Tag Info

New answers tagged

0

I mean the voltage across inductor is Vsinωt - Vc = L di/dt, so based on derivative being positive, the current increases.But what makes the current go to infinite? After all for a simple L only circuit, the voltage across inductor is Vsinωt = L di/dt and it is not getting infinite In an L only circuit, the voltage across inductor is $$ ...


3

Can anyone explain how does the current amplitude go to infinity, at resonance? The brief answer is that the AC steady state current isn't infinite but, rather, the circuit has no AC steady state solution. Recall that one of the assumptions justifying AC (phasor) analysis is that the circuit is in AC steady state, i.e., that all transients have ...


0

The real world application varies upon the application and the type of the signal. If linearity is not a concern, then the above circuit diagram can be improved to remove the 47(ohm) resistor and the circuit will act as a digital circuit, with almost no sound when there is no light. On the other hand if linearity is a concern for analog applications, then ...


2

That is a really crappy circuit since it depends on the gain of the transistor being just right. Real transistor gains vary widely, even within the same production lot. Competently designed circuits work from the minimum guaranteed gain of the transistor to at least 10 time that, preferably to infinite gain. I would start with the transistor stage setting ...


5

I understand the impedance goes to zero,which explains the steady state current as infinite. But can you explain the same by simple circuit analysis?I mean the voltage across inductor is Vsinωt - Vc = L di/dt, so based on derivative being positive, the current increases.But what makes the current go to infinite? Yes, the net impedance falls to ...


1

This is known as an emitter follower. There is no voltage gain in the circuit, this acts as a impedance buffer i.e. it can drive heavier loads that the CDS cell could on it's own. Light on the CDS cell generates current which goes into the base of the transistor, this current is multiplied by the transistor (this gain is known as Hfe). This current when ...


1

It is a simple non-inverting transistor amplifier. The resistor to ground is what allows it to be non-inverting. When the transistor is off (the CDS cell not conducting), the Headphone output at the emitter of the Transistor is pulled to ground level through the 47Ω resistor. When the transistor is on, the headphone output gets pulled up to 3V, with the ...


0

You could pull a known amount of current out of the inverting input and measure the output voltage. Say -1 Volts through a 100k ohm or 1 meg resistor.


0

Here you are the correct FSM you asking for!


0

The flip flop is used for change states over the time using a clock. If you go on wikipwdia you can easily find Flip Flop D description. You can read this link for learning FSM: http://www.cs.umd.edu/class/sum2003/cmsc311/Notes/Seq/fsm.html First you learn sequential logic you have to know basic combinatory logic rules. If you don't I reccomend to read ...


0

If you want to test the AD5241 it's a lot easier to disconnect one leg and use a meter. Having a small link that can be fitted between op-amp (CMP1?) output and the AD device makes total sense to me. You are possibly aware that if you expect a decent frequency response from this circuit that you'll be disappointed - the self capacitance of the AD device ...


3

Since the - input of the opamp will be at GND, (zero volts) no matter what, the opamp's output voltage will change in order to force enough current through the AD5241 to keep the - input at zero volts, no matter what the diode's doing. So, if the AD5241 is sitting at, say, 1 megohm and everything is stable, then if you shunt the AD5241 with 1 megohm, the ...


0

With help of Latching switch and LDR we can do that. Latching Switch . Load Circuit Replace Load with the circuit as shown above and push button (S1) with LDR.


0

You blew it on this one. If your intent was to build the controller from scratch, then you need full data on the extruder. However, the seller clearly states: Pls kindly note that the manufacturer doesn't provide any English manual for this item. However, there are ways to get clues about the hookup, at least. The seller's page says that the ...


-2

Entered by truthtable: F1 = A' B C' + A B' C; Minimized: F1 = A' C' + A C;


1

you have to read theory of DC/DC (buck or boost) converters. http://www.learnabout-electronics.org/PSU/psu31.php


2

Despite several requests for detailed requirements, not nearly enough has been forthcoming, and it has been requested "What I am looking for is feedback regarding the circuit itself, addition of decoupling capacitors, impedance matching problems, timing problems etc.. not the models themselves". Well, I'll do my best. What requirements are known are -ESD ...


2

Here's one viable approach.. the buffer simulates the micro output. The output in this case is 0-Vdd = 0-5V and that is filtered and buffered and amplified by 1.6 to give 8V. I used 1kHz for this, but you can adjust resistor and cap values to suit. For precision, you can use an actual buffer and give it a separate reference supply. The op-amp type will ...


0

works well after simulation as well. Well, no. Your op amp is configured as a non-inverting gain of 7. Its inputs (pins 5 and 6) have, according to your last schematic, 2.5 and .5 volts on them. Does this not tell you that you have a problem? For an opamp operating normally, the difference between the inputs will be microvolts. Your simulation appears ...


