# Tag Info

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A good place to start on how to connect gnds is the following link: http://www.analog.com/library/analogDialogue/Anniversary/12.html

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I would use a dual opamp and connect it as a voltage follower i.e. the - input is connected to the output, each output goes to the separate loads. The + inputs go to the output of the original opamp.

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Start with power supply requirements - does the prospective replacement work on the power rails supplied? Then, on a similar theme, determine the amount of ripple or noise on those power lines to make sure the replacement has power supply rejection figures suitable (or maybe better than the current op-amp). Look at data sheet graphs for this. They should ...

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They might become loose over a couple of years. You should use a latching connector, but that really depends on your application and requirements. Check the usual suspects, 3M, Tyco, Molex, Hirose and tons of others. It is usually the work of the assembly house, either SMT or TH (Contract manufacturers), PCB manufacturers produce PCBs, not assembled ...

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The simple answer is that is a just an example circuit. LTSpice needs you to have ground path for each node, easiest way to do this is put the standard grounds on both sides. To simulate isolated grounds, put a very large (1,000,000,000 ohms or so) resistor and a very small (1 pico farad) capacitor in parallel. This gives you an isolated ground as you ...

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First off, the precision of your resistance value is going to be limited to the resistor's rating. Consider the list of standard values for resistors with a 5% tolerance. Any number between 10 and 91 is going to be within 5% of one or two of the preferred, E24 numbers. Your perfect resistor might even be in your supplies, labeled as one of these. E24 ( ...

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You've come up with a neat, creative idea for a science project, and I think it's great that you're interested in electronics. However, in this case, I think you've bitten off more than you can chew - you don't seem to have enough experience in basic things. Regardless I'll try and give you some starting points: Calculating the Theoretical Resistance of ...

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A good transformer designer will left no space in the "window". The sizes of cores are standard dimensions for standard powers. You always need an ammount of iron and ammount of copper, regardless what is the ratio and current/voltage. Higher voltage means more turns, but thinner wire, lower voltages require less turns but thicker wire. Comparing cross ...

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Assume you put N turns of copper wire through a 'window' in the iron core. The length of that winding will be more or less independent on the area of the wire used. This means that the resistance of the winding (a bad thing) will be inversely proportional to the copper area. A good transformer has as low a resistance as possible, so will have the largest ...

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The fundamental RMS value of the output waveform for a square wave is given by: Vrms = (4 X Vdc) / ( Sq-rt-2 X Pi) However a square wave produces 33% 3rd harmonic, 20% 5th harmonic, 14% 7th harmonic in addition to the fundamental. If the positive and negative square pulses are only 120 degrees instead of 180 degrees, the fundamental will be about 86% of ...

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So once again, in an inverter design without a transformer, what DC input voltage would result in 120VAC naturally? If you use a H bridge output then you need a DC supply capable of delivering the peak voltage of a 120 V AC sinewave. That's about 170 volts but you'll probably need a couple of volts more to account for volt drops.

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DRAKE radio encyclopedia 1928 edition .Fascinating to see what they knew then .If you look up "computer " there is nothing.Very entertaining book .

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the op-amp is correctly wired up with in an inverting circuit configuration. because of the negative feedback through passive components, the "-" terminal is a virtual ground. the node equations ($V_2$ is the voltage at the node where are $R_1$, $R_2$, $C_4$, and $C_3$ are connected) are: $$\left(\frac{1}{R_1} + \frac{1}{R_2} + sC_4 + sC_3 ... 1 I find that there is a bit of a gap between a board passing DRC and a board successfully being manufactured. Some PCB houses have an automated online tool which analyses your Gerber and Excellon files and let you go through what the board will really look like in a "what you see is what you get" fashion. Just upload a .zip file, wait for 2min, and play ... 2 I use Electronics Design Checklist by Hank Wallace. It sums up the most important points to take care of in your schematics, PCB design, PCB assembly, ensuring testability, maintainability and more. 1 Figure 1 in the data sheet: - It also says 0.8 mA on page 1. The better option would be to use a DC (average) milli ammeter. -1 I'd like to recommend EasyEDA.It's a free web based PCB tool for anyone involved in electronics design and able to share the work with others. I have been using EasyEDA for over a year now with great results. The most startling things about EasyEDA are not just that it is free up to the point where you want to physically buy a (low cost) PCB directly from ... 0 Your transfer function is NOT correct. The numerator must be $N(s)=-H \omega_0^2 s$. Therefore, the midband gain is $A_m=H \omega_0 Q$. For $A_m=1$ we get $H=\frac 1 {\omega_0 Q}$ (note that H is given in seconds.) Using your components we have $A_m = \frac {R_5} {R_1 + R_1C_4/C_3}$. EDIT: I have to modify my statement that the function as ... 0 If you set$$s = \omega _0A(s) = \frac{-H \omega^2}{(2+\frac{1}{Q})\omega^2}$$and if A(s) = -1, then$$H = 2 + \frac{1}{Q} Note that the - sign implies that the output will have a phase shift of 180 degrees.

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Power in electrical and electronic systems is voltage x current. Voltage and current are both signals but power is one signal x another signal. In other words, power doesn't exist as an individual signal. However, if you had two voltage signals with resistive loads connected to each of them that dissipated the appropriate power, then you could attenuate one ...

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You need to make the [closed loop] output impedance equal RL. If the speaker is inductive, as in real life, it gets more complicated as you want power factor correction essentially by having an output impedance that is the complex conjugate of the load. See http://en.wikipedia.org/wiki/Maximum_power_transfer_theorem

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It's all about the context of the larger application. "Clock divider" implies to me that the output will also be used as a clock, or maybe a clock enable signal. "Preamble generator" implies that the output will be used as data. This is a fairly trivial example, but in more complex projects, meaningful names can go a long way toward making a design ...

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I would like to point out that a clock divider would imply multiple clock domains, something you might not want to do in high speed design. In the high speed design, logic changes state on a clock edge. Adding what can be viewed as another clock can make routing and performance more difficult. The preamble generator is more an accurate description of the ...

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