# Tag Info

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Your 1K resistor is way too low value. A high value resistor will also reduce LED current changes due to reduced current changes in transistor due to supply voltages. Experiment with 10k-100k

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When transistor is in active state the Base-Collector junction needs to be zero or reverse biased. When $V_{BE}=0.7V$ and $V_{CE}=0.3V$ you get $V_{BE}-V_{CE}=V_{BC}=0.4V<0.7V$. It's not important for Base-Collector junction to be forward biased hence your confusion.$V_{CE}$ can be as low as $0.2V$ ($V_{BC}=0.5V$), and transistor would still ...

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$A$ is the area. $I_S$ is the device's saturation current, which is proportional to $A$. The device with area $nA$ would have saturation current $n I_S$. So your second line should have been: $$= V_T \ln \left( {{I_C}\over{I_S}} \right) - V_T \ln \left( {{I_C}\over{n I_S}} \right)$$ (Note $n$ in denominator instead of numerator) The rest ...

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Mark - what do you want: A "demonstration" (calculation) or an explanation of this effect? Let me start with an explanation: The MILLER effect concerns the input impedance of your circuit at the base node. Here, we see two pathes: (a) into the base and (b) into the 10pF cap. The input impedance is defined as the ratio of the input signal voltage divided by ...

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Well, we have the following circuit: simulate this circuit – Schematic created using CircuitLab When analyzing a transistor we need to use the following relations: $$\text{I}_\text{E}=\text{I}_\text{B}+\text{I}_\text{C}\tag1$$ Transistor gain $\beta$: $$\beta=\frac{\text{I}_\text{C}}{\text{I}_\text{B}}\tag2$$ Emitter voltage: $$\text{V}_\text{BE}=\... 0 @vtolentino already got it right in the comments: in DC analysis all capacitors are open-circuited, it should be fairly simple to find what you want after that. I'd like to add that this BJT is not a Common-Emitter BJT, it's a Common-Base BJT because the base terminal is common to both inputs and outputs, so you're dealing with a CB NPN-BJT. 1 The current through the base-emitter junction of a transistor is governed by the diode equation: $i_e = I_S \left(e^\frac{v_{be}}{n\ V_T} - 1\right)$. At room temperature, $V_T \simeq 26\mathrm{mV}$. So unless your peak-peak input voltage is significantly smaller than $26\mathrm{mV}$, you'll get distortion. On the bright side, you have lots of ... 0 Say I choose a bias point on an I-V characteristic curve at the IC = 16mA. If this transistor has a beta of 200 then that means a IB = 80uA. Lets also say I want 1 volt across the emitter resistor R2. So first what we do is to select R2 resistor value:$$R_2 = \frac{1V}{16mA} = 62\Omega  Now we want a "stiff" voltage divider which means that $R_{... 0 (EDIT: I've completely rewritten my answer to hopefully provide a better answer.) Consider the application of Kirchoff's Current Law for the two cases where the transistor's base is and is not connected to the midpoint of voltage divider R3 and R4. Let$V_X$be the voltage at the midpoint of voltage divider R3 and R4. CASE 1 - The Transistor's Base Is ... 1 Usually the current in the voltage divider is 10 times the base current because a transistor part number has a range of beta. Your transistor might have a beta that is 100 minimum, 200 typical and 300 maximum. You get whatever is available. 0 VTH is a little greater than 1.7V. R3 & R4 values would be chosen to give 1.7V at the base of Q1 taking into account the base current which is drawn through R4. This base current increases the voltage drop across R4. Now if we disconnect the transistor, the voltage at the junction of R3 & R4 will rise to the value of VTH which is a slightly higher ... 1 "As the base current increases the slope of the line increases, resulting in an increase in output impedance with increase in base and collector current" They got it wrong - as the slope increases you get more amps per volt so conductance increases and conductance is the inverse of resistance. 4 That resistor speeds up turnoff and increases maximum allowed collector circuit voltage. The turnoff speeding is based on the fact that the charge which is stored to the BE junction is dissipated in resistor R2. The effect is substantial in pulse circuits. There's some leakage from C to B and it can cause substantial unwanted base current at high operating ... 0 Transistors are not perfect switches. They have parasitic capacitances. When a base current is applied, some of them charge up and and when base current is removed, some will discharge ($Cπ$). A similar behavior happens when a voltage is applied to the collector ($Ccs$). This takes time, and thus the RC curve you see in your graph. Here is an small ... 0 There is already a similar question, however the design procedure is still misssing there. Now to your questions: I can't really find characteristic curves for transistors on datasheets which makes it difficult. Although there is no output characteristic curve in the datasheet, you can find specific biasing points which might already fulfill your needs. ... 0 That 2.2uF is connected the wrong way around. The +ve should be towards the higher voltage level. With no signal from the mic all three terminals of the pot are at 0V. With a mic output signal the signal at the top of the pot wiggles about 0V and the wiper signal also wiggles about 0V but is a reduced amplitude version of the signal at the top of the pot. ... 1 The 2.2uf is WRONGLY installed. god eye for catching that, Jeffrey. Since the VDD is Positive, the base of Q2 will be about 0.6 volts above ground, thus the 2.2uf cap needs its (+) terminal to the base. ============================= Unfortunately, during the power-on transient as VDD quickly rises to +12 volts, the top pin of the volume control will ... 1 The two capacitors are DC blocks, conducting the (AC) signal while blocking the DC so that different circuits can work at different bias levels. The pot is just a volume control. 2 The 47k component appears to be a potentiometer, using a somewhat nonstandard symbol. The center/side connection going to the capacitor is the wiper. The capacitors are being used as DC blocking capacitors, preventing any DC link between the stages of the amplifier and decoupling the microphone's bias voltage from the signal. As @BrianDrummond and @... 1 Just draw the small signal model of the circuit to calculate the impedance seen at the emitter. Assuming a resistance$r_b$is connected to the base, the impedance is calculated as: simulate this circuit – Schematic created using CircuitLab Apply KCL at the emitter to give: \beta i_b + i_b +i_x = \frac{v_x}{R_E} Apply KVL in the emitter-base ... 0 Well, we have the following circuit: simulate this circuit – Schematic created using CircuitLab When analyzing a transistor we need to use the following relations: \text{I}_\text{E}=\text{I}_\text{B}+\text{I}_\text{C}\tag1 Transistor gain$\beta$: \beta=\frac{\text{I}_\text{C}}{\text{I}_\text{B}}\tag2 Emitter voltage: \text{V}_\text{E}=\... 1 I think that expression should actually be:- ((R1//R2//Rs)/Beta) + re where Rs = signal source resistance In the ideal case of Rs = 0 the left hand term disappears and the equation reduces to re. 1 Looking from the emitter terminal, collector is a current source whose current cannot be changed. It offers infinite resistance and acts as an open circuit:$R = \dfrac{\Delta V}{\Delta I} = \dfrac{\Delta V}{ 0} = \infty$Thus$R_C\\$ doesn't affect the impedance seen from emitter side. You can obtain the resistance of a block by measuring the voltage ...

