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What's a step by step way of thinking about the flow of electrons in such a circuit? How can we tell it would provide constant current for instance?
The first step is to forget about flow of electrons. Think about conventional current flow from positive to negative or ground. This is the way we (nearly) all do it and why we draw the positive rail at the top ...
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There are downfalls with choosing very large resistors and very small resistors. These usually deal with the non-ideal behavior of components (namely Op-Amps), or other design requirements such as power and heat.
Small resistors means that you need a much higher current to provide the appropriate voltage drops for the Op-amp to work. Most op amps are able ...
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The current in any 2 terminal device is always the same, there's no way for a 2 terminal device to provide "extra" current. You can't have 0A in and 4A out.
What it does do is force the current to be the given value regardless of the impedance across it. (So therefore you can't put in an open circuit an ideal current source, the voltage would go to ...
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Photovoltaic cells and betavoltaic cells are examples of devices that produce more-or-less constant current up to some maximum open-circuit voltage, just as a battery produces more-or-less constant voltage up to some maximum short-circuit current.
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R2 (10k R4 in my diagram) is there to form together with C1 (1nF capacitor) a Miller Integrator to prevent unwanted oscillation.
And yes, this circuit will sometimes oscillate, mainly due to poor PCB/breadboard design.
And here you have a real world example (the breadboard one).
Without the Miller capacitance:
And after I add the Miller capacitance into ...
15
The circuit employs negative feedback and utilizes the very high gain of the op amp. The op amp will try to keep its non-inverting and inverting inputs at the same voltage \$V_{\text{set}}\$ due to its very high gain. Then by Ohm's law
$$I_{\text{set}} = \frac{V_{\text{set}}}{R_{\text{set}}}$$
Negative feedback causes the op amp to adjust the transistor ...
15
A current source is a voltage source which - in ideal circumstances - creates as much voltage as needed for the specified current to flow. Think of it as a power supply with an adjustable voltage and a fairy watching the current and adjusting the voltage rapidly to preserve the constant current.
An ideal current source would suppply the required voltage in ...
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Most likely just to simplify the schematic. They would likely be implemented with various scaled and interconnected current mirrors as well as some sort of reference current generation circuit. One complicating factor is that this is 1 channel out of 2 on the chip, and it's entirely possible that the biasing circuitry is shared between the two channels in ...
13
This is what your circuit looks like:
simulate this circuit – Schematic created using CircuitLab
When you close the DIPSW the LED will be shorted thus the current will flow through the DIPSW contacts instead of the LED. So it will turn off.
If you want to turn the LED on when the DIPSW is closed then you should hook up the circuit according to this:
...
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What's the intuition for reading transistor/resistor circuits?
More precisely speaking, "reading" means "understanding" here. So, the question is, "How do we intuitively understand circuits?"
What understanding is
But what does it mean to understand a circuit? For this it is not enough just to see that, for example, "when ...
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Do constant current batteries exist?
["Battery" used for "cell" in many places below]
Essentially, no.
A battery could easily be made that had that characteristic by adding electronics in the battery housing, and 'for extra points' one could be devised which electrochemically did a good job of current limiting at a designed maximum current, which would ...
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Russell is bang-on for conventional electrochemical batteries, AFAIK, which makes sense because electrochemistry works on voltages.
However, there is something that is considered a battery that inherently produces a constant current. See this presentation.
Betavoltaic cells are based on radioisotope sources such as tritium. There are even some commercial ...
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Constant current sources ... there are no such devices in nature; almost all of sources are (constant) voltage sources. So we have to make them.
I think you are correct. Due to physics chemical batteries (DC) and electrol-magnetic (AC) almost always (wriggle room!) create voltage sources. Note, however, that if the load resistance is very low in comparison ...
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I can speak from some small experiences on a part of this. I'll leave the "highly constant and highly stable" current source design to those who know this better than I. But I was involved in creating LED-based light sources intended as "lamp references" for optical work. Your description suggests the same to me.
(1) Even if you have a perfect current ...
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The 12AX7 double triode is connected as a differential amplifier (long-tailed pair). In the Wikipedia article you see a similar circuit implemented with BJTs, but the principle is the same: consider the cathodes to be the emitters, etc.
For optimum performance the AC resistance on the common cathode should be high. This is achieved using an LM334 connected ...
answered Dec 21 '17 at 13:46
Lorenzo Donati -- Codidact.com
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The beta value you are getting is rather typical for a transistor in reverse active mode. That is, using the collector as an emitter.