2

You got the right answer, but I didn't check your algebra, LOL.


8

It depends a lot on the volume of production. If the volume is small (or not sure) it's best to have a header or spring clip. You can always not populate it later, and the cost is minimal unless the board is really cramped. I always like to see well-marked ground and power supply test points, even if they're just holes or pads. There's no point in testing ...


1

It looks to me like a CMOS 7200 wristwatch chip would do what you need. According to the CMOS Cookbook, if you connect a 1.55 volt input through a 10k resistor into the TEST pin on this chip it will put out a 1 Hz square wave across the resistor. That's intended for calibrating the time, but in your application, you could skip all the other hookup to the ...


0

I think you'd be better off dropping your 36 volt battery output to 12 volts and running from there. 36 VDC is a fairly common golf cart battery voltage, and there are very cheap 12 volt buck regulators available for this application. Typically they have quite a wide input range. Try eBay.


0

If the switching regulators have common ground between the input and output, you cannot put them in series. If the input and output grounds are isolated, you can put them series. What is limiting your 42V input to a single regulator? If it is the input capacitor, replace it. If the switching device, MOSFET mostly, cannot handle 42V across its source and ...


0

The Potentiometers that Andy aka mentioned are overkill. You will probably need to disipate around 30 watts - still too much for a nice, small pot. Stick with PWM, but you might want to use a dedicated circuit instead of the arduino. Something like the PWM-Circuit shown here would be about right. It gives you a pot to control the brightness, but it uses ...


1

Given that you want to shut current down to maybe 25mA (100:1) reduction you might be looking at a pot resistance of about 250 ohms ohms. Point 1 Point 2 is the power rating - with the pot at a setting that maybe controls half the amperage (1.25 amps) it might be dropping 3 V across it and that means a resistance of 2.4 ohms and a power dissipation of about ...


2

Your analysis is incorrect. For one, the 6V source is not all across the 3k resistor so the current thru the 3k resistor is not 2 ma. The easiest way to solve this is to first find the thevenin equivalent of the 6v source and the 3k and 6k resistors. The equivalent voltage is simply (6/9)*6 or 4 volts. The equivalent resistance is 3k in parallel with 6k ...


1

About the charging circuit: The circuit connected to the ADJ pin controls the output voltage of the LM317 regulator. The output voltage is 1.25V above the voltage at the ADJ pin, which is set by the R3/R4 divider. R2 and Q1 are used to limit the output current. When the voltage across R2 is high enough to turn on Q1, Q1 pulls down the voltage on the ADJ ...


0

I wanted to comment to Spehro, (But I lack in reputation.) As a first pass guess, assume Vdiode ~ zero. Then there is 1 mA into the diode node from the 10 V supply and 2mA drawn from the 20V supply. So there's something like 1mA going through the diode. In practice that would be enough. For your class you'll have to work out the "exact" answer. (for ...


1

As Spehro said, I would prefer combining the 2 sources and resistors into a single one. This can be done by noting that the two sources + resistances are in parallel branches. The formula for equivalent Voltage in parallel branches is V_eq = ( V1/R1 + V2/R2 + .....Vn/Rn) / ( 1/R1 + 1/R2 + .......1/Rn) Be careful about the polarities of the batteries ...


0

I don't know what the complete diode model you refer to is, but for an ideal diode: Consider the circuit without the diode. What is the voltage between the resistors? 15V falls across each resistor, so it will be 10-15 = -5 volts. So when we add the diode back into the circuit it will be forward biased (or "on") and the cathode will move to just below zero ...


-1

What about a high-wattage Zener across the bulb? Simpler and does the same thing (in this case) as the crowbar.


4

To get useful data points it's helpful to look at what the fuse will likely do from a datasheet: At the standard temperature (usually 25°C) and 1A, a 600mA-rated fuse will typically blow in about 15 seconds. The time will be less if the ambient temperature is higher, and greater if the temperature is cooler than 25°C, but even a +/-10°C ambient change ...


3

You did not state the current of the bulbs, so we can't predict whether they (alone or collectively) will blow the fuse. Hence the only sure way left to blow the fuse is to connect it directly (no lamps) to one of the batteries (assuming they can deliver enough current, which might no be the case, especially for the 9V battery). Note that a 0.6A fuse means ...


3

As mentioned, you are more than likely going to damage the bulb before enough current will flow to blow the fuse. You might want to consider a crowbar circuit. http://en.wikipedia.org/wiki/Crowbar_(circuit)


4

You can blow the fuse by allowing more than 0.6A to flow through the circuit. The voltage doesn't blow the fuse, but the current. To be able to predict the current for a specific voltage you need to know the resistance of your bulbs, then calculate the total resistance of the bulbs in parallel, then apply Ohm's Law to calculate the current. You're far ...


