Podcast #128: We chat with Kent C Dodds about why he loves React and discuss what life was like in the dark days before Git. Listen now.
17

The opamp in the circuit is actually a voltage comparator, it's designed to output only 2 voltage levels - high and low. The positive feedback makes it Schmitt trigger -circuit (=a comparator with hysteresis) to make the comparator a little faster and to increase noise immunity. The transistor is pulse amplifier. The capacitor makes it have low pass filter ...


13

D1-D4 form a bridge rectifier that will charge C1 up to about 1/20th of the peak voltage of the incoming AC power. (R1 + R2 are 440K, R3 is 22k 440/22 = 1/20). During this time that C1 is charging the current will flow through C1 and through D5. This will reverse bias the base emitter junction of Q1 so it will not pass any current. For a short period at ...


10

Q4, R40 and R44 form a pullup and buffer that can tolerate quite high voltages for a logic low (up to about 3.6V depending on temperature) from a relatively weak sink of perhaps < 1mA. C22 and R46 form a noise filter (-3dB at about 3.4kHz). U7 and surrounding circuitry is a comparator with significant hysteresis that also limits the output high voltage ...


6

You actually have two problems right now, but let's just talk about one first (the one you are somewhat aware that you have. You seem to be oblivious to the other one right now but we'll get to it eventually). It boils down to this: You cannot know the directions of currents or voltages through a circuit before you have solved it. You might have an idea for ...


6

The gain is defined as the output signal divided by the input signal. The amplifier has a voltage output. It's a virtual ground input configuration, the input will stay at 0v by feedback action. The input is therefore not a voltage signal, but a current signal. The gain therefore has dimensions volts/amps, otherwise written as ohms. This configuration is ...


6

I am learning electronics myself and Kevin's answer confused me to the point where i was wondering what i was misunderstanding about the circuit, so i simulated it and found that my initial instincts and node analysis were actually correct. I will explain: The following is my copy of your circuit in LTSpice, ignoring the isolation aspect, which should be ...


5

Sorry, I don't see any paradox. I don't think you understand how to use superposition. If there are \$N\$ independent sources you construct and analyze \$N\$ circuits, where each circuit has just one of the independent sources while the other sources are deactivated. If there is just one source you have a trivial case: you analyze one circuit with that ...


4

What's the top rectangle circuitry is for with negative voltage? Note that U1 and U2 require negative voltage so their outputs can swing to 0 V. ... and whats is RL1,RL2? Relays 1 and 2 are used to select the lowest transformer voltage that will supply the load. This keeps the power dissipation in the output transistor, VT low. Why there's two ...


4

When you apply a fixed voltage to an inductor for a set period of time the current through the inductor rises from zero (if previously uncharged) to some positive value. That is what your circuit does; you apply a pulse to the base of the transistor and the collector current rises from zero to some positive value. I'm ignoring whether a 100 kohm or 100 ohm ...


4

I seriously apologize for the horrible formatting but I don't know LaTex OR MathJax and I have to leave town in the morning. You could solve this using voltage and current divider equations, but I find it easier to just consider the current flows. Let \$I_1, I_2, I_3\$ and \$I_4\$ denote the current through \$C_1, C_2,\$ etc., and let \$V_1, V_2,\$ etc., ...


4

Equivalence of circuits means that if each was hidden in a black box (no other information from inside) with only their terminals exposed you could not discriminate between them; no matter what you connect to the terminals. If you provide a certain voltage to the terminals the current through the terminals will be the same and and vice versa, i.e. if you ...


4

You need significantly more capacitance from each rail to ground. Try 470uF or something like that in parallel with C3 and in parallel with C5. But if you're using a fancy high performance amplifier like that one, you're probably going to want to regulate the supply rails in reality. So that would mean a positive and a negative regulator, in addition to ...


4

Here's the schematic drawn up using the on-board schematic editor: simulate this circuit – Schematic created using CircuitLab STEP 1 Before I delve more deeply, a quick check is worth doing. Assuming that the first voltage divider on the left is unloaded, \$Q_1\$'s base voltage is the Thevenin source voltage of \$V_\text{TH}=V_\text{CC}\frac{R_{\...


4

I would say no. The effects of temperature coefficient can be expressed as $$ R= R_0 \left[1 + \alpha\left(T_0-T_{\text{ref}}\right) \right]$$ So, only the change in resistance (not the resistance) will be proportional to the temperature difference, and that holds best for small temp changes about a working point. To be even more specific, $$ R=\frac{\...


3

No, it isn't a Darlington configuration, because T1's collector is tied directly to ground, rather than to T2's collector. The advantage is that T2 can fully saturate, reducing its C-E voltage drop and power dissipation. R2 serves to keep both transistors off when the input connection is open, and you should keep it. C1 probably doesn't do anything very ...


