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0

Check any electrolytic capacitors in the audio amp section. Could be they have dried up and are killing your audio.


1

The simplest calculation is to assume the base current is 0, then Vb = \$Vcc\frac{R2}{R1+R2}\$ and Ve = Vb - 0.7V. Once you know Ve you know the emitter current Ie= Ve/Re, and since the base current is assumed to be zero, you can find Vc = Vcc - Ie * Rc. That is valid if \$\frac{R1R2}{R1+R2}\$ << \$\beta Re\$ If you want more accuracy you ...


0

You would really need to know the output resistance of the input voltage source to Vb. Without it you can't really know for sure. If an op-amp or a similar amplifier stage drives Vb, then the output impedance should be quite low. If it's low, then it's as if you're applying a true voltage source to Vb. In that case, all of the bias resistors really don't ...


1

You must analize the Thevenin's equivalent circuit: where $$ R_{th} = \dfrac{R_1\cdot R_2}{R_1 + R_2} $$ and $$ V_{th} = \dfrac{V_{CC}}{R_1+R_2}\cdot R_2 $$ then, the equivalent circuit where \$V_b\$ is \$V_{th}\$ and \$R_b\$ is \$R_{th}\$ Then, you can write KVL, for the input circuit: $$ V_b = i_b\cdot R_b + V_{BE} + i_b\cdot (\beta + 1) \cdot ...


5

instead it offers a resistance called ac resistance. why it is so ? why dont it rectify ? Let the voltage across the junction be of the following form $$v_{BE} = V_{BE} + v_{be}$$ where \$V_{BE}\$ is positive and constant while \$v_{be}\$ is AC and is, in some sense, small (I'll later clarify what that means). Now, recall the equation for the base ...


2

You will never be able to get accurate amplification of a 1mV full scale DC signal with a simple circuit using discrete components (I do know a way, but it's not simple). Even a crappy LM324 will do better (because the transistors are matched and on one chip). Tell us what you have and we can, perhaps, help. An instrumentation amplifier can be made with ...


3

The full operation of the circuit is analyzed in two parts: the small-signal model and the bias circuit. In the bias circuit, one considers the nature of the rectifier base-emitter base-collector junctions to establish the operating point (point Q), together with external components needed. In the small signal model, considering all the factors that alter ...


2

It is rather a property of the small signal model than a property of the diode you are applying the model to. It is the very purpose of the small signal model to look at any non linear device as if it was a linear device, i.e. a resistor. Of course it can only be applied with restrictions, e.g. that the function is differentiable at the operation point. ...


3

Because if you look closely enough at a small enough portion of a nonlinear curve it looks pretty much like a straight line. There's a detailed mathematical treatment here, but I'll also reproduce the Wiki graph to illustrate intuitively (I hope) what is going on: In the small signal model we are looking at very small changes from the bias (operating ...


0

A circuit using a phot-interrupter would look something like this: simulate this circuit – Schematic created using CircuitLab D1 and Q1 are inside the photo-interrupter. There will be a gap in the package between them so something can interrupt the light path between the LED and photo-transistor. If light can pass from the LED (D1) to to Q1, ...


0

The easiest approach is to use logic-level output photointerrupter like this one. There is no need for two power supplies, connect the transmitter side through a dropper resistor to set IR LED current (usually 50mA at 1.2V). Output of this particular series is open collector, so if your load draws less than 50 mA,you can connect it directly between output ...


3

The variable g is used for what are called the inverse hybrid network parameters. It is indeed defined as \$g_{22}=\dfrac{v_2}{i_2}|_{v_1=0}\$ The inverse hybrid parameters are very rarely used in practice, and they don't have conventional names except "the inverse hybrid parameter \$g_{22}\$". The z parameters are conventionally called the impedance ...


1

Your first answer may be what they are looking for, if one assumes Vbe and \$\beta\$ are constant. The second is slightly off, as you failed to take into account the base current. The third is way off. It's approximately 100K/\$\beta\$ but I'll let you work out the exact value.


1

No, the output impedance is not zero. To figure it out, try increasing or decreasing the value of the current source by some small increment, and then calculating the resulting change in output voltage. ΔV / ΔI gives you the effective impedance of the resistor+transistor circuit, which is also the output impedance, since the impedance of the ...


1

When we specify output impedance of a two port network, does it imply open circuit output impedance or Thévenin equivalent impedance(short circuit output impedance)? If an output produces a 1V signal when open circuit and can drive a 10mA current signal when short circuited, the output impedance is 1V/10mA = 100 ohms. It's the same as have a ...


1

It's the equivalent of sticking a resistor on the output of a zero impedance output


2

BJT transistors can't switch AC. They are also not suitable for switching the large currents you're planning on using them for. I'm not an EE either, but here's my understanding: A transistor is essentially a diode, the resistance (and thus the current flow) of which can be controlled. It limits the flow of forward current by acting like a variable ...


3

I'm impressed with the thought that must have gone into your circuit, but unfortunately it is fundamentally wrong. For one: it is not the base voltage that you want to control / increase, but the voltage across base to emitter or even better the current flowing from base to emitter. The transformer and the load (lamp) prevent this. Another thing that will ...


