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You can detect overflow after the CSA using the two most significant carry-bits, denote $C_o$ (the bit that is discarded), and $C$, and the most significant sign-bit, $S$. The addition will overflow if $C_o = S = 0$ and $C=1$, i.e., all positive numbers in, but the sum can not be represented in the corrsponding two's complement represenation ...

0

It sounds like what you're building is a power bank, or external recharger battery pack for a (smart)phone. These usually consist of a USB input jack, battery charger, lithium-ion or lithium-polymer battery, battery-to-5VDC regulator, and USB output jack. The phone then gets powered from 5.0VDC on the USB output jack of the power bank, and since the phone ...

2

Voltage at the node where R2 is connected is Vo, consider voltage at other node is $V_1$ By KCL: $$\frac{V_o}{R_2} + \frac{V_o-V_1}{R_1} + \frac{V_o-V_2-2V_o}{R_3} = 0$$ To find $V_2$ here it is simply $R_4 \cdot I_1 = 48V$. Now put the values and you will get your answer. However you can do same using the loop analysis and KVL. The thing is keep ...

4

Typically, no. The USB charging standard has the Data Pins tied to ground or each other through resistors to indicate what type/capacity charger the phone is plugged into, to let the phone know how much current it can pull. While connecting the phone to a regular USB Host also does not have that information by default (phones using non-standard drivers other ...

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If you mean do they put out a radio signal ie connecting to the mobile network when charging via USB, then the answer depends on whether it is switched on or not. When the phone if switched off and is being charged there should be no communication with the network. If you want to make doubly sure, switch it into airplane mode before switching off and ...

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Mobile phones use Lithium based batteries. Mostly Li-Ion. They are voltage limiting battery. For more on such batteries pl. see http://www.cellphoneshop.net/usbdacaforsa59.html Since we understand that is the voltage that needs to be sensed for perfect charging do see some voltage sensing circuits. Mobiles, such as old Nokia 100, would give a "fully ...

2

Short answer: Use a single common node for the whole circuit. Long answer: When one part of the circuit sends a signal to another part of the circuit (for example one IC sending a digital signal to another IC, or an op-amp sending an analog signal to a filter) there has to be a complete circuit for the signal to be transmitted. If the signaling scheme is ...

1

1) The total impedance of your circuit is $$Z_{total}=R_1+(-j{{1}\over{10 C_1}}||j\cdot 10L_1||R_2)=R_1+{1 \over j\cdot (10C_1-{1 \over 10L_1})+{1 \over R_2} }=1+{1 \over 1+9.9j} \approx 1.01+j\cdot 0.0999 \Omega \approx 1.01504 \cdot e^{-j\cdot 0.098}\Omega$$ From that, you can easily calculate the complex amplitude of the current: I={V_1 \over ...

1

To get complex power, you first need to get voltage on the three components after R1. You can calculate it like a voltage divider, when you first combine impedances of R2,C1,L1. $Z_{L_1}=j\omega L_1=10j\Omega$, $Z_{C_1}=\frac{1}{j\omega C_1}=-0.1j\Omega$, $Z_{R_2}=1\Omega$. Make inverse terms to get admittance, add them and invert again. ...

0

To find the Thevenin equivalent resistance you need to turn off the independent current and voltage sources. A current source that has been turned off has 0A and is therefore an open circuit. A voltage source that has been turned off has 0V and is therefore a short circuit. For this circuit, therefore, $R_1 + R_2 = 20\Omega$ is in parallel with $R_3 = ... 0 Well, if you want to find the Thevenin equivalent, I would suggest using source transformations. In this case, start with the 3A current source and 16 ohm resistor. This pair forms a Norton equivalent source. Transform it to a Thevenin equivalent by replacing them with a series combination of a voltage source and 16 ohm resistor. The voltage source will ... 0 It looks like your Vtest in this case is the Vo across the 10 ohm resistor. It might make more sense to you if those leads are extended further out: simulate this circuit – Schematic created using CircuitLab From there, you can use your regular Thevenin methods to find the equivalent circuit. CircuitLab tells me that Vtest is 6.40V; if I add a ... 4 From a more practical point of view, this is what you have: simulate this circuit – Schematic created using CircuitLab A voltage divider on the input, a unity gain buffer and a voltage divider at the output. Voltage divider output is calculated as follows:  Vo = \frac{Vi \times R2}{R1 + R2}  Where Vi is 7.5V for the first divider. 4 Golden Rules. With negative feedback in place... No current flows into the input terminals of the opamp The voltage at the plus terminal equals the voltage at the minus terminal. Because no current flows into the positive terminal, current must flow from the voltage source, through the series resistors 16 ohm and 24 ohm. So you should be able to figure ... 1 Depending on the current and and quantity of bypass capacitors, the secondary voltages can 'fall' at different paces. Sometimes, for sensitive components, Schottky diodes are placed between power supplies for avoiding inversions, which could occur in your example if the 5V crumbles faster than the 3.3V. simulate this circuit – Schematic created ... 2 As far as I can make out and Looks like an 8 pin job of some description. 1 Get 5 V and 3.3 V regulators that have enable, sometimes called shutdown, pins, then sequence them accordingly. Properly sequencing the shutdown lines is not as trivial as it might appear at first glance. Probably the simplest way is to use a tiny microcontroller, like the PIC 10F200. This would have a single shutdown input, which then causes the ... 1 There is no bypass cap in sight. Put 100 nF to 1 µF ceramic directly between each power pin and ground. Showing only the pin numbers doesn't show us the pin function. It's impossible to see what you have connected to what pin without looking up the pinout in the datasheet. This information should be in your schematic. 22 µF is about 6 ... 2 Voltage across$R_1$, V_{R_1} = I_TR_T'\times\frac{R_1||R_4}{R_1||R4 + R_2||R_3} Then current through$R_1$, I_1 = \frac{V_{R_1}}{R_1} Since you have already calculated the values of$I_TR_T'$,$R_1||R4$and$R_2||R_3\\$ the calculations would be easy.

