# Tag Info

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Using a 4.5V pump with a 12V power supply has probably burned out the pump. The voltage of the power supply needs to match the voltage of the pump. I suggest you find a 12V pump to match your power supply. Or: a get a new 4.5V pump and find a 4.5V power supply to match your pump. (Or 3V. The 3V/4.5V pump will run slower on 3V than on 4.5V) I don't think ...

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It's not really accurate to refer to steady-state responses as transients if nothing is changing. What may be happening here is the instructor is thinking of the different common types of simulations: DC AC Transient As a simulation type, transient actually just means something like "time-domain." It is the type of simulation which is required to see the ...

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You're correct in noting that "transient DC" is a contradiction. I think this is just a somewhat ambiguously-worded syllabus title. They probably mean "responses of {RL, RC, and RLC} circuits to {DC, sinusoidal AC, and transient} inputs".

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A picture may help: In this example, a load is connected to a power supply for a while and then disconnected. This causes two transients, at the connection and disconnection. DC analysis techniques work for the steady state, the flat periods of the graph. For the pointy bits we need transient analysis.

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But then if I put that output to a voltage doubler, the voltage should be 18V, not 9V right? Correct. Voltage doublers are easy enough that if you sorta-kinda understand how a capacitor and a diode works you can work this out for yourself, on paper. Nothing does a better job of understanding circuits than working it out yourself. ...but something ...

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The PI Zero seems to take 120 mA max. The PI Zero is not supplying power to the LED strips so you should be totally fine. If anything the TLE4271 is probably overkill for this task. my only concern is that the 24v input power supply that came with the led strips has no margin for the extra 120mA for your PI. You should check the 24V power supply's current ...

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1) Yes it looks good but it depends what are you using the ADC for and how big currents use the analog groun. If it is only for battery reading, it's overkil 2) Application notes suggest one small bypass capacitor per power pin pair and one large for digtal and one large for analog. Placement of large bypass near certain pair is sometimes mentioned. 3) ...

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My understanding is that you want to detect the presence of a low-resistance load on a headphone audio output. simulate this circuit – Schematic created using CircuitLab Figure 1. Detection of load connection using weak DC pull-up. This is just presented as an idea with no promise of support. R1 provides a weak pull up to NOT1. If nothing is ...

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Ya'll are making this way to complicated. Here's the easiest & quickest wig-wag relay design out there. Go to Autozone, pepboys, Advance or wherever & they will have what you need. Get a SPDT relay(novita rl45 is what I use) & get a heavy duty flasher relay(novita long life 555 20amp is what I use) I use a 2 prong flasher relay. this design only ...

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Here's the internal schematic of the uA776 opamp, pulled from the datasheet: Note that Iset (pin 8) is connected to a group of high-side current mirrors (Q12, Q13, Q16) that set the bias current for a bunch of the middle stages of the opamp (and also Q19, Q20 on the low side). Note also C1, which is added to provide negative feedback at high frequencies in ...

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As I can remember proteus can copy the complete subcircuit and past it anywhere else including the child sheet but if it doesnt work you can simply use a feature of proteus called "Project Clip", this works by selecting the complete circuit in Isis and the complete circuit in Ares, the you go to menu File > Export Project Clip, this operation saves the ...

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If you are doing this with sockets connected to the same power lines and you connect live to live and neutral to neutral then it is nothing which can theoretically misbehave. But if you flipped the wires, a good short circuit awaits for you.

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Yes, it does. It's a simple oscillator that can be turned on and off by the microcontroller. An oscillator that switches Q3 on and off is obviously required for the inductor-diode-capacitor circuit to function as a boost converter (you can read plenty about their operation anywhere), the only peculiarity of this circuit is that the oscillator is built from ...

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It’s one of those ‘just because you can, doesn’t mean you should’ things. File it under ‘Darwin Award Candidate’, cross referenced to ‘Hold My Beer and Watch This’. If - and this is a very big If - you had the hot/neutral phases right, and you were connecting to the same hot phase, it would be possible to switch over using this method. After all, you are ...

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Best case, a loop. Worst case, short circuit. The simple way is with a suicide/widowmaker cable. The other way is with a hotplug field kit. Often used by forensic teams to confiscate computers and servers without turning them off. Which is a really unique bit of kit. The correct way is the get a UPS for the TV and just swap the UPS plug.

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You're confusing unlicensed with legal I think. It is not allowed to make and use an intentional radiator, even an unlicensed one, that hasn't been certified to comply with FCC Part 15 limits for the frequency it uses. More here: www.arrl.org/part-15-radio-frequency-devices From this doc: "Low-power intentional radiators: Part 15 rules also permit ...

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any frequency that it is not illegal to transmit at There's no such frequency, unless you either get type-approven equipment and operated in a band specified for that type of transmission (i.e. mostly ISM bands; LoRa might be an interesting technology for you) or get a license to some spectrum. What you describe seems to be covered by amateur radio usage, ...

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L5 is a high value inductor and is used therefore as a radio frequency choke i.e. it blocks RF at the normal frequencies your device uses. However it will pass DC and normal intermediate (IF) frequencies thus terminating the mixer correctly in 300 ohms at the IF pin. C10 and C11 form an output attenuator on the clock signal and this is done, more than ...

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Why do main on/off switches in electronic circuits/appliances (any small, low power electronics) are always placed on the positive wire (in case of DC circuits) between the power supply (batteries, adapters, etc.) and the circuit regardless of the low operating voltage? Mostly, convention. You connect ground to the negative supply, because that's what all ...

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it's a snubber. MOV is the crucial part, it dumps the energy from the winding. The cap is an additional type of snubber, it takes part of the energy before the MOV starts conducting. The MOV has to have lower brakdown voltage as capacitor rating. From the datasheet the MOV is has a breakdown voltage of 56VDC.

