Hello ladies and gentleman,  I'm trying to figure out how this circuit works :   I was thinking it is a kind of charge controler where Rx is the load. Any help would be pleased
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I have used the above simple circuit for a constant source where \$V_{\text{in}} = -5\$V, \$R_{\text{in}} = 390\Omega\$, \$R_f\$ is the load which ranges from \$110\$ to \$170\Omega\$....would using this circuit for a constant current source for the load be recommended?  As current $$i = -\frac{V_{\text{in}}}{R_{\text{in}}}$$ $$V_{\text{out}} = -V_{\text{in}}\frac{Rf}{R_{\text{in}}}$$ so it should serve as a constant current source, but while browsing the Internet for a constant current source I typically found these sorts of circuits:        Is there a compulsion to include the transistor in my circuit? If yes, why wouldn't my circuit serve the purpose for producing a constant current?
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I've implemented above circuit and it was successful.  The given circuit needs 12VDC to run properly.  I want to run this circuit on 5-6VDC or 9VDC whichever is possible. I know I have to change resistors but not confirmed which to use.  Also if I use 9V battery only to activate 6V relay for 8hrs daily, for how many days will that battery work?
98539
Does anyone know of a Voltage Controlled Current Source IC? Or do you know how I can build one with as few other components as possible taking up a minimal amount of space?  current range needed is from 0mA to 350mA
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I think I may have found an instance of "hill-billy" wiring in my circuit box.  I have a 15A GFCI breaker.  Strangely, it is hooked up to the up stairs bathroom AND the outdoor outlets.  Recently it has started to perform poorly.  It either trips easily or doesn't work after reseting the breaker.  I opened up the panel and was surprised to see what looks like two circuits wired into the breaker.  That is, you have a two pairs of black/white (hot &amp; neutral I presume).  The hot lines are going into the breaker together (on the top) and the same with the neutrals (on the bottom).  Is there any valid reason for this? It seems like the installer was lazy and because there isn't much room left on the panel they wired these two circuits into one breaker.  UPDATE:  This is a Bryant circuit breaker box from the late 80's.  The GFCI breaker was a 15 amp Sylvania which raised a bunch of eye brows when I brought it into the electrical supply place. Now it sounds like everyone makes "BR" compatible breakers including Eaton.  I ended up getting an Eaton breaker from Home Depot and only wiring in one pair of wires (the wiring to the outside sockets).  Seems to work fine.  I am going to buy another breaker and put it in one of the open spots in the box.  My guess is that whomever was last in this box didn't want to spring $40 for a second GFCI breaker.    UPDATE 2:  Thanks for the comments!  Giving that the house was built in 1986 it seems like it was standard to use one GFCI for multiple outlets.  In my case - this certainly caused problems over time and I am replacing with multiple GFCI breakers.
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I bought this new battery really cheap (it is really small and i needed a small 12v battery) but i am not sure of some things. First of all, it says it is 1.3Ah/20HR. Does that mean that it can deliever 1.3Ah for 20 hours? Like, i could power a 12v 1A light bulb (estimately...) for 20+ hours from it? Second of all, is this a deep cycle battery? ( i do not speak polish, so i do not understand the label). If i fully discharge the battery (9-10 volts) will it lose its capacity to store charge, or there is no problem? What is its optimal voltage? I used some rough google translate and it turns out that the "standby voltage" is 13.6-13.8 volts and the "usage" voltage is 14.5-14.9 volts. Isn't that too much for a 12v battery? Right now it is at 12.6V. Is it undercharged? Also, what is the voltage needed to charge it and is it a bad idea to charge it with a stepped up 5v power supply? (Phone charger, stepped up to the battery voltage required for charging)
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Well, I have the following circuit:         &ndash; Schematic created using   When the switch is closed for a long time, the current that is flowing trough the resistor \$R_1\$ and the coil is given by:  $$\text{I}_\infty=\frac{\text{V}}{\text{R}_1}=\frac{10}{100}=\frac{1}{10}\tag1$$  With that given I can solve what happens when I open the switch. The current trough the coil will stay in the same direction as when the switch is closed. Now, at the moment when I open the switch the voltage across the coil \$\text{V}_{\text{Y}_1}\$, is given as:  $$\text{V}_{\text{Y}_1}=-\text{V}\cdot\frac{\text{R}_1+\text{R}_2}{\text{R}_1}\cdot\exp\left(-\frac{\text{R}_1+\text{R}_2}{\text{L}}\cdot t\right)\tag2$$     So, the voltage over the coil will start (when I open the switch) at \$-\text{V}\cdot\frac{\text{R}_1+\text{R}_2}{\text{R}_1}=-10\cdot\frac{100+100}{100}=-20 V\$, which is (absolutely seen) higher than the supply voltage. How is that possible? Or is the formula \$(2)\$ wrong?