Assuming every thing else is ok (vendor, supplier, and manufacturer) then confusing the pin-out is your most likely issue.
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If something has infinite resistance wouldn't nothing happen because everything is blocked?
I also don't understand what an "open circuit" really means (to my eye it just looks like a broken circuit that isn't going to do anything). It's all quite horribly confusing to me and all the online guides and answers just repeat the same stuff. Can anyone ...
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The maximum ratings are usually about two things, broadly speaking. One is about local dissipation -- the port pins are often arranged near each other on the die and if you dissipate too much in a small area within the micro, then that area will become a hot spot and there is a limit beyond which you shall not go. Another is about the aluminum interconnects. ...
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To drive 50 mA through 330 ohms, you need 16.5 V.
For 50 mA through 1330 ohms you need 66.5 V.
In the constant current mode, the maximum output voltage of the LM317 circuit is \$V_{in} - 3\ {\rm V} - 1.25\ {\rm V}\$. With 9 V input, that means you won't get more than about 4.75 V output.
Your circuit should work pretty well for load resistance below ...
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D2 generates a stable voltage, while the forward voltage of Q1 varies with temperature (by about -2 mV/deg. C). You can compensate for this by putting a standard diode in series with D2 -- e.g. a 1N4148. Another transistor like Q2 with its base-collector tied together would also work well. Then you need to change R1 as the current is then 2.5 V (from D2) ...
8
The current rating of a power source (your battery) is the maximum current the supply can provide.
The current rating of a power load (your IOIO board) is the amount of current it can demand, for proper operation.
Thus, if the IOIO board just requires 10 mA, and operates at a specified voltage range (5-15 Volts as mentioned), then the power supply can not ...
8
Will a capacitor bank and high current relay do it?
Ignoring the inductive load for the moment since that raises an entirely different question, simply assuming that you need to supply 20A @ 12V for 5 seconds, the energy required is:
$$12V \cdot 20A \cdot 5s = 1200J$$
Let's do a quick calculation to get a feel for the size of capacitor bank to deliver ...
8
It would probably make more sense if the label were adjusted sligthly:
Those -5V and -12V outputs really are negative voltages (not the ground returns for +5V and +12V) and they do sink current.
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You don't need a feedback resistor and neither do you need C1. I guess the "designer" has some strange perception that the circuit will oscillate without them but it won't.
Oscillation will occur if Q1 provides gain - it won't because it is a source follower.
Oscillation will occur if Q1 produces significant phase shift and this is more of a possibility ...
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Since this is quite within reach I will try to go through the whole story.
Let's take a generic network N and spot one generator (either current or voltage), for now we will not consider it as dependant.
I will name quantities with capital letters, i.e. V, I, meaning DC values, AC phasors, Laplace's variables but also -pushing a little- time domain ...
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I don't understand current sources! Batteries are voltage sources, but
how does one actually implement a current source?
Here's a current source circuit that uses an op-amp: -
This type of current source relies on Vset being applied to the non-inverting input of the op-amp. Due to the op-amp having a massive open-loop gain, the inverting input can be ...
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In your specific op-amp circuit, the voltage on junction of Rf and Rin is the same as the voltage on the non-inverting input. This has to be so - it's called a virtual earth. Given that fact, this means that your signal (Vin) sees an input impedance of exactly Rin. It also means that your output (without connecting to anything else) has to drive an output ...
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"Output current" should be read as "current into the output pin". As such, negative values indicate that the pin is sourcing current, and positive values mean that it is sinking current.
The floating issue can be remedied by using a device with bus hold such as the SN74LVCH245A. The bus hold feature holds the input pin at the previous detected logic level, ...
answered Mar 23 '14 at 2:25
Ignacio Vazquez-Abrams
47.3k2 gold badges69 silver badges97 bronze badges
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The opamp is acting as a unity-gain buffer, though it may not be obvious:
The rule for opamps is that the output does whatever it has to to keep the two inputs equal, provided that it doesn't clip of course (run into its own supply and stop there).
The transistor is used as an emitter-follower, in which the emitter voltage follows the base voltage minus a ...
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When we say an ideal current source has infinite resistance, we are not talking about series resistance, we are talking about shunting resistance. The model of a non-ideal current source looks like this:
simulate this circuit – Schematic created using CircuitLab
In this diagram I labelled the resistor as "G" instead of "R" to emphasize that this is ...
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