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After replacing the 1N4148 diodes with the much stronger HER305 diodes, the clipping disappeared, even though the 1N4148's seemed in fine working order when tested with a multimeter. Still not sure why this worked, but it did. I was able to push the output wave up to 13.7V @ 3 MHz. And for those who were asking, this is what my amplifier looks like (...

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The 2N2369 used to be popular for 500 picosecond risetime pulses, in relaxation circuits. Given the dominant spec on that transistor was the 1.6 GHz Ftau, th precise dopings and thicknesses would vary from vendor to vendor. Back then either Motorola or Fairchild likely would have provided the device.

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Q1 emitter-base short circuit. Check out Q1 or replace it! The effect is also apparent in the absence of 15 volts of the Q1 collector.

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I'm not sure why you AC-coupled your input and output, given that you have balanced power supplies, which makes it unnecessary. The point that the commentors are making is that your bias network doesn't do what you think it does. You're assuming that the junction between the two emitters is close to zero volts with no signal. But in fact, the actual voltage ...

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Heating IS produced inside the transistor. You can easily calculate how much (voltage × current, averaged over time), and show that it is well within the transistor's ratings. You can't substitute a zener diode for the transistor, because it doesn't have the negative resistance characteristic that the transistor has.

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In your circuit, you are still not driving the BJT into deep saturation region (switching behavior), since the base voltage (760mV) is slightly larger than the diode's forward voltage (650mV). If you sweep the base voltage, you would get the following curve. Here you still see that the output current still has some strong dependency on the base voltage ...