5

you have not connected the body/bulk connection properly in your schematic. PMOS to VSS. But that circuit leaves the output node floating for the A=B=1 condition. Scan through the various solutions here, there is a robust TG version of XOR and XNOR, it uses 4 transistors though. If you look at all the possible states, you have 2 transistors each in 2 ...


3

It's been a few years since I last did this kind of thing, so I thought I'd try to figure it out. Here's the transistor diagram: Here's the transistor diagram with the intermediate nodes labeled 1-4, and with the Euler path drawn in for both the PUN and PDN: Here's the stick layout:


2

You've mixed up a few of your transistor types. Here is the actual circuit taken from Arxiv Generally you want to size the PMOS width to be 2.2X wider than the NMOS but here P! and P2 should be looked at closely. P2 will fight against N3 in series with N2 and P1 will fight against N1 and the low drive form the previous gate so I'd say that P1 and P2 ...


4

You can do this to drive your motor. The 1k resistor must be tuned based on your motor's expected current for a given voltage. Usually you have a spec like 500mA@12V or something of the like. Basically, to get the 500mA spec, you want collector current of 500mA. The base current is always Ib=Ic/Hfe. Tip122's Hfe is min 1000, so you get 500µA for Ib. ...


2

In a nutshell, bipolar junction transistors work because of the physical geometry of the two junctions. The base layer is very thin, and the charge carriers that are flowing from the emitter to the base do not recombine right away — most of them pass right through the base altogether and enter the depletion region of the reverse-biased base-collector ...


1

For switching purposes you should be using the saturation voltage guarantees rather than the DC current gain. For example, the 2N4401 is guaranteed at Ic/Ib = 10, so a base current of ~26mA will typically result in 220mV or so drop at 260mA (eyeballing the below plot). The reason they say "pulsed" is that they are not accounting for the self-heating of ...


3

You are missing one easy thing: voltage across the resistor won't be 5V. Transistor side you have a \$V_{BE}=V_\gamma\approx0.7V\$, so the base of the resistor is about 0.7V lifted from ground. Microcontroller side, when the output is high you don't have full \$V_{CC}\$ but a voltage that is somewhat lower. For a cmos chip that can be quite low, some ...


2

The quote is poorly worded in my opinion. Of course, there must be a forward voltage across the base-emitter junction for there to be a current through. However, once 'on', the current through can change drastically for a relatively small change in base-emitter voltage. Thus, one must have some series resistance such that the current cannot exceed a safe ...


1

The point of the circuit appears to be to demonstrate that you can switch something (the lamp) on and off by using a transistor. In this case, they just use a mechanical switch to trigger the transistor, but in the case of most electronics, one switch may be triggered by other transistors (logic gates). In that way, you can have a small electrical signal ...


2

Rule 1 isn't a "good idea", it isn't a "guideline", it is a fundamental tenet of transistor physics. If for any reason (during normal usage) it is unable to hold then the circuit will not operate. As for the lamp, it is a purely resistive element. It should have 10V across it, but thanks to the transistor it won't. So the transistor gets 0.2V and the lamp ...


7

So as you mentioned it says that the transistor is essentially two diodes. You should, but may not, know that the typical voltage drop required over a diode to make it conduct is ~0.7V but can vary depending on the diode of course. So if you just 'stick' a voltage across the terminals as when you increase the voltage over the diode current flows: As the ...


0

The transistor is a current controlled device. The emitter current is related to the base current as I_e = (B+1) * I_b ( B = beta ) In the forward bias mode for a diode ( using exponential characteristics), as soon as the voltage crosses a threshold ( about 0.7 V or so for silicon), the current value shoots up dramatically. So if you directly ...


6

Well, you are indeed applying voltage across B-E, and you also must limit current with a base resistor. You can find the max base current of a transistor in its datasheet. Same story for diodes. If you want to power a LED you must include a current limiting resistor in your circuit.


4

Even though the body diode in MOSFETs is often drawn as a physical diode next to the MOSFET, it's not actually a discrete building block of a MOSFET. It is a 'parasitic' element that exists inherently because of the silicon build-up of a MOSFET. There are no MOSFETs without body diodes, it's a fact of life. That also means that this body diode isn't ...


0

I think you can't answer this with currents as above, instead you have to use charge flows, i.e. the drift of specific carrier populations. Let's take an NPN transistor for example, with a supply connected between C and E to reverse-bias the CB junction. At the micro level, what is ICBO composed of? Currents in reverse biased diodes are from random ...


1

If a 0 is applied to a P gate then think of it as a short-circuit. If a 1 is applied to a P gate then think of it as an open-circuit. Each input, A and B are both going into 1 P gate and 1 N gate each. Now let's walk through the truth table. A=0, B=0. Both P gates are short circuits and both N gates are open circuits. Therefore C is tied directly to the ...


1

You could think of this as two gates- one is an AND gate with inverted inputs that can only drive high (AND with inverted inputs is logically the same as a NOR). That's the two p-channel transistors in series. Both inputs must be low for the output to be driven high (open drain, so it cannot drive low, only high). The second is a NOR gate. That's the two ...


1

A goes to both the P and N transistors. If A is low, then the top transistor conducts and the second transistor connected to A does not. If B is also low, then its upper transistor conducts but its lower one doesn't. The upper transistors pull C up, and the lower transistors do nothing. Now, let A be high. A's upper transistor doesn't conduct, but the ...


3

jippie's right. D2 is your problem. On the reverse cycle, you've got no regulation on the voltage/current from D2.L3's top is at "virtual ground". L3's bottom is at positive 23 volts or so. This forward biases D2. The output is the cathode side of D2 I assume. That means you've only got regulation on 1/2 of your cycle (through D1). Your regulator sees the ...


2

As drawn, your circuit should be able to hold it's output voltage for several seconds. The potentiometer P1 does provide a leakage path though, with a time constant of about 50 s. This means you'll see the voltage droop noticeably in just a second or two. It should take about 2 min to get down to 1.5 V, though. Once it drops far enough, Q1 should be ...


5

The transistor's base is tied to ground, which means that the current through R1 depends only on the output voltage (along with the base-emitter drop of the transistor). The transistor sets the current through the 1.24K resistor (R2?), but keep in mind that there's also a significant voltage drop (roughly Vin - 1.24V) across the transistor itself, which ...


1

Look at Q2 - the parasitic diode just takes all the charge away unless you keep the input voltage at a high enough level to keep that diode from being forward biased. BTW, there is no such thing as "pure DC". Try sticking a diode in series with Q2 but somehow this is probably messing with what you are trying to achieve (which you haven't completely come ...


1

It is straightforward* to write the bias equation for this circuit using KVL as so: $$I_C = \frac{V_{BB} - V_{BE}}{\frac{R_1||R_2}{\beta} + \frac{R_E}{\alpha}} \approx \frac{V_{BB} - V_{BE}}{\frac{R_1||R_2}{\beta} + R_E}$$ where $$V_{BB}= V_{CC}\frac{R_2}{R_1 + R_2} $$ For bias stability, we want the right-most term in the denominator of the bias ...


1

First let me lead with a redrawn partial version of your circuit, which might be sufficient to convey the idea: simulate this circuit – Schematic created using CircuitLab Maybe that makes it easier to think about R1 and R2 in parallel. If not, read on... Forget the transistor for now, and consider just a basic voltage divider, connected to ...


1

It is the task of the emitter resistor Re to allow current-controlled voltage feedback. However, this works only in case the dc voltage at "other side" of the B-E path is kept constant (indpendent on temperature changes, tolerances and other uncertainties). That means that a "stiff" base potential is desired. This would require a very low-resistive voltage ...


1

Looking into the base terminal we see the equivalent of a resistor of value Re *hfe, so if hfe is 200, it looks like a 1.5M resistor to ground. They are saying we can ignore that if R1 || R2 << (Re * hfe), where they consider an order of magnitude to be close enough- so a reduction in swing of Vcc/20 is considered insignificant. There's nothing ...


1

R2 is not effectively in parallel with RE. It is uncoupled to the extent of the hFE of the transistor. Sure changes in current flow in the emitter are a direct result of current flow in the base. But the idea being described is to make the base resistors be an order of magnitude lower in size than the effective RE reflected back to the base. Said another ...


2

Circuit looks OK for not too demanding use. At extreme extremes it may stutter. Frequency response to input signal and acceptable rise and fall times were not specified and if important need to be known. Vbe of Q1 will clamp base at ~= 1V max. Ibe can be limited by using say two diodes from R1-R2 junction to ground and a small resistor (say 100 Ohm) from ...


4

I designed a very similar circuit myself once when I needed some "rugged" inputs. However, I used R1 = R2 = 100k (rather than 10k). It really doesn't take much input current to saturate Q1 with R3 = 10K. Reduce C1 by the same factor if you want to keep the same corner frequency. If you want some hysteresis to improve the switching characteristics, you might ...


5

Looks good to me. The inverse diode D1 is a good idea. If you have a minimum of 12V available you may wish to reduce R2 somewhat. This circuit has a threshold of maybe 2V, you could easily halve R2 or double R1. In the case of momentary extreme over-voltage, the base-emitter voltage (forward biased) will not rise above a volt or so, even with 100mA. It ...


0

Step 1. ICBO flows from collector to base. It is essentially just the leakage current through the reverse-biased diode, the C-B junction, so it increases as Vcb increases (as well as temperature, etc). Step 2. If the base connection is open circuit, there is only one place this current can go ... the emitter. This forward-biases the base-emitter junction, ...


0

1/ ICBO is reverse leakage current going from the Collector to the Base. This current is then amplified by β to produce additional Collector current, thus the "1+β" term. 2/ Both currents exist simultaneously, but ICBO is included in ICEO. 3/ The reverse current ICBO is amplified, just like external Base current would be.



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