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Just wondering if someone can give me an explanation for this capacitor layout - what is it's function? The chip to the right is an ADC.  Circled in blue.    I'm guessing it is well known but i don't know how to search for it.
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I have seen countless times the following diagram for a Schmitt trigger oscillator.    As shown in the picture, the slow-rise wave-form generated by the charging and discharging capacitor is translated to a square wave-form at the output, as a result of the Schmitt trigger. However, I have no idea how the capacitor even charges up in the first place. I don't see any "input voltage" other than the upper and lower bounds on the actual Schmitt trigger. How is the capacitor charging up in the first place?  I'm sorry if this is a very dumb question.
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There is a circuit shown below where we have a sub-circuit I marked in red. I know that it is a source follower with the AC behavior shown below. As you can see the circuit has DC voltage input only. What is the DC behavior of the part I marked with the red arrow?
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Can anybody explain the necessity of the RC network shown in the figure. I found this in circuit for a logic power supply using Fly-Back converter?
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What will happen if motor requirement due to more load is high(lets say 5A) and we limit current with current limiting circuit to lets say 2A. what will happen.. will it work without anything getting damaged?If not what should i do? Thanks..
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I simply can't figure out what the capacity of this circuit is.  I can do the math myself, I just need a hint how to create an equivalent circuit where it is obvious what is parallel and what is in series...
35623
Ok here's the problem. I have a power antenna in my car. The antenna goes up when a voltage is supplied to the voltage sensor in the antenna and goes down when the voltage is removed. I used this fact to install a switch allowing me to listen to CD's without deploying the antenna.   The problem is: Whenever I start the car when the radio is on, the voltage drop is enough so the sensor registers a loss of signal and begins to withdraw the antenna into the car only to redeploy three seconds later when the car is running and the voltage returns to normal. This is hell on the gears and the torque switch. My plan is to install a capacitor in the signal circuit so that the voltage sensor no longer registers a temporary loss of signal when the car is started but will still withdraw the antenna (after a bleed time ) when the car is off or the switch is turned off.   Two questions: Since I need about a 10 second bleed time how big a cap do I need? And  I install the cap in parallel across the signal leads, correct?  \$Update\$  Got it done tested, soldered, boxed and installed it works great!!!  Thanks to  @Davetweed AND @RedGrittybrick for your help and insight.
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I've been working on a data logging device that will eventually go into my personal vehicle.  I've scoured the interwebs and built what, in theory, should be a fairly stout power supply capable to handle fast transients and undervoltage and the fabled load dump.  My problem is, though... I want to actually test the power supply.  In previous questions that I've asked related to the design of the power supply portion, I've been told that the "right way" to simulate a load dump is with a proper, preofessional testing device.  I'm curious, though... what's stopping me from rigging up something that is "close enough"?  My idea is simply this: get a low-voltage DC power source, such as an old PC power supply.  Get a DC-DC converter to step that 12V up to something in the 80V range.  Assemble a bank of capacitors to get me in the neighborhood of a theoretical 12V load dump..  which from what I can tell is in the hundreds of joules range.  Once charged, simply connect it to the battery input of the device - which wouldn't be connected to an actual power source lest I want to potentially screw that power source up - and see if explodes, and if not, see if it works normally afterwards.  Does this sound like the dumbest idea you've ever heard?  Is it a somewhat decent idea that simply needs a lot of planning and number checking to in order to work roughly as intended?
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I've read the , and I don't get it. I am really interested, but I've read the explanation there a couple of times and I still don't get it how the Duff's device works.  What would a more detailed explanation be?
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I would like to know about the workings of mobile phone battery chargers. I know the basic principle, you take a transformer to bring it down to 9/6/3.7 volts then a bridge rectifier circuit and a capacitor input filter to rectify it into DC. Then you apply it to the battery. But I know this should not be enough, I need at least two things,   Protection against back current flow Switch off power when the battery is fully charged.   Can someone point me to something simple (apart from this )
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In my book there is no explanation of the phrase short circuit but at many places the author has used it. I had googled it. Some explain it as the flow of charge along a high potential difference while others explain it as the flow of charge along a low resistance path. What exactly is a short circuit? An explanation along with a diagram would be very helpful.