4

You need a 24 VAC transformer to power the thermostat, and a 24VAC (coil) 120VAC (or more) (contact) 1amp (or more) (contact) relay. The 24VAC supply goes to the thermostat and one side of the relay coil. The other side of the relay coil goes to the switched terminal, and the (24V supply) neutral goes to the thermostat. The hot supply of the 120VAC goes ...


1

I see two glaring errors: The Zener is a 1N4735A, and is specified as having a Vzt of 6.2V +/-5% with an Izt of 41 mA through it. With 6.2V across the string comprising R2, R3, R4, R5, R6, and with R2 set for minimum resistance, the current through the string will be 459 microamperes. With R2 set for maximum resistance, the current will be 264 ...


1

The potentiometer appears to me as a way to adjust the clamped voltage by adjusting the current flow through the resistive path. More current through the zener diode and the voltage will raise a bit, less current and the voltage drops. It's likely used as an adjustment to fine-tune the reference voltage seen at the head of the resistive divider. For a ...


2

The zenner limits the input voltage to the resistor ladder to 6.2V. The resistor ladder divides the voltage up into discrete steps for comparison. For each step sum the resistors above and below, then use the 6.2V as the input voltage (\$V_{IN}\$) to the standard voltage divider formula: $$V_{OUT} = \frac{R_2}{R_1 + R_2}V_{IN}$$ \$R_1\$ is the sum of the ...


0

This will minimize "pops" on plug-in, and will keep the amp quiet when nothing's plugged into it:


0

Usually the tip is the right channel, the ring is the left channel, and the sleeve is ground. Therefore on your diagram, 1 is the (-) input, 2 is the (+) if you want to use the left channel, 5 is the (+) if you want to use the right channel. Pins 3 and 4 are connected to 2 and 5 when there is nothing connected and disconnected when something is plugged in, ...


4

If you want both the left and right channels to provide audio you need to make a little mixer. Two resistors (around 10KΩ) connected to the left and right inputs (pins 2 and 5), then joined together, will make a simple mixer. Then you link that to the LM386's IN + through a capacitor (around 10µF is usually OK). Pin 1 links direct to the IN - You ignore ...


4

There is no one single strategy, and I don't think you can expect to get the design absolutely right without some trial and error. Some blocks may need buffering from others. It is common to minimise external influence on the LO, and that often means a buffer between it and the mixer. An emitter follower (or source follower with a FET) running at fairly ...


-1

Designing stuff by blocks is all very well but at the end of the day it is one circuit and it should all work together. Adding buffers between blocks just to make the "block diagram" meaningful is pointless in the main. Live with the less-than-perfect loading of the mixer on the local oscillator and if necessary alter the L.O. design to give a slightly ...


1

I don't know any part numbers off the top of my head but they do make Current Sense IC's which basically measure the voltage drop across a very low value (0.5 ohms or less) resistor. You set the threshhold at one of the IC pins with a voltage divider and another pin gives you a TTL high or low output. We used these to monitor current levels in an RF ...


0

One way would be to use a microcontroller, if you're familiar with them. 3 output pins of a microcontroller (ANODE1, ANODE2, ANODE3) would get connected to 3 anodes of all RGB LEDs. Then 3 more output pins of a microcontroller (SELECT1, SELECT2, SELECT3) to 3 bases of NPN transitors, connected to 3 cathodes of the LEDs. simulate this circuit – ...


4

Generally there is no modification necessary. Two of the 1x8 headers can be used side by side to mimic the 2x8 configuration. (Note that: Yes it is possible to find long tailed dual row headers in the market). You could probably use a small amount of epoxy or silicon adhesive to glue the bodies of the two single-in-line headers together to create the 2x8 ...


0

Thevnin voltage gives the open circuit voltage across two terminals (capacitor removed in this case). And\$V_{th} = \cos(4t)\$ in here. When a load (capacitor in this case) is connected across these terminals, the voltage across the terminals A-B will change and you can use voltage division rule to obtain the output voltage. $$V_{AB} = ...


1

The input of the circuit is a 2 bit number (max value = 3) and the hence the maximum value at output will be 27 (a 5 bit number). The truth table of the circuit is: $$\begin{array}{cccccccl}&\mathbf{x} & &\mathbf{b} &\mathbf{a} & &\mathbf{Y_4} &\mathbf{Y_3} &\mathbf{Y_2} &\mathbf{Y_1} &\mathbf{Y_0} & &x^3\\ ...



Top 50 recent answers are included