# Tag Info

6

Yes, BJTs will work in reverse mode (collector and emitter switched). Most types will have very low gain (5 or 15 rather than 100 or 300) and the breakdown voltage is generally very low, less than 10V. A few are made to be symmetrical and don't much care which way you connect them.

5

It is simply a test circuit for you to test and observe. It is not a "complete" circuit in the sense you expect - it is not a "switch." The point of the exercise is for you to carry out the experiment and develop an understanding of how to assemble the circuit and test setup, and then to gain an understanding of how the base current of a ...

5

As we know the signal is supposed to apply at the base of the transistor, This is true for a common emitter or common collector stage, but for a common base stage, the signal will be applied at the emitter. "with no input signal applied" means the base voltage is zero at some point, This is not correct. "No signal" in this context ...

4

When is a MOSFET more appropriate as a switch than a BJT? Answer: 1) a MOSFET is better than a BJT when: When you need really low power. MOSFETs are voltage-controlled. So, you can just charge their Gate once and now you have no more current draw, and they stay on. BJT transistors, on the other hand, are current-controlled, so to keep them on you have to ...

3

The TTL input multiple-emitter transistor works in such a reverse mode when a logical "1" (+5 V) is applied to the input emitter. Then, in contrast to the usual common-emitter configuration, the emitter is connected to Vcc and the collector to ground. As far as I can remember, this mode has been also used in the past in transistor switches due to ...

3

A differential amplifier generally has two transistors for symmetry, because in many applications, it is desirable for both inputs to have similar characteristics, and for them both to be referenced from the same node (e.g., ground). You can create a differential amplifier from a single transistor by applying the signals to the base and the emitter, but the ...

3

You should know that the current amplification factor $\beta$ or $h_{fe}$ is of bipolar transistors almost never has a very predictable value. Go look in the datasheets of these transistors (or any other bipolar transistor!) and look at the minimum and maximum values of $h_{fe}$. For example in this datasheet when $I_c$ = 15 A, $h_{fe}$ can vary ...

2

Theoretically you could use the transistor's current gain to calculate the required Base resistor value. However this is a poor method of determining Collector current because (due to process variation) the current gain may vary widely between individual parts, as well as with temperature and (to a lesser extent at medium values) Collector current and ...

2

You can derive this from the Shockley equation: *Vbe = Vt(ln(Ic/Is)) so at two different currents Ic and Ic0 we have $V_{BE} - V_{BE0}$ = $V_T (\ln(I_C/I_S) - \ln(I_{C0}/I_S))$ = $V_T \cdot \ln (I_C/I_{C0})$ At room temperature Vt = kT/q is about 25mV. By using two currents, the saturation current Is cancels out, so you can predict the behavior ...

2

You don't need to prove it if you accept that an op-amp (even quite modestly priced op-amps) that use negative feedback, seek to ensure that the two input voltages (+Vin and -Vin) are identical. In effect, what ever voltage you apply to +Vin (Vi in your example) results in the same Vi appearing on -Vin. This inevitably means that: - I_{R_3} = \dfrac{V_i}{...

2

I think you are misreading the logarithmic Y axis (which is simplified for readability) Take that hFE reading at 15A. Starting at 10. the SECOND line is labelled 20 (the first is approximately the right place for 15). Then the next three are 30,40,50 ... and the curves intersect 15A approximately on the 40 line, as the datasheet says. At Ic=1A, both curves ...

2

Your signal sources are never ideal, nor are they always floating. Connecting them in series will cause them to distort each other at best, and short each other out at worst. The transistors are there so the sources can drive their own higher impedance input so that they do not affect each other, nor are loaded down by the input as much. It also makes it so ...

2

You can examine the distortion of a bipolar transistor under the combined DC_bias and the AC_peak_peak excursions. If I recall rightly, 4 millivolts PeakPeak will cause 10% distortion. Understanding this will be a big step forward in applying bipolars in your circuits.

2

The minimum β is 75, but the simulator model probably uses a "typical" value instead of the minimum value. In fact, your simulator says it is "powered by LTSpice", and the LTSPice 2N2222 model has a forward $\beta$ of 200: In the real world, the $\beta$ parameter is not very well controlled from device to device, so we usually try to ...

2

I believe such symmetrical bipolars were once used as analog multiplexors. =================================== Regarding the use of 2N2484 (?) transistors, what was crucial params of that part number?

1

Trying to estimate the actual beta the BJT is using in my circuit gives me something like 197, it makes no sense at all looking at the datasheet. Using the data sheet graph at 11 mA collector current: - According to the graph β is a shade under 200. Using a minimum figure is bound to be creating a discrepancy with a simulator that will use typical values. ...

1

Constant input impedance implies a FLAT FREQUENCY RESPONSE, very useful in amplifiers with feedback where high_frequency poles causes poor phase_margin.

1

Is there any reason MOSFETs in Colpitts oscillators seem so rare? The most basic problem with a MOSFET in a colpitts oscillator is that the gate-source junction is much more capacitive than either a JFET or a BJT (like a hundred to a thousand times more). So, that gate-source capacitance throws a big wrench in the works. I mean, they can work (sub 1 MHz for ...

1

Your question is about the so-called "compliance voltage" - a very important property of current sources (despite the respectful name "current mirror", here Q2 is just a simple transistor current source... or, more correctly, a sink). When you begin increasing RL, the voltage drop VRL across it will begin increasing as well. To compensate ...

1

With 10mA thru the transistor (eyeball Ic), the transistor is in saturation. I would do this. Rbase_upper = 10K Rbase_lower = 5.1K Rcollector = 1Kohm Remitter = 1K ohm Expect about 7 volts on the collector Expect about 3 volts on the base. Expect about 2.3 volts on th emitter. That gives 2.3mA Ie and Ic. That gives 'reac' of 11 ohms. If we could ignore the ...

1

I think the real purpose of this setup is to measure and draw the input IV curve of the transistor - the function Vbe = f(Ib), which is actually the IV curve of a diode (p-n junction, as you noted). "Ib vs Vbias" does not make much sense because practically, it is the IV curve of the resistor Rb. "Sweep the value of Vbe" is not so correct ...

1

Q point is the point where a non-linear circuit (like diodes, transistors, etc.) can be approximated by a linear circuit, and we get a DC set of voltage and current of that non-linear circuit.

Only top voted, non community-wiki answers of a minimum length are eligible