1

I would attack this as a control problem. No doubt the actual mirror position lags the current or voltage the unit is being driven with, and may have overshoots or ringing, depending on how the system as a whole is compensated. I would try to do as little in analog as possible. For something this slow (10 Hz response), you can easily close the loop or ...

1

Andy and gbulmer are great answers. Adding this as a long comment: I keep mine consistent because I'm a neat freak. Looking at my display board, I have 0.3mm SPI trace (10MHz) because the level shifter pads were 0.3mm. They then run about 2mm apart (miles and miles!) because it was hand etched and soldered and I had the space. On another board, the traces ...

4

On the Internet there may be 'specification wars' between PCB manufacturers. Track, space, drill holes and vias are a differentiators. Some companies might be conservative and quote figures which are well within capability, and others may be at the edge of their capability. Laen of OSHpark has run some tests on several services offered over the Internet, ...

2

Are wider traces better or doesn't matter? When it comes to 0V or signal 0V, a ground plane is always better hence the width is as wide as the board. Yes, in this example wider traces are better. On other tracks (such as SPI) you don't want wide tracks because the capacitance to ground will be high and more power supply current will be needed and ...

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If this is to be manufactured by a PCB house, you can choose track/space width as you like, as long as you stay above some minimal value. For manufacturing, it is good to have high copper/non-copper area ratio, and to have non-copper with uniform width and uniform distribution across the board. That is especially true for inner layers, in order to be ...

2

Think everything you have on the circuit a resistor even the wires to connect different component. In ideal condition the resistance of a wire is 0 Ohm. We make short circuits with wires. Therefore the resistance of the short circuit is 0 Ohm. Now if you calculate equivalent resistance of the 20 Ohm and 0 Ohm resistor then you get 0 Ohm. The theory here ...

2

There is no way to tell whether you circuit will "work" since you haven't said what it should do. "Working" means the actual behaviour matches the specification. Obviously that requires a specification, which is missing in your case. What this circuit will do is to blow out the LED when the light is interrupted. Whether that is the desired behaviour and ...

2

Note that my answer, below, pointed out defficiences in the question's original circuit. Your op-amp has no negative rail and your signals, power and feedback use 0V as the reference voltage. This will mean your output will be a 5 Hz sinewave half wave rectified. Try creating a -12 v rail OR bias up your signal and feedback grounding resistors to midrail ...

0

Put a resistor between the INPUT and the 24V output to deliberately draw some current, just for testing. For example, at 2.4 kΩ resistor will cause 10 mA to flow in the loop. Note that it will also dissipate 240 mW, so it should be at least a "1/4 W" resistor, but "1/2 W" would be better. Now you have a much lower impedance circuit and any ordinary ...

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You will have to test the output of each switches in reference with the ground, and starting by the first switch. If your voltmeter reads 0, it means the switch is open. Prepare some cables you can carry around, connect one side to ground (or zero volt) and the other to your voltmeter common. Example: You measure the output of sw1, it's ok. Then you ...

6

I'm sorry to say that your circuit will not work: The MOSFET RDS(on) (2.8 Ω) is too big compared to your load resistance (2.5 Ω). If your load needs 2 V to work properly, be aware that it will only have around 1 V (assuming the MOSFET was fully on, which as mentioned above, it won't be). The LED (indicating if the load is powered) has no current limiting ...

5

In addition to the too low supply voltage there is another issue. If the LED is intended to light when the load is activated you will need to turn the LED around the other way and place a resistor in series with the LED to limit it's forward current. In the end the LED may not even light if the supply voltage is kept lower than the forward voltage drop of ...

