(Welding) Why Is "earth Clamp" Called so When It Originates From the Negative Terminal?

It is conventional, for safety's sake, to connect one side of a high-power electrical circuit to a point close to earth potential. This process is called earthing or grounding. The earth clamp is conventionally on the negative side of a DC circuit, but it does not have to be. If the circuit is not completed by the clamp, the welding current will not flow. There is, however, no chance of a shock unless you touch the work while the welding electrode is also applied

(Welding) Why Is

1. if I get a really short seat post would it work in me eastern growlers intagrated clamp?

Na its not going to be too close to the ground

2. D.U.I can clamp for income tax?

Of course not! Fines and penalties are NEVER tax deductible! How could you POSSIBLY think that you should be rewarded for a DUI??! I did not get one, so where's my "bonus" for staying sober??

(Welding) Why Is

3. Just curious... how do sign stealers pry off those dual-sign clamp things?

why are u asking something thats illegal?

4. How to make your own piercing clamp?

You can get clamps for around $6. Just buy some

5. If you were to add "DC offset” with a clamp, can you easily remove it afterwards?

It is interesting to see what coupling capacitors really do in AC amplifiers. The usual viewpoint at the capacitive coupling is that capacitors prevent affecting the DC biasing current from the previous stage; thus the name "DC blocking capacitors". For example, Wikipedia says:"In analog circuits, a coupling capacitor is used to connect two circuits such that only the AC signal from the first circuit can pass through to the next while DC is blocked. This technique helps to isolate the DC bias settings of the two coupled circuits."This explanation is true and it does good work for conventional "teachers" that do not like their student to "trouble" them with confusing questions-:) But it is a formal and sketchy "explanation" since it actually does not explain how the AC signal passes through the coupling networks and what they really do. The frequency-domain explanation where the CR coupling hetwork is presented as a high-pass filter is even more formal...Here, I would like to suggest to your attention more four fresh and not so usual time-domain explanations of the legendary capacitive coupling technique; I have been using them at the lectures with my students through years...The key point of understanding and explaining the capacitive coupling is first to see that the capacitor is "hanged", as a kind of bridge, between two sources: from the left side - by the perfect input voltage source that has to ensure a galvanic path to ground; from the right side - by the (more) imperfect voltage source built by the biasing voltage divider and the power supply. Then, assuming the input voltage is zero, it is easy to see that the capacitor charges to the bias voltage of the right source (the left is absent) and the amplifier is correctly biased. Finally, wiggling the input voltage fast enough (so that the average voltage across the capacitor stays almost constant), we can use one of (and why not the all?) viewpoints below to explore the circuit operation in an attractive manner.PERFECT VOLTAGE SOURCE IN PARALLEL TO IMPERFECT ONE. First, we can see that two voltage sources are connected in parallel to the amplifier input. The left of them is composed of two perfect voltage sources in series (the input source and the charged capacitor); the right is the biasing one. So, in this arrangement, the perfect left source (Vin Vbias) dominates over the imperfect biasing source and imposes its full voltage on the amp input.CURRENT STEERING. The BJT emitter-grounded amplifier is a special case of this arrangement since its input (the base-emitter junction) behaves as a constant-voltage nonlinear element (a voltage stabilizer). Now we have two (almost) perfect voltage sources connected in parallel and this combination is supplied by the biasing current source (the base resistor and the power supply in the attached picture below). So, when the input voltage varies, the biasing current vigorously diverts (is steered) between the input source and the amp input.SERIES VOLTAGE SUMMER. Of course, we can see the series voltage summer (according to the Kirchhoff's voltage law) needed for the biasing - it sums the input voltage Vin and the biasing voltage Vbias across the charged capacitor so their sum Vin Vbias is applied to the amp input.VOLTAGE SHIFTING. And finally, we can imagine the capacitor is a kind of an electrical "shock-absorber" shifting (translating, moving, displacing...) the input voltage variations.Let's consider the three typical situations in the circuit operation:1. CHARGING THE "BATTERIES" (ZERO INPUT VOLTAGE). For clarity, first assume in the beginning, when we switch on the power supply, the input voltage is zero and the left C1 terminal is grounded through the DC zero source resistance. The capacitor C1 and the base-emitter junction are connected in parallel.The capacitor is initially empty and diverts all the biasing current flowing through the loop E > Rb > C1> ground > E. The base-emitter junction is shunted by the capacitor; it is cutoff and does not affect the charging process. When the voltage across the capacitor reaches 0.6 V, the base-emitter junction begins diverting a part of the biasing current and finally diverts all the current. The capacitor does not play any role since there is no current flowing through it; the biasing current is determined completely by Rb. Thus, at the end of this initial stage, the input capacitor C1 is "automatically" charged to the needed bias voltage 0.6 V.Similarly, the output capacitor C2 is "automatically" charged through the galvanic load to the quiescent voltage 1/2Vcc (it is interesting to see how). Until this initial stage, when the capacitors charge, the amplifier does not work properly and the load can be even damaged if the capacitors have too big capacitances.We can present this initial stage in an impressive way to readers, if we made them think of the capacitors as of slowly stretching shock-absorbers (the red voltage bars Uc1 and Uc2 give some impression of such an representation).2. DISCHARGING THE "BATTERIES" (POSITIVE INPUT VOLTAGE). When the input voltage rises above the ground (becomes positive), the total voltage of the left composed voltage source (Vin Vc) exceeds the base-emitter voltage (0.6 V). So this source adds an additional current to the initial bias current (set by Rb) through the base-emitter junction thus discharging. But because the capacitance is large enough, the voltage across the capacitor decreases slightly, and the input capacitor conveys the increasing input voltage variations to the transistor base (they are "lifted" with 0.6 V as the transistor "wants").In a similar way, the output capacitor descends the collector variations with 1/2Vcc thus conveying them to the load as it "wants".In the impressive mechanical analogy, the shock-absorbers convey the fast input/output movements up.3. RECHARGING THE "BATTERIES" (NEGATIVE INPUT VOLTAGE). When the input voltage descends below the ground (becomes negative), the total voltage of the left composed voltage source (Vin Vc) becomes less than the base-emitter voltage (0.6 V); so the composed source begins diverting a part off the biasing current thus recharging. Again, because the capacitance is large enough, the voltage across the capacitor increases slightly (it restores the charge), and the input capacitor conveys the decreasing input voltage variations to the transistor base ("lifted" with 0.6 V as the transistor "wants").In a similar way, the output capacitor conveys the increasing collector voltage variations to the load as it "wants".In the mechanical analogy, the shock-absorbers convey the fast input/output movements down.

