# Tag Info

2

As others have stated, the block diagram you have drawn is physically unrealisable, so our physical intuition does not apply here. What you have found by $\frac{Y}{X}=\frac{A}{1-AB}$ is a DC equilibrium point of your "feedback" system. It's true that if we feed in an input X and compute the output Y using the aforementioned equation, we obtain a ...

3

There already are two answers, I'll tackle the part that confuses you. What you have there is not a "system" -- you have a mathematical equation described, visually, as a block diagram. And because it's represented visually, you're interpreting it as a feedback system. It is not. As I mentioned the $G$ and $H$ "blocks" are simply ...

3

I believe that this system, where the loop has net gain and the response functions G and H are instantaneous, is a physically-impossible and mathematically ‘pathological’ case. I am speaking a bit colloquially with this term, but will explain what I mean by it later. The upshot is that all we need to do is add an infinitesimally-small amount of memory to get ...

2

Why can't the equation include that information? I don't see anything wrong with the derivation and block diagram. It won't and it never will, because for you to know the output of any time varying system, you must know two things: The initial condition of the system (t=0) OR the current state and current time the transfer function of how the system ...

3

As ErikR noted, you only need 4 GPIO's to control up to 12 LEDs using a technique known as charlieplexing. simulate this circuit – Schematic created using CircuitLab To turn on LED D9 pinMode(0, INPUT); // tristate as not used pinMode(1, OUTPUT); pinMode(2, INPUT); pinMode(3, OUTPUT); digitalWrite(1, HIGH); digitalWrite(3, LOW); In real code you'd ...

3

Actually chips that can drive bar graph LEDs based on input voltage do exist. They are called bar display drivers so you can search for them.

2

You could use a shift register like the SN74HC595 SN74HC595 to expand your port numbers.

2

Use a circuit like this to measure current: https://circuits4you.com/2016/05/13/dc-current-measurement-arduino/ You can measure the current, feed it into the ADC of the micro and then use any manner of logic to shut down the processor. Many voltage regulators can be purchased with enable lines, you could have the voltage regulator enabled by a push button ...

2

Well, a general transfer function is given by: $$\mathscr{H}\left(\text{s}\right):=\frac{\text{v}_\text{o}\left(\text{s}\right)}{\text{v}_\text{i}\left(\text{s}\right)}=\frac{\text{k}}{\text{n}+\text{ms}}\tag1$$ This implies that: $$\text{v}_\text{o}\left(\text{s}\right)=\frac{\text{k}}{\text{n}+\text{ms}}\cdot\text{v}_\text{i}\left(\text{s}\right)\tag2$$ ...

1

After some time, I sort of found out what was wrong above and why the circuit wasn't "working". So I'll just answer my question. It was not problem with the circuit but the LED used. On Proteus, you can select the kind of LED. Above I unknowingly was using a digital LED. This means it either turns on or turns off. I don't exactly how the forward ...

3

Because the Laplace transform of $V_o(t)/V_i(t)$ is not the ratio of the Laplace transforms of $V_o(t)$ and $V_i(t)$. A little deeper dive... Everything is correct up to this point: $$\frac{V_o(s)}{V_i(s)} = \frac{1}{1+sRC} = \mathcal{L}\left( \frac{e^{\frac{-t}{RC}}}{RC} \right )$$ and taking $\mathcal{L}^{-1}$ of both sides yields:  \mathcal{...

3

I have no evidence for this but I suspect that the chassis symbol is a cropped version of a mechanical cross-hatch representation of a metal chassis.

1

Here is a way to calculate the return ratio T (or modern loop gain) analytically. I use your small signal circuit with the Vs and Rs. Delta is the network determinant. Delta0 is the network determinant with gm=0 (the active element). The return difference F = Delta / Delta0 and T = F - 1 See e.g. Wai-Kai Chen "Active network analysis" page 226. Now ...

8

It actually does seem like a relatively appropriate use of "chassis" ground. In the datasheet you linked to: Note: Please solder the corner pad and the bottom thermal pad of the QFN package to the GND pattern of the PCB. (p. 2) Note: All the grounding wires of the TB67H420FTG should run on the solder mask on the PCB and be externally terminated ...

0

Yes, I bought about 10 circuitts MSGEQ7 from various dealers in Taiwan. But not a single one worked, even the oscillator did not swing. Only the purchase of original circuits from UK led to succes. The fault wasn't mine, and the circuit and software work exactly as described everywhere. Very simpli and fine, no critical timings, very easy. It is essential to ...

50

Chassis ground. See: simulate this circuit – Schematic created using CircuitLab (Illustration from Ground - Ultimate Electronics Book)

27

This from IEEE Std 315 for that symbol: "A conducting connection to a chassis or frame, or equivalent chassis connection of a printed-wiring board. The chassis or frame (or equivalent chassis connection of a printed-wiring board) may be at substantial potential with respect to the earth or structure in which this chassis or frame (or printed-wiring ...

7

That is one of several symbols that may be used to indicate "Ground". I have seen lists defining specific applications for each symbol, but most often the different symbols seem to be used at random.

39

Normally it would mean chassis ground. But it would probably mean the substrate in the case of an IC schematic. The substrate is usually connected to the ground pin or the most negative voltage (depends upon the IC design). In the specific example shown it connects to the back-gate of the FET and the ESD diode so it is pretty certainly the substrate that is ...