3

In your case, the transfer function is easily cobbled out. (I've seen H and G used interchangeably, so don't get bogged down on some imagined foolish consistency.) $$G_s=\frac{R}{R+s\,L+\frac{1}{s\, C}}$$ Moving towards a standard form of some kind (and I'm sure you can handle the algebra for it), this becomes: $$G_s=\frac{\frac{R}{L}\,s}{s^2+\frac{R}{L}\,...


3

Your circuit is indeed a tone control circuit. For low signal frequency (Bass booster) we can ignore \$C_2\$ and \$C_1\$ because for low-frequency capacitance reactance \$X_C\$ is high. So we have this situation (max bass boost): And bass boost gain is: $$A_{Vmax} = \frac{R_2+P_1}{R_1}$$ And as signal frequency increases \$C_2\$ capacitor reactance ...


3

What you have there is a classic, simple data slicer. D2 and \$C_{p2}\$ form an AM demodulator. What goes to C6 and R1 is no longer RF. C6 and R1 form a sort of high pass filter/differentiator. R2, C7, and R3 are there to slightly delay the edges of the detected low frequency signal. U3 is a comparator. It compares the signal (V+) with the delayed ...


2

Additional to previous answers that in forward bias, the diode has about 0.7 voltage drop and very low impedance which ignores the parallel resistor, then according to capacitor charge, time=R*C, it will charge quickly because of low R (impedance of diode). If you remove the diode the charge time will be long because here R is the resistor which is high.


2

Basically you need to add up all the Vf (forward voltage) of each LED at your design current and subtract that from the supply voltage, then apply Ohm's law to figure out the total resistance required. You can then split that into as many resistors as you want, say if you want to use small SMT resistors and spread the heat dissipation out physically across ...


2

First, you don't need a resistor between all LEDs, one resistor is enough. To calculate it use the following rule: rule (voltage = current * resistance) V = I * R However, because of the forward voltage use (V - Vf) = I * R Thus (assuming 24 V), and 5 LEDs: (24 - 5 * 2.7) = 0.03 * R <=> R = 350 ohm If you don't have a 350 ohm resistor, use ...


2

The circled point is not tied to ground, the two wires are just crossing. Wires that join are depicted with a dot on the junction as can be seen in several places in the diagram. If you think about it, there would be no point in joining the wires inside the red circle as it would short out the 1.2V LM113. What may be confusing you is that there is no need ...


2

"Any electrical component or circuit" is far too broad. The elements being discussed only apply to lumped-circuit models, in which the connections between elements are assumed to be ideal — no resistance, no electromagnetic field (parasitic) effects, no delays, etc. Such models are by definition only an approximation to real-world circuits to begin ...


2

As @Huisman points out, this statement applies only to linear ideal elements. Specifically, the statement does not apply to non-linear elements or, in general, to real elements. The structure of the article suggests this distinction but the single sentence, taken out of context, can be a bit misleading. Having made that qualification, I think the statement ...


2

My Laplace transform math is pretty rusty, but the neat trick behind a Laplace transform is that the test waveform that the transform compares your input function to changes amplitude over time. Examine the Fourier transform: \$\hat{f}(\omega) = \int_{-\infty}^{\infty} f(t)\ e^{-i \omega t}\,dx,\$ Essentially, it takes a test sine wave (\$e^{-i \omega }\$)...


2

simulate this circuit – Schematic created using CircuitLab You can use this diagram to understand the current direction and accordingly get correct loop equations. Decide the loop current direction. Then mark voltage drop according to the current direction (current entering at a point is considered as positive). Apply KVL, in the direction of loop ...


2

As with all perfect current sources, their effective impedance is infinite so it, and all the components to its left are redundant in the calculation of the time constant. The effective resistance seen across C1 is R5 in parallel with (R3 plus R4).


2

What is the purpose of those two schottky diodes? isn't it causing voltage drop which we have to compensate in software? No, because the current and voltage are being by PI controllers; any offset from the diodes is compensated by their integral action. How the opamp in voltage section works, is it comparing OUT+ with OUT- or OUT- as ground because of ...


2

Observations: With the bridge rectifiers in series, if (say) V2 was quite active and (say) V1 was close to zero volts, the DC output would be 4 diode drops down on the AC voltage and this is worse than just collecting energy from V2 and its bridge rectifier. Having the bridge outputs in parallel means you always collect "decent" energy even if only 1 source ...


2

Note new REFDES (..reference designators) allowing for slight dimmer with low battery @ 8V and < =20mA max at 6.2V nominal for two Blue LEDs with some wider tolerance possible depending on parts and current. Collector R's determine the current limit say at 20mA or maybe you prefer 10mA. Let R4 = R2= (9-6.2V nom)/ 0.02A = 140 Ohms (assuming Vce=0) ...


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