0

I know all the Simulview does is make one pair "double left eye" and the other pair "double right eye" Pressing the button on top of the glasses makes it double left or double right. You take an actual pair of PS 3D TV Glasses. Normally to turn it on, hold the power button for 2 seconds. But you quickly tap the button to change between left eye, right ...

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It won't work, your MOSFET has a VGS(Th) of 3 V. Your supply is only 2 V. What is this circuit for; if we know we can help you better?

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(This should perhaps just be a comment, as it only provides a partial answer, but I wanted the drawing...) You don't need one of the LED power supply modules you lined to for the photo-interrupter. Electrically, the photo-interrupter is two independent parts - the IR LED, and the photo-transistor. You should treat the IR LED as you would any other LED - ...

0

Inject a high-frequency AC that rides on the DC carrier, and "sniff" for that AC signal. One simple means of doing so would be to clip a small FM transmitter's antenna connection to one end of the wire (no other connections) and drive it with a simple oscillator. At each switch along the line, an FM receiver tuned to the same frequency should be able to pick ...

6

C3 turns the circuit into an oscillator. To understand this you need to think hard about the role of the emitter and the thing you have to think hard about is this - the emitter (as well as the base) is an input to the transistor that is amplified by the transistor. An ac voltage (superimposed on a dc voltage) between base and emitter is amplified by ...

0

I could be wrong, but I strongly suspect that your better use of the device is as a parts supply - scavenge the display & learn how to hardware-hack it, scavenge the button switches, maybe grab a few other internal parts if they look handy.

2

There is no guarantee that the manufacturer put any way to program the MCU onto the PCB. MCU's could be programmed before assembly. It may be infeasible to program the MCU even if you could get at the right pins. Most MCU's have built-in mechanisms to allow a user (in this case camera manufacturer) to prevent a chip being reprogrammed, or the program ...

0

You will have to find a JTAG adapter suitable for this MCU, and, obviously, the relevant JTAG connection on the PCB. Most consumer products cannot be programmed or debugged through the USB connection.

0

It's not a great circuit. The PNP transistor would not switch on fully so I'd use an NPN with the emitter grounded - this way, current into the base turns the transistor on to about 0.1 volt across collector and emitter and wastes less power. Having said all that I'd use an N channel MOSFET that maybe switches on to a few millivolts.

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To dim the running lights, they could be switched from being parallel across the 12V supply to being in series by using a relay with two sets of change-over contacts. When high beam is not selected, the relay contacts are as shown and the two running lights are in parallel across the supply and therefore at full brilliance. When high beam is selected, ...

1

Just to add to my comment with the circuit diagram. The first circuit shows the DRL(day) connected to the common pole of relay 2. (The way I think you have wired your circuit) With relay 1 unenergized the ACC(+) has a path through relay 1 and relay 2 switches to DRL(day). However, when relay 2 is energized all that happens is DRL(day) and DRL(night) are ...

3

This DPST NO (2 form "A") relay solution will work:

0

You have a serious input and output voltage range problem : the input signal must stay within input voltage range (V- to V+ -2V) and the expected output voltage must stay within output voltage range (V- to V+ - 3.5V). Here is what I suggest : simulate this circuit – Schematic created using CircuitLab This way : The input voltage swings around ...

1

Could I interest you in building a bridge rectifier with Schottky diodes instead? Then polarity would be immaterial.

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Thinking only the positive side of your circuit: assuming you use a normal diode with a voltage drop of 0.7V, GND and +12 reversed, the voltage across thq 1k resistor is U=RI => all remaining voltage will be over the resistor. Thus the protection doesnt work. One way of doing this (better or worse) would be remove he 1k resistor and using a diode with as ...

1

The diodes will have no effect to speak of on reverse polarity, and will provide no protection. If the device is a chip, it will act like a forward biased diode, and will take almost all of the current compared to your parallel diode with 1K in series. Edit: You can play with simulations below. With the left one I've simulated the 12V source (reversed) with ...

5

The most obvious problem in the circuit is that the + input of the amplifier is floating; it is not biased to any DC voltage level. You need to convey a reference voltage to this input. (This has to be done through a reasonably large resistor; if you just tie the input to a stiff reference voltage, the input will have no impedance to work against.) ...

0

You have no path for bias current. Add 1Mohm resistor from +3V to in+. Also the ring of audio input must be referenced to half of supply voltage, so add voltage divider consisting of two 10k resistors between +3V and 0V with the centre connected to ground (the grey line).

2

In general it is bad practice to apply input signals to a device before applying power. When you do this, you can cause semiconductors to 'latch up'. The semiconductor can act due to its physical build as a SCR (Silicon Controlled Rectifier), which can cause it to short out its pins. Subsequential application of power can cause excessive currents which ...

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