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Application of Operational Amplifier As Differential Clamp Diode in Comparator
The influence of internal differential input clamp diode on operational amplifier when operational amplifier is used as comparator. I ask a question - will these clamps affect the operational amplifier circuit? If the voltage between the two inputs of the operational amplifier should be about zero, then these diodes will never be forward biased in the standard operational amplifier circuit... Or will they be forward biased? A little reminder, we are discussing some differential clamp diodes that may appear in some operational amplifiers, see Figure 1.Generally, the influence of operational amplifier circuit can be seen in the basic non inverting amplifier configuration structure (including a simple g = 1 buffer amplifier). Let's look at a forward input step. The output cannot immediately follow the change of surge input voltage. If the input step is greater than 0.7V, D1 is conductive, affecting the non inverting input. When the operational amplifier is turning to its new output voltage, the current at the input of the operational amplifier will suddenly increase to a higher peak value, as shown in Figure 2. Finally, when the output "catches up" with the input, everything gets better again.Many applications themselves process slow or band limited signals, which is much lower than the conversion rate of operational amplifiers, so this will certainly not happen. In other applications, even if the input voltage changes rapidly, the input current transient will not adversely affect the operation of the circuit. However, in some special cases, the input current pulse will cause many problems. One notable case is the multiplex data acquisition system. The following figure shows a simplified case of this system with only two input channels.In this example, the multiplexer switches between channel 1 and channel 2, so it is required that the output of U1 can be quickly converted from - 5V to 5V. The forward bias of D1 and the resulting input current transient pass through the multiplexer switch, thereby releasing the voltage of C2. The R / C input filter is usually used to maintain a stable voltage during channel switching, but the current pulse partially discharges C2. Now, C2 needs more time to recharge to the correct input voltage, which reduces the multiplexing rate, that is, the accuracy. The solution is to select an operational amplifier without differential clamping for U1. FET input amplifiers such as opa140 have low input bias current (to reduce the burden of MUX series resistance) and no differential input clamping, which is very suitable for multiplexing input. Opa827 performs well in most applications - FET input, very low noise, high speed, stable and fast. However, it has some differential input clamping, so opa827 may not be the best choice for operational amplifier multiplexers. Previous posts focused on differential clamping and introduced the general principles of using various types of operational amplifiers. Refer to "operational amplifier as comparator" for details. I don't want readers to have the impression that differential input clamp operational amplifiers are risky and should be avoided, but this is not the case. In a few cases, they will affect your circuit. But if you know this, you won't make blind choices. Do you find that differential input clamping affects your circuit in other ways?
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