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Here is an LTSpice XVII circuit which (at least in simulation) does what you want: When the 3.3 voltage source is connected (representing a HIGH on your microcontroller pin) there is 17mA running through your 5-LED string. When the 3.3 voltage source is disconnected (by deleting the little bridge inside the circle) the current running throught D1 is ...

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Based on my testing (see comments), the glow can be caused by transistor leakage. In my setup, the glow starts when the voltage is 26 volts (could easily vary depending on the transistor). Capacitive coupling would have a similar effect. One 10k resistor across all 5 LEDs removes the glow in my setup. This works up to 31 V, the highest that my power supply ...

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This is likely not due to a flaw in the driver, or in the signal source, since only some flicker. Because LED's are not absolutely identical, the voltage across each in a series string will not be equal for DC due to differing leakage currents, and for AC this would be exacerbated by differing capacitance wihin each device and from device to ground. ...

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this prevents the voltage divider's output from lowering under loading conditions This is correct. R1 and R2 act as biasing resistors -- they're meant to generate a DC voltage that biases the transistor in a suitable operating point. The signal itself is injected via C1. You could think of the four currents going into or out of this node: those from/to C1,...

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When calculating output impedance why are RB and rbe neglected? Because the lower end of the current source and rE are grounded. End of story. The current through Rc is modified / dependent on external load currents.

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Well, we are trying to analyze the following circuit: simulate this circuit – Schematic created using CircuitLab Using Mathematica, I wrote the following code: In:=FullSimplify[ Solve[{Ix == I1 + I2, I1 == Ix + I0, I2 == Iy + I6, 0 == Iy + I0 + I5, I6 == I3 + I4, I5 == I3 + I4, I1 == (Vx)/R1, I2 == (Vx - V1)/R2, I3 == (V1)/R3, I4 == (V1)/...

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My question is how these two seemingly contradicting characteristics can be reconciled. It's all covered or implied in another curve called the safe area of operation and this curve factors in time or duration: - So, for example, with a current of 1 amp and a collector emitter voltage of 10 volts the power dissipation is clearly 10 watts but this can't ...

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In 1975 when I graduated working for Bristol Aerospace, a gold medalist EE Bill Whitehead was my mentor and he showed me these RLC Impedance graphs. Since we used a slide rule before the LED calculator came out in the early 70's I understood how to do log math on a slide rule. Since I assume you have some Impedance math skills, you ought to be able to do ...

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The problems that I see with this circuit: No power supply decoupling. The gain at the collector is 12 at most (1200/100), which may be too low. Loading at the collector by the feedback pi-filter, (and the oscilloscope!) will further reduce the gain. No adjustable bias voltage. The emitter bias current controls the gm (transconductance) of the transistor. ...

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Where is your VDD Cbypass cap? You need a 0.1uf from the top of R1 to ground.

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My question is how do you determine which ones are in cutoff mode and which one is on When you are first learning, use the method you were taught for analyzing any non-linear circuit: Guess and check. First guess the state of each nonlinear element. Then analyze the circuit with those assumptions. Then check that there's no logical contradiction (for ...

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https://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem What is the Thevenin-equivalent voltage at the 2-resistor node? Hint: Disconnect the base. What is the Thevenin-equivalent resistance? Re-connect the base. Solve.

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1. Block diagram. 2. Operation. Here is how I have visualized ECL operation in three steps: Input logical '0'. Transition 0 -> 1. Input logical '1'.

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To understand a new circuit means to see well-known circuit concepts and building blocks in the new circuit solution. Let's apply this technique to the present circuit. Structure. This is a digital circuit - a voltage level converter, operating as a switch with two states. The input voltage of 0...3.3 V is converted to output voltage of 0...12 V by two ...

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Noise margin is the reason. Tho the high value of R11 reduces the noise margin. However the R11 responds well to FET gate charges splitting into gate R , and also coming thru the bipolar Cob (base-collector). this will appear as "snappy" switching, or as outright oscillation.

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You have to use a power amplifier or maybe a line driver. These type of opamps can supply more current than the standard opamp. (You can find some e.g. at TI website under Amplifiers/Line drivers or Amplifiers/Power Opamps). If you are sure that you don't need more than 20mA, you might also find some standard opamp. Take care, that the higher the output ...

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