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Please give an intuitive explanation of difference between how a signal or data is carried using voltage and current?

I'm wondering even though current and voltage inevitability co-exist, why do we use the term for one concept(current or voltage)? If there is a signal out there somewhere, it is both current and voltage at the same time. But we name only one of its property.

We see in some circuits like heart rate signals or some other small signals which have to be in form of voltage in order to transfer the whole signal it to an amplifier (which has high input impedance). Even if it is voltage signal, we still have current passed to the amplifier.Why it is mentioned that voltage is transferred to input of the amplifier and why not current (when voltage causes current to pass)?

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  • \$\begingroup\$ please do not repost a question and then delete the old one ... improve the old question instead \$\endgroup\$ – jsotola Jun 20 at 21:46
  • \$\begingroup\$ it most likely has to do with the usage of an oscilloscope to view the signal ... voltage is read very easily, current no so much \$\endgroup\$ – jsotola Jun 20 at 21:48
  • \$\begingroup\$ @jsotola I'd say the exact opposite – a current is easily sensed through a transistor, a voltage is brittle and must be met with high input impedance, which opens up the whole can of noise and stability worms. \$\endgroup\$ – Marcus Müller Jun 20 at 21:59
  • \$\begingroup\$ High speed current signals are more immune to parasitic capacitance. No change in voltage across a low impedance input means no signal lost to parasitics. That’s why current feedback op amps are so fast! \$\endgroup\$ – DavidG25 Jun 20 at 22:13
  • \$\begingroup\$ @DavidG25 Does inductance at that scale just work out to be less significant than capacitance? \$\endgroup\$ – DKNguyen Jun 20 at 22:16
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Usually it has to do with what the author thinks is the best way to sense the signal that's on that pair of wires. If they say "voltage signal" then they (probably) feel that it should be applied to a high-impedance amplifier that responds to voltage; if they say "current signal" then they feel that it should be applied to a low-impedance amplifier that responds to current.

An example of a current signal is a 4-20mA current signaling loop. A device that uses this expects to be fed a wide range and possibly varying voltage, and it will impose a current proportional to the quantity it's measuring on the power supply.

An example of a voltage signal would be the output of a typical op-amp. The op-amp imposes a voltage which remains largely unchanged regardless of variations on the current, at least up to the point where the amplifier can no longer coerce the output to be correct.

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  • \$\begingroup\$ Don't you mean 4-20mA? \$\endgroup\$ – Kevin White Jun 20 at 22:03
  • \$\begingroup\$ @KevinWhite Thank you. Edited. \$\endgroup\$ – TimWescott Jun 20 at 22:15
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There can be benefits to both in different situations. And in reality, as you said, we never have a pure 'voltage' or 'current' - you have both. The key is the impedance: A high output-impedance source will have it's output current change very little if the load changes, but the output voltage will change a lot - you want your load (say an amplifier) to measure the current. A low output-impedance source will have a very small change in output voltage when the load changes, but the output current will change a lot - you want your load to measure the voltage.

Take for example the inside of an integrated circuit. If we have a voltage signal, what we actually mean is 'the voltage between two points' - usually this is between the signal wire and the ground. However, the ground is not necessarily stable - other currents are also flowing through this ground and might cause voltage drops, which will lead to errors.

A current signal will not suffer from this moving ground - a current of say 1 mA is still a current of 1 mA if you have a bit of difference in ground. (this is incidentally the same reason why industrial sensors like using current mode sensors - the long wires can cause a significant voltage drop that leads to errors if you were to use voltage-mode signalling).

There is also the matter of impedance. Some sensors have a high output impedance, such as a photo-diode. Because of this, these sensors are usually used as a current-source, and not a voltage source.

On the other hand, at low speeds, a MOSFET appears like a high-impedance input voltage-to-current (the 'correct' name is transconductance) amplifier. You want to drive it with a voltage, as it will turn this voltage into a current at its output. So when driving a MOSFET, you want a voltage signal.

Things change when you go to systems with matched characteristic impedances at higher frequencies. At that point, every device has the same impedance, and you don't really think about voltage and current anymore, and instead consider power.

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We use the different terms because in the case of a voltage signal, the observed quantity is voltage, and in the case of current signals, current.

That's really all there is to it.

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  • \$\begingroup\$ Really that’s all, what about CML? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 20 at 21:48
  • \$\begingroup\$ That's a signalling standard. If I had a guess, the receivers of that observe a voltage signal and use external 50Ω termination. To be honest, this is a bit like asking whether a TV cable transports a current or voltage signal; the answer would be "yes", and also that Maxwell links these two, but the receiver observes either the current, or the voltage signal aspect. \$\endgroup\$ – Marcus Müller Jun 20 at 21:52
  • \$\begingroup\$ I think in terms of implementation. CML uses emitter inputs to maximize BW at the expense of no current gain in the 1st stage and complementary current sources for drivers. Unlike CMOS and TTL however uses voltage gain for receiving and saturated switches for driving, whereas video analog uses linear mode 75 Ohms source and load impedance. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 20 at 21:55
  • \$\begingroup\$ well, doesn't that answer the question about which quantity is relevant for the input stage? \$\endgroup\$ – Marcus Müller Jun 20 at 21:57
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Data transfer especially with controlled impedances is often done with either voltage or current sources with source and load impedances matching the cable transmission line. Voltage source amplifiers are true for single-ended or differential drivers like RS-485.

In Current Controlled Logic ( ECL,PECL,CML) gain in speed by using Emitter Couple speed advantages and also uses differential line currents to reduce EMI and skew.

So both Voltage mode and Current mode drivers are used depending on the link speed (or rise time in ps,ns or us )and distance. The overlapping bit rates commonly used may be in the 100MHz to 1GHz range but are not hard limits but for >>1GHz , CML is preferred and is standard up to 10Gbps with higher levels possible.

1Gbps Ethernet links use differential voltage drivers with controlled impedance lines and the current into the matched load is used by voltage amplifying receivers. But it also uses a bandwidth compression form of modulation and compensation to make the cable less sensitive for signal integrity/ quality. They also with BALUNs and transformers to magnetