# Tag Info

0

Yes, lower base doping: Increase emitter injection efficiency (ratio of emitter:base doping) Increase base transport factor as lower doping means less recombination for minority carriers injected from the emitter Increase early voltage - because the base-collector junction depletion region will deplete more into the collector than the base- therefor the base ...

1

Here are two answers to your questions: 1.) Sallen&Key topology: When an RC network is combined with a (positive) finite-gain amplifier (gain of "one" or "two" or any other value <3) and if there is single-element signal feedback. 2.) Center frequency fo: For finding the center frequency (for maximum output) you must not ...

-2

First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course). Well, we are trying to analyze the following opamp-circuit: simulate this circuit – Schematic created using CircuitLab When we use and apply KCL, we can write the following ...

2

Nodal analysis is the way to go. You can work on s-domain to make the life easier. According to your schematic, $\mathrm{v_b=V_o}$. Using the properties of a closed loop operational amplifier, you'll get $$\mathrm{ V_i+(1+s\ RC)V_o=(2+s\ RC)V_a }$$ where $$\mathrm{ V_a=V_o\frac{1+2s\ RC}{sRC} }$$ So the gain function can be written as  \mathrm{ G=\...

0

$CMRR = (A_{dm}-A_{cm})[dB]= 80~ dB ~, A_{dm}= 40 ~dB,~ thus ~A_{cm}=-40~dB$ $V_{IN_{cm}}= 79 mV ~RF???$ This CMRR only applies to DC to < 10 Hz and reduces 20 dB per decade to some small value like < 20dB Considering the question says RF noise, we can safely assume they're is not much differential or CM gain so the output is likely only ...

0

If the RF noise on the leads from the thermocouple sensor to the data logger is 79 mV, what will the noise level be on the amplified signal in mV? Here's where the question isn't tied down enough - does it mean that the RF noise on both leads is simultaneously 79 mV i.e. the common-mode interfering noise is truly 79mV or is it trying to tease out something ...

0

I suspect that your notes have two standard formulas, one for a circuit where RLC are all in series and one where they are all in parallel. This circuit is neither of those cases, so you may need to work a little harder. Notice that the series RL and the capacitor are in parallel. This means that the voltage across the RL and the capacitor must always be the ...

0

First, I will present a method that uses Mathematica to solve this problem. I know that this approach is not 'smart' but this method will work all the time, even when the circuit is way complicated than this one. This makes my answer valuable in my opinion. Well, we are trying to analyze the following circuit: simulate this circuit – Schematic ...

1

The datasheet includes a complete circuit diagram for a monostable (oneshot) circuit. Directly underneath the schematic is a section that describes how to calculate the length of the output pulse: The value for $R_A$ is in ohms. Use farads for the value of $C$. The value of $t_W$ is in seconds. You are given a requirement of 10 seconds - that's $... 1 Here's the circuit with the mesh currents drawn as you described them, all clockwise: simulate this circuit – Schematic created using CircuitLab These are the correct equations applying KVL with mesh currents: Bottom left mesh:  24 (I_1 - I_2) + 4 (I_1 - I_3) - 22 = 0  Top Mesh:  15 \cdot I_3 + 4 (I_3 - I_1) + 3 (I_3 - I_2) = 0  Some points ... 0 You can do a star->delta transformation to simplify first. Then try again by resolving parallel resistors etc. should become straightforward this way ... 0 It is not the resitor that is negative. It is the polarity of the voltage drop expressed as V=IR. It might be more intuitive to write: Loop 4: 0= 5.6(ı3-ı4)+10ı4 0 This kind of circuit is called a PD3 (parallel double, 3 phases -> q=3, so n=6). (unless "writing" errors). General assumption made is that Current load Ic is assumed constant =$Ic$. It has some formulas to describe it. (for 2 diodes current only allowed) Mean value of voltage on load : (Vm is peak voltage);$ Uco=2 * q /Pi * Vm * sin(Pi/q)\...

0

1)You can apply KCL at the negative terminal of V1 to find the current flowing through R2. Finding Vx then is just applying Ohm's law. 2)You can then apply KCL at the negative terminal of V2 to find the third current(two of them are known). 3)Finally you can apply KVL through the loop consisting of the two voltage sources to find out the current through R3. ...

0

Here is the basic approach: Look at the information you have. If you know the voltage across or current through a resistor, use Ohm's Law to find any missing value. Use KVL around any loops. If you have an equation with just one unknown, solve for it. Use KCL at all nodes. If you have an equation with just one unknown, solve for it. Repeat this process ...

3

The 741 can't get even close to the voltage rails: Source: TI 741 datasheet You need another opamp or another voltage source to power your circuit. The op-amps im allowed to use are TL074CN,CA3140EZ, LM358AP and the LM741CN The LM358A is slightly better but none of them have rail-to-rail inputs and outputs: Source: TI LM358 datasheet Since it is just a ...

2

You can't really use an LM741 op amp at low voltage, like 4.5V. The output is not rail-to-rail, so at low voltage you will only get a couple volts swing on the output.

1

The general strategy for problems like this is to combine resistors until you are left with a single resistor in parallel with the voltage source. Then you can use Ohm's Law to find the current through that single resistor. Now that you know the current, take that value back to your next-to-last circuit simplification. Use the newly found current to find the ...

-1

The left side with the power supply and Ra & R8 can be simplified by creating the thevenin equivalent circuit. Ra & R8 are really in series. And you can apply thevenin equivalent circuit again on Ra8 and Rb7, which are really on parallel. That should be able to give you just resistors in series. If you have issues seeing this check that your Thevenin ...

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