ignore \$ \alpha \$. Typically \$ \beta \ge 100 \$ and \$ V_{be} \approx 0.7 \text{V} \$. We also know \$ I_E = \dfrac{V_E}{...


0

You have too many resistors. Most of them can be 0Ω (short circuits). simulate this circuit – Schematic created using CircuitLab Since this has the appearance of a homework problem, I'll leave it to you to explain how this circuit works and what its limitations, if any, might be.


0

I am wondering if there is a negative to negative (and/or positive to positive) switch. simulate this circuit – Schematic created using CircuitLab Figure 1. A double-pole, single-throw (DPST) switch can be used to connect negative to negative and positive to positive. I know mechanical relays work, but I need an electrical way of doing it. Is ...


0

Page 37 has clear limits for max current. You will have to reduce your current by /8 to drive 40 outputs.


1

Estimate the variability of \$V_\text{BE}\$ If you pulled and tested 100 adjacent BJTs from the same reel, at low enough currents that the Ohmic base and emitter resistances weren't much of an issue, you might see a spread of \$20\:\text{mV}\$ (or \$\pm 10\:\text{mV}\$) for the \$V_\text{BE}\$ across the group. And that's an exceptional situation. If you ...


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