2

Unfortunately, not every L298N module follows the recommendations put forth in the datasheet. I have this particular module: (Source: https://forums.parallax.com/discussion/156410/how-to-use-a-l298n-dual-h-bridge-with-a-microcontroller-quickstart-board/p2 ) It has the bypass cap for pin 9 but not for pin 4. Fortunately pin 4 comes out as the VMS signal so ...

3

what does PMID actually stands for? It stands for "power-middle" as per this explanation from TI: -

1

This is just a very high current gain frontend with no load or feedback to regulate the DC output and the output resistance to convert voltage to current and thus voltage gain. The common mode Bias current is defined by Vb1 and Vb2 and thus the emitter load Ve/Ie. so the DC must be near equal and then a load on Vo to get a thing any AC without being ...

2

In order for Q3 to mirror the current in Q4 both Q3 and Q4 must operate at the linear(forward active)region How is that even possible? Correct, Q3 and Q4 must operate in active mode. Q3 will be in active mode as it is connected as a "diode". Then Vce = Vbe and the transistor must therefore be in active mode. Q4 will be in active mode as long as it ...

1

To answer your question, no, you can't make a USB to serial adapter with simply onnectors and TTL-to-RS232 tranceiver chips. And most likely no DIP chips exist unless you build a software bit-bang implementation with a DIP MCU, which I also don't recommend. But you don't have to maka a USB serial port yourself. You can just buy a ready made cable that ...

0

You are correct, the term isn’t used consistently. In general it refers to the smallest feature that is used in the design. Since many or most features are larger than this it is not a precise term.

3

Your mistake was in the original expression for the voltage gain (which you called $A$). The voltage gain in Miller's Theorem is the voltage gain of the circuit including the effects of the feedback impedance. Miller's Theorem is valid when the output voltage is proportional to the input voltage with some constant of proportionality that I will call $K$ ...

2

From Wikipedia article (https://en.wikipedia.org/wiki/Miller_theorem#Explanation): The Miller theorem implies that an impedance element is supplied by two arbitrary (not necessarily dependent) voltage sources that are connected in series through the common ground. (The emphasis is mine). As there is a current source in the drain branch of your circuit and ...

1

Here, the way you are doing it is correct. In $A = -g_m r_o$, $r_o$ is the impedance seen at the collector. It is the impedance of the load in parallel with the output impedance of the transistor. In this case it is the output impedance of the transistor $r_o$ (you get confused when you use the same letter for two things), $Z_f$, and the impedance ...

1

IThe equivalent circuit for the noise source by itself is found by zeroing out all of the other voltage sources (principle of superposition). Consider how RC2 and Q4 are connected. The collector of Q2 can be ignored because its effective impedance is much larger than RC2. The effective resistance of this combination (hib4) is RC2 divided by the current gain (...

0

There is no easy answer to why different capacitors stabalize different circuits. But as far as power supplies and SMPS are concerned. The primary metrics of concern are equivalent series resistance, voltage / dielectric stability and finally frequency response. You can use much lower value ceramic capacitors due to their relative frequency stability ...

2

The output capacitor affects ripple and transient response so it doesn't need to be too exact. Larger = less ripple but slower response. The feedforward capacitor is the one I would be more worried about. I suppose you could just plot out the output voltage vs feedforward capacitor and interpolate the curve. The capacitances for tantalum and electrolytic are ...

0

Is Vt given at 33nA? If so then Beff from Ids is correct.

0

When you connect pin2 to the GND you should then disconnect it otherwise the internal RS will be always set and the output will stay high. Try to close and directly open the switch and wait. In fact, closing the switch is just set the output then you must open it to wait for the reset. Since you mentioned in comments that you de-soldered the capacitor from ...

1

ESP32 has Wifi, so it is remotely controllable, and it has an excellent PWM with 16 outputs including HF dithering. TLC59711 - seems to be best, but not really hand solderable and current output (so not suitable for the led amplifier) You can hack the current output into a voltage output, but it will be inverted. You can invert it back by connecting a ...

2

According to smdmark it is a ADP3040ARM, switching regulator. This post on the ADI engineer zone website details that is a customer specific part with no documentation freely available.

3

Digital design Digital circuit design is usually done using a HDL language like Verilog or VHDL. This can be simulated on digital level with tools like Modelsim / Questa. https://eda.sw.siemens.com/en-US/ic/modelsim/ A free alternative for VHDL would be GHDL in combination with GTKwave to view the (digital) waveforms. These softwares basically run the ...

1

First look at the equation. You have Y,P and a desired range (upper bound) of length (l not L). It's trivial to convert that into a range of available frequencies with a lower bound, so do that first, and choose an achievable frequency ( >= that lower bound). If that is achievable (the lower bound encompasses the frequency range you want) you can compute ...

1

After adding current limiting resistors change the 7-segment setting from common anode to common cathode.

2

You need a resistor for every segment of your LED display. Otherwise the LEDs are shortening the driver outputs.

-1

Tinkercad is not a tool that would be used by many of the professionals on this site, and your cartoon diagram is hard to decipher. So, I think it will be up to you to debug this. I'm guessing that Tinkercad gives you some way to check the logic level or voltage at a signal node. Begin by verifying that everything that should be connected to power or ground ...

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