New answers tagged

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Confirm DC signal is not clipping Balance Bias for Vce= (Vcc-2V)/2 Increase DC gain with more stages Your circuit has only 1 stage of voltage gain (Q2) divided by the R ratio feedback gain , this excess gain is the feedback towards nulling the quadratic non-linearity of Ic vs Vbe. It’s not enough but could be a lot better at lower input signals and ...


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No. When \$V_E\$ increases, \$I_E\$ decreases, because \$V_{BE}\$ decreases. Any extra current that flows through \$R_E\$ is supplied by the signal source, not the transistor.


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Q5/Q6 could just as easily be a darlington and it would work fine although it would slightly reduce the maximum negative output swing as it would saturate at a slightly higher (less negative) voltage. Usually the diode labeled as Q3 would be a diode configured transistor which then, with Q4, forms a matched current mirror. This current mirror then acts to ...


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let’s consider linear vs non-linear differences. We know from datasheet typical curves and min/typ/max specs how hFE varies with Ic, Temperature and Vce. since Zout is a direct function of hFE, and impedance applied to the base it is important to understand some variables and how impedance control rise fall time into some fixed charge , Q or estimated Ciss ...


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First look at the equation. You have Y,P and a desired range (upper bound) of length (l not L). It's trivial to convert that into a range of available frequencies with a lower bound, so do that first, and choose an achievable frequency ( >= that lower bound). If that is achievable (the lower bound encompasses the frequency range you want) you can compute ...


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First the \$I_{CBO}\$ current will flow into a "positive" direction in the collector (electrons current in NPN). Thus, \$I_C = I_{C_F} + I_{CBO}\$ where \$I_{C_F}\$ = "forward current". But for the base terminal, the \$I_{CBO}\$ current is a "negative current" (holes current) \$I_B = I_{B_F} - I_{CBO} \$ And now after we ...


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The BJT is typically operated in one of three broad domains: cutoff: The base-emitter voltage is below some assumed threshold. For example, one might assume values below \$V_\text{BE}\le +400\:\text{mV}\$ for a silicon NPN. (\$V_\text{BE}\$ may be forward-biased, but only a little. It can also be reverse-biased.) The collector pin floats, in the sense that ...


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The base emitter voltage VBE causes an emitter current to flow in a BJT. In a perfect transistor, this would all flow to the collector. In a real transistor, some of this current flows to the base as well, due to various things like the finite thickness of the base. The ratio of collector/base current is often called β, or hFE. It can be in the hundreds for ...


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It is not a surprise. Notice that in rule 3 you neglecting not only the Early effect (\$r_o\$). But also a fact that: \$\frac{1}{g_m}\$ << \$r_\pi\$ Aditional notice that $$\frac{1}{g_m} || r_\pi = \frac{1}{ g_m + \frac{1}{r_{\pi}}} = \frac{r_{\pi}}{g_mr_{\pi} +1} $$ Also, did not forget that : \$ g_mr_{\pi} = \frac{I_C}{V_T}\frac{V_T}{I_B} = \frac{...


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The transistor is a voltage-controlled current source, moderated by the voltage typically around Vbe=0.6 for Ic=1mA and it is exponential. The smaller turns of coil to the base is like a step-down transformer which also lowers the impedance to the base or conversely raises the low base-emitter incremental resistance when conducting that the load sees on the ...


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The schematic is roughly this: simulate this circuit – Schematic created using CircuitLab This is the same as (using a Thevenin divider approach): simulate this circuit From this it is easy to work out (from KVL) that: $$V_\text{TH}-I_B\cdot R_\text{TH}-V_{\text{BE}_1}-I_E\cdot R_\text{E}=0\:\text{V}$$ But also you know that \$I_E=\left(\beta+1\right)...


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It can be shown - as far as DC bias point stability against tolerances and temperatur changes is concerned - that the best bias method is to use a resistive voltage divider at the base node for providing a bias voltage Vb as "stiff" as possible. For this purpose the resistor niveau should be as low as possible - depending on some limits set by ...


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With an BJT transistor in a linear circuit, you need to make the collector current (Ic) not depend on the HFE (used to be called Beta) of the transistor. The usual method is to make some resistor on the emitter have a lot more voltage drop than the Vbe (Base to Emitter) of the transistor. This can be done by having a positive and negative supply voltage and ...


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One thing is simple which you have to fix in your mind that we bias transistor in order to make Operating point (Q-point) stable so that we could get faithful Amplification. Now as Operating point depends on two factors, 1) Collector Current Ic 2) Voltage a=across collector and emitter Vce. So in base bias we we analyze input part we get a base current, we ...


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TThe dynamic resistance \$r_e\$ is nothing else than the resistance of the parallel combination of \$r_{\pi}\$ and the current source (nomenclature as in the figure): Test voltage: \$v_T=i_T r_e\$ with \$i_T=i_b + \beta i_b\$ and with \$i_b=\frac{v_T}{r_{\pi}}\$. From this you can solve for \$r_e=\frac{r_{\pi}}{\beta+1}\$ By the way (physical interpretation):...


