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I simulated your circuit in LTspice using a BC547B. Here is the result:-
Before t = 0 the transistor's C-B and C-E parasitic capacitances are charged to 5V, and the B-E parasitic capacitance is uncharged.
When the 1V step is applied the B-E junction is effectively short-circuited by its capacitance, so all terminals jump up by ~0.33V as Rb and Rc||Re ...
4
You say you want precise numbers (I think you mean accurate) and you are not willing to assume that \$I_B \ll I_C\$. At this point you should use a good SPICE simulator instead of hand calculations. However, the model parameters for the transistor will also be approximations so those results won't be 100% accurate either.
You could build the circuit and ...
4
There's an easier way to find the DC Q points and R value starting the 5V for Vce1. (or UCE1 )
It is simpler because we were permitted to ignore I(R) thru T2 as it is only 0.5% more (hFE=100 * 10uA vs Ie2~2mA ). 10uA/2mA*100%=0.5%
answered Jan 20 at 0:41
Tony Stewart Sunnyskyguy EE75
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4
Miller input capacitance makes a RC low-pass filter with the internal resistance of the signal source. If you change the capacitance to negative in a RC low-pass filter, it seems still to have the same low pass function because the sign of the capacitance vanishes in the formula of the attenuation vs frequency due the 2nd power. My circuit analysis program (=...
3
Negative capacitors can generate energy...that is, they are active devices. So...depending on the surrounding circuitry, you may find your gain increases, or your circuit breaks into oscillation.
Another way of looking at this is to notice with positive Av, the feedback is positive, and positive feedback can be very dangerous ( or helpful, if it’s well ...
3
Let's try to find some sense in this formal classification...
Transistor vs circuit input. The input of the bipolar transistor is its base-emitter junction; so there we have to apply the input voltage. In the general case, the base-emitter junction is floating... but we prefer to deal with single-ended (grounded) voltages. Then we present the floating base-...
2
For high current loads. BJTs are a current-controlled device, so one might not provide enough current amplification. This is called the beta value and is normally around 100-300 for general purpose transistors
The base current is being amplified and projected through the collector to the emitter (in NPN). If it's common base, the current is the same through ...
2
Simply follow the current path. Look at this example:
Where each transistor has a current gain equal to \$\beta = 99\$
As you can see the input base current is very small due to the BJT's current gain.
And the input signal source see the \$R_E\$ resistor as much large resistor
\$R \approx \beta_1 \times \beta_2 \times R_E \approx 100k\Omega \$
2
Here is my answer for the Darlington combination (input resistance hie,D):
hie,D=hie,3 + hfe,3*hie,1 (the input resistance of Q1 appears at the base of Q3 multiplied with the current gain)
We have hie,1=hfe,1/gm1 with gm1=gm,3*hfe,1 (because Ic1=hfe,1*Ic3).
From this: hie,D=hie,3 + hfe,3/gm,3= 2*hie,3
Hence, the input resistance of the darlington ...
2
To look at schematic and decide if it looks right or not, first examine the DC Q points for each terminal voltage and current and consider the impedances that affect input loading, voltage gain and symmetry.
For example when you choose Rb,Re,Rc all the same value in a CE amplifier, there is no voltage gain (0dB).
A more useful ratio of values might have ...
answered Jan 4 at 0:18
Tony Stewart Sunnyskyguy EE75
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2
What you describe is the common situation where an emitter resistor provides negative feedback in the common-emitter configuration. Do you have an actual question?
Your language is confusing. What does it mean to say that "B is open circuit"...an open circuit requires the specification of two nodes. The upper limit for \$I_C\$ is determined, to first order, ...
2
I think your question has too many questions. So I'll answer what I'm motivated to answer. Also, you don't write much at all about the context for these questions. And that's important in allowing me to focus what I say. So I will be somewhat terse, as well, in reply because you've not motivated more from me. I hope I strike a balance you find acceptable.
...
2
D101 along with D105 do protect the receiver input from huge transient events, as OP surmises. They also limit amplitude of the transmitted signal going into the receiver. But D103,D104 are DC-biased with current from the RF GAIN control, to attenuate merely large input signals while receiving.
Common-emitter RF amplifier transistor Q101 has resonators (27 ...
2
Is there a chance that the change in the base voltage will decrease Vbe and force the transistor into the cut-off region?
Yes this is possible, and it would result in non-linear operation of your amplifier.
Is there any chance this might happen before first entering the saturation region?
Yes it's possible to (mis-)design an amplifier so this happens.
...