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Here's an LTSPICE simulation of Woodward's circuit done by someone on DIYAudio: Obviously R5/R5A represents the potentiometer. It's interesting (in theory anyway).

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You can use an optocoupler if the motor is small, otherwise, you can use the output of the optocoupler to turn on the motor driver. Here is the circuit of the optocoupler, notice the LED that turns on the transistor on the other side. Here is the circuit with the optocoupler substituting the LED: simulate this circuit – Schematic created using ...

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Figure 1. Industry standard solution. Source: http://Rakuten.com. You could consider making a clapboard and this may be of use synchronising the audio too. For the data connection you can add a switch or make one by pinning stranded wire to the moving and fixed halves of the clapboard so that the circuit is closed when the board is closed. If you really ...

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If you only have a push-to-make switch:- simulate this circuit – Schematic created using CircuitLab

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You could build a circuit like this with SPDT switch simulate this circuit – Schematic created using CircuitLab

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Try using a change-over type of push-button, also called a single pole dual throw (SPDT) switch. Connect it like this: simulate this circuit – Schematic created using CircuitLab

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All switching circuits will require some current flowing through the switch. It can, however, be kept small so that a miniature switch can control a larger switch rated to connect and disconnect the load. simulate this circuit – Schematic created using CircuitLab Figure 1. Simple relay circuits. Figure 1a shows a simple 12 V relay switching the load....

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You could use a device like the TI TPS25940A eFuse to improve the circuit. It has current ramp rate control for cap charging and programmable current limit. This particular part has a minimum current limit of 0.6A it looks like. But you can probably find a similar part with lower current specs. It will be cheaper too. The eFuse voltage drop should be very ...

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The article says:- Positive feedback from A1's output to P's CCW terminal results in constant current drive to the pot... the differential voltage seen by A3... From this I figure that each half of the pot is driven with a current proportional to the input signals. The voltage across each half is proportional to its resistance, so as the pot is turned ...

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Perhaps a simple example with real numbers may help to explain the difference: Lets assume that for high gain we need a dynamical collector resistance of rc=20k. In case of an ohmic (passive) part we have rc=Rc=50k with a DC drop of 20V for Ic=1mA. In many cases, this is unacceptable (Large supply voltage). Using a BJT as an active load it is not a problem ...

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FET's are voltage controlled Resistors according to (Vgs-Vt) over the range of Vds to sink current from Nch Enhancement mode FET's when used as open-drain loads to a power source. The resulting Rds is not inverse linear with gate voltage but can be made linear with feedback. BJT's are exponential Vbe controlled current sinks or with a base resistor, ...

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There appears to be a couple of problems with your design: D5 needs to be a Schottky diode (and as noted in another answer needs to be able to withstand the string voltage). I doubt that D1 is reverse avalanching since you don't report excessive heat, but you should select a more appropriate diode. The chip draws pulses of current up to 750mA, but with L1 ...

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If you use a resistor as a load, then its slope resistance (change in voltage drop / change in current through) is equal to its resistance. If it's being used to bias an amplifier, then its voltage drop is given by the bias current times the resistance. If you want a higher slope resistance, then you have to suffer a corresponding higher voltage drop at any ...

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Active loads are used because transistors are cheaper and easier to make on a silicon die than resistors are. Therefore, you find active loads more in integrated circuits while discrete circuits tend to use resistors more for simplicity (and less availability of matched transistors). I cannot understand why dc voltage drop across them is lesser ...

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Please use a diode with higher voltage ratings. When the regulator is switching, SW is pulled to GND and then released. C116 charges up to 10V, but in the next cycle D5 reverse breaks down, and activates the overcurrent protection of U23. Soon both D5 and U23 will be very hot and D5 may even die. If it does not work even after you replaced D5 with a higher ...

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My (quick and dirty) AM Tx design for a low power personal use requires you know how to generate the input frequency amplitude exactly or at least close enough for personal use. Using the low end of the AM makes it easier to listen to and use a 1m wire as a loop antenna in series with the 10K resistor to limit power to 1mW with a 9V battery and a low ESR e-...

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Unfortunately it's not easy to find a monostable multivibrator IC that will give you a 10 ns pulse. Want you want to do is purposefully synthesize a glitch. Like this: simulate this circuit – Schematic created using CircuitLab R1, R2, and C2 debounce the switch. Using a Schmitt trigger inverter will help with debouncing as well. The inverter drives a ...

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Your best bet might be to start with a 100 MHz oscillator, a dual flip flop, and an AND gate as shown below. There are other combinations of flip flops and logic that will do the job. You will have to make sure you de-bounce the switch and watch your layout carefully. You'll probably want ECL parts to keep your pulses square, otherwise you will have rise ...

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This is not an answer but just to help with terminology. simulate this circuit – Schematic created using CircuitLab Figure 1. In switches "open" and "closed" refers to the gap between the contacts.

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A 2.2mH inductor + 736pF capactor is one hell of a Quality factor (would be about 10k with 10 Ohm serie resistance - physicaly impossible). $$Q = 2/R*\sqrt{L/C}$$ with L,C & R the pure inductance, pure capacitance and total serie resistance of the tank. High quality factor doesn't make your life easier : extremely narow bandwith makes it easy to ...

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Does anyone have any idea which way I should go here? Consider initially the 3 op-amp amp Instrumentation Amplifier (InAmp): - Gain is set by one resistor ($R_{gain}$) and, if you throw away the final stage op-amp circuit and take your differential output directly from the two op-amps to the left you should get what you want. You do need to spend a ...

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