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Ohm's Law only applies to ideal resistors. It doesn't apply at all to non-linear elements like transistors and diodes, and can have significant errors when applied to real-world resistors. Ohm's Law is an approximation...a model... of real behavior. And George Box said it best: "All models are wrong, some are useful."


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The term for what you're looking for is probably a VCXO. These exist in commercial forms, but one common arrangement is to use a variable capacitance in parallel with the crystal as follows: In practice, varicaps can be used; they are a diode, constructed to have a capacitance that varies with the DC bias across it. One realization described in this source ...


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I would add a simple intuitive explanation to the detailed explanations in the links provided. It is based on seeing well-known electrical building blocks in this electronic circuit. Voltage divider With a little more imagination, we can see two cascaded voltage dividers here: 1. High-resistance voltage divider made by R1 and R2. It is a static voltage ...


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40 dB is 100:1 is close enuf, but that CE/CC amp won’t give you the current gain with feedback that you haven’t specified. Typically a speaker power amp has a damping factor of 100 which means the woofer is damped if you have one , such that back EMF current is reduced 99% or the driver output impedance is 1% of the speaker or 80 mohms roughly. ...


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The chosen value for RE depends on the value of RL. If RE is too big compared to the value of RL then, when Vout swings negative, at some signal amplitude the current through RL will equal the current through RE. This will reduce the current in the transistor's emitter to zero mA and negative clipping of the load voltage will occur. So you need to ensure ...


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This is a classical emitter follower. The base voltage is set by Vcc*R2/(R1+R2). Due to the negative feedback loop, the emitter voltage "follows" the base, but with a diode drop. This in turn determines the voltage across RE (call it Vout_int), the DC value of which determines the NPN's bias current. It also determines Vce, since Vce = Vcc - ...


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1mA is just a typical and convenient value for the collector current. The higher the chosen collector current, the lower can be the value of the collector resistor to achieve mid-supply dc biasing. Since the collector resistor nominally defines the output resistance of a common emitter amplifier it means that a lower value collector resistor results in less ...


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I was able to fix the circuit I posted above by simply changing the value of the capacitors from 1uf to 1nf Though if I added any value of resistance ( to the VCC) , it completely destroys the output. What Can I do about this ?


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You cannot connect the emitter of a PNP transistor to +5V and the base to ground. That will burn out the transistor. Instead, you must include a resistor, either between base and ground, or between the emitter and +5V. I have only seen the later today, when researching I2L logic. (Actually it seems pretty interesting). Although I don't have a circuit for a ...


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There are a few quick techniques you can use that do not ignore the base current. But I'll discuss just one here. My recommendation is that you put a little time into it, think about the results and whether or not it's worth it for the future or if you now want to decide that the technique the author used is "good enough." Before I discuss it, let'...


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You have already answered your own question. When performing hand calculations we usually simplified a bit to speed up the calculation process. Thus, the author decided to neglect the \$I_{B2}\$. Also, the \$I_{B2}\$ current will be small compare to the \$I_{C1}\$ current. And such a simplification is justified because in the typical application when BJT's ...


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Answer Let us study the circuit design of a real High level trigger 5V relay KY019 shown below. The relay characteristics, operation, and photo are shown in Appendices A ~ C below. The KY019 relay uses a Songle 5V relay switch, with nominal current of 90mA (measured 70mA). GPIO High is 3V3. For the OP's 12V relay switch with smaller nominal current of 40mA,...


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The current guaranteed to switch a relay is amplified by a saturated NPN here that only has about 10% of it’s maximum linear hFE and current drive capacity far greater than the load to support a low Rce/Rcoil ratip to dissipate the same ratio of load power e.g. a PN2222A has an Rce = Vce(sat)/Ic ~2 Ohms , so a 40 Ohm coil means the transistor uses 5% of the ...


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The base current, if the GPIO input is actually at +5V, will be approximately (5-0.7V)/R. You should pick the R to provide 1/10 to 1/20 of the DC current that the relay draws, according to the relay datasheet. That's a rule-of-thumb with a typical jellybean NPN transistor. If you study the transistor datasheet you can usually find Vce(sat) values at Ic/Ib = ...


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Martel, try this: simulate this circuit – Schematic created using CircuitLab It should be fine. I think \$Q_3\$'s base-emitter junction might be placed under some momentary reverse voltage stress. So I added \$D_3\$. As indicated in comments, I used \$R_1\$ and \$R_2\$ in order to limit shoot-through currents and also to make the behavior more ...


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Even using a much simpler circuit, with both transistors working as voltage follower, and a collector resistor to limit the initial current, this attempt almost reaches your goal: You have not mentioned the ripple that is acceptable:


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Bipolar transistors had a head start of about ten years over MOSFETs so they were adopted first. They were, and remain today, somewhat less expensive than MOS to manufacture, at least for discretes, because they require fewer process layers. Although eclipsed by MOS for logic some time ago, bipolar devices and techniques are still widely used in power, mixed-...