2
Power dissipation ,Current ,Voltage and Safe operating Area .On the Ap note circuit life is good because the input /output differential voltage is only 2V7 so power wasted is manageable with the Husky MJE4502 on a heatsink .You want to increase max current by 40% Which is not a show stopper .However your expected input /output differential will be more ...
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I guess this question is more about vocabulary than about BJTs.
"AC small signal analysis" or "small signal conditions" is a useful approximation to simplify the analysis of a circuit.
This approximation assumes the AC signal voltage wiggles around the DC operating point in a "small enough" interval that the operating point and all characteristics of ...
2
General Forward-Biased Solution
Here's the equivalent schematic you need to analyze:
simulate this circuit – Schematic created using CircuitLab
The solution of the KVL equation is:
$$\mid \:I_\text{B} \mid\:=\:\frac{V_T}{R_{_\text{TH}}}\cdot\operatorname{LambertW}\left[\frac{I_{_\text{SAT}}\cdot R_{_\text{TH}}}{V_T}\cdot e^{^\frac{I_{_\text{SAT}}\...
1
I'm ignoring C3, R3 and R (2 Mohm) because they are largely unrequired.
In a common emitter/base configuration, it is the collector resistor (\$R_C\$) that sets the output impedance. In your circuit it has a value of 100 kohm and this is in parallel with the collector (acting as a constant current source) so, if the BJT were ideal, your output impedance ...
1
Well, we have the following circuit:
simulate this circuit – Schematic created using CircuitLab
I used Mathematica to solve it, using the following code:
Solve[{Ix == I1 + I4, I4 == I7 + I8, I2 == I5 + I3, Ix == I3 + I6,
I1 == ((Vx - V2)/(R1)), I2 == ((V3 - V4)/(R2)), I3 == ((V4)/(R3)),
I4 == ((Vx - V1)/(R4)), I5 == ((V4 - V5)/(R5)), I6 == ((...
1
Here's a different approach. It's the way I thought about the problem:
simulate this circuit – Schematic created using CircuitLab
In the problem text, they suggest that \$I_{B_1}\$ and \$I_{B_2}\$ can be ignored (\$I_{B_1}\ll I_{C_1}\$ and \$I_{B_2}\ll I_{C_2}\$) for the purposes at hand. But that's obviously wrong. If it really were true, then there ...
1
Yes the AC signal superimposed on the DC bias voltage can take the base-emitter voltage into the cutoff region
Some amplifiers are even designed to have the signal always do this, these are called class B or class C amplifiers. Amplifiers that have the singal always be in the linear region are called class A
To know whether your amplifier is operating as a ...
1
Given;
uncompensated discrete amplifier with closed loop gain of 20 dB , (R1+R2)/R1= 10x
high Q resonance at 5MHz,
~20dB/decade rolloff >10MHz
0dB gain at 50 MHz which becomes the GBW product.
We could compute the open loop gain
and determine the ω/RC breakpoint for some circuit Req and added Miller C to achieve 0 dB gain
and then expect 45 deg phase ...
answered Jan 8 at 19:16
Tony Stewart Sunnyskyguy EE75
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1
Your output is saturated open loop due to offset between sensitivity to R5, R9 values and hFE assumptions. I can tell by your choice of R9 = 595 Ohms and Vout = 14.3V
Increase R5 to say 4k7 until you get in the linear output range and test Aol with 1uVpp input. You don't have to null it.
Also your 4 Vbe output bias circuit needs to be more than 3:1 R ...
answered Jan 7 at 0:37
Tony Stewart Sunnyskyguy EE75
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1
From the considerations above, we can conclude that, during the transition, the transistor can be thought as something slow... rather an integrator than an amplifier. So, to imagine how the transition is going, we can place ourselves in its place (empathy). The conceptual picture below can help understanding in this way. There I have visualized the voltages ...
1
1. Why do we use Darlington pair?
simulate this circuit – Schematic created using CircuitLab
The Darlington pair can be seen as a new single transistor with modified parameters.
The dynamic input resistance is rather high: hie,D=2*hie,1
The "beta-factor" (hfe) is large: hfe,D=hfe,1*hfe,2
The gain (transconductance) is reduced: gm,D=0.5*gm,2 .
...
1
How to make a better linear power supply. (Merry Xmas)
Assuming you had a bank of matched NPN 2N2955(PNP)'s on a 200W heatsink and a 40V Op Amp that senses to 0V input (PNP type) with any voltage reference like 1.25V bandgap ref. with a log scale 1k Pot to fine tune the more sensitive high V range near 0 Ohms.
This is what I imagined it to look like.
The ...
answered Dec 26 '19 at 18:33
Tony Stewart Sunnyskyguy EE75
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