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First of all, MOSFETs are voltage controlled transistors while BJTs are current controlled transistors. MOSFETs form an excellent insulator Silicon Dioxide SiO2 during the manufacturing process. SiO2 is used as an insulation layer between the gate and the conducting channel and provides an excellent isolation. Yey, it's a good dielectric material. MOSFETs ...


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When I first started (in the 1970's) MOSFETs were brand new and had limited power and voltage capability. Engineers were comfortable with BJTs so they persisted with them as MOSFET performance improved. Also, the industry was tooled up for BJT manufacture, and MOSFETs required new capital investments. BJTs still have their place in specialized ...


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According your description you dont want feed fan with pulsed voltage but continous DC, right? Than you need an amplifier, or even better controlled current source. How do you generate the PWM, because for both solution is better a voltage signal (not PWM). If you want PWM to controll DC you must use a RC filter after PWM to obtain DC controll signal.


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Linear Technology (now a part of Analog Devices) has a ton of voltage monitors and hot-swap controllers that can handle positive and negative voltage combinations, have on-board MOSFET drivers, etc. There is a catch-22 (metastability) issue with the task, that is solved by delays. Since at power-up no two supplies come up at exactly the same moment, there ...


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If you put the equations correctly: point C : io vo ve ib point E : vo ve ib ib def : ve ib 3 unknowns : (vo OR io) AND ve AND ib to get vo = function(io) OR io = function(vo)


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How about something like this? When the voltage difference between the rails is lower then the zener voltage, no current can flow. This means no voltage drop across R1/R2, keeping both transistors off. Otherwise current can flow trough the zener, when the drop across R1/R2 is larger then 0.7V both transistors turn on. R3 limits base current for both ...


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The BJT is technically working as a booster, it amplifies in power (or current, to be exact), in this case. The opamp senses the current thru the emitter resistor. It then drives the base so that the current is equal to the 'programmed' value (on the non-inv input). Thanks to retroaction this works independently of Vbe, hFE or whatever other condition. ...


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A typical opamp can’t supply more than 25mA (obviously this change from manufacturer) That’s the purpose of the bjt: supply the extra current


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In order to check the proper functioning of the circuit, I recommend the following steps: 1.) Connect both input nodes to ground and verify if there is proper DC bias point 2.) Set one of the inputs to zero and check if the output signals at both collector nodes are equal in magnitude but opposite in the phase. 3.) Do the same with the other input node(s). 4....


1

To top up on Jasen's answer and previous comments: i) the differential amplifier is, namely, an amplifier, with a gain that is defied for differential mode and for common mode; it is usually used for diff mode, so the Vout (difference of collector voltages) is ratioed to the input voltage (the voltage applied to the two bases, from which the comment from ...


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The differential pair is one amplifier, not two. The two sides are linked by sharing a single emmitter resistor, because of this a signal on one side has an inverse effect on the other side.


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Technically, this is not an answer, but an extended comment. It would be superfluous to add one more answer: @Neil_UK gave a comprehensive answer. The aphorism 'All models are wrong, but some are useful' is enlightening, and Wikipedia has a more straightforward article related to the subject matter, Bipolar junction transistor. The paragraph Voltage, current,...


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The BJT is a physical device - and, of course, it is possible to describe how and why it works. It is not a problem to show that and why the BJT is a voltage-controlled device following the well-known exponential Shockley equation Ic=f(Vbe). That is not a "model", it is a description of physical properties - however, somewhat simplified because the ...


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All models are wrong, but some are useful Neither of the expressions you give is correct, as they ignore the collector voltage, β is not a constant, and the temperature sensitivity is rarely accurately known enough. However, they are both useful. If the base is being substantially current fed, so from a high impedance, then the β model is most useful. If we ...


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It's better to go back to the original papers to understand something well. In this case, "Effects of Space-Charge Layer Widening in Junction Transistors," by J. M. Early, 1952, from the Proceedings of the I.R.E. Here's the first diagram from that paper (illustrating an NPN BJT): While Shockley\$^1\$, in 1949, and then Shockley, Sparks, and Teal\$^...


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Question: "Since now the base is narrowed and there would be less chance of recombination, so most of the emitter pumped electrons will go to the collector, so does it mean the base current will reduce?" Yes - for constant Vbe the emitter current Ie will be constant and because of Ie=Ic+Ib the base current will reduce for rising Ic. Nevertheless, ...


3

The chance is less because electrons "flight time" in the p region is short. The base transit time is of minority carriers (electrons) in the base, at low current injection densities, is: Tn = Wb ^ 2 / Dn Where Wb is the base thickness and Dn is the electron diffusion constant. Source (equation 5.22): https://www.sciencedirect.com/topics/...


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This is the simplest circuit I can think of using BJTs that will provide under-voltage lockout. simulate this circuit – Schematic created using CircuitLab How it works: If Vin is higher than the the Zener voltage of D1 plus 1 diode drop (Vbe of Q1) then D1 will conduct, providing base current for Q1. Q1 will conduct providing base current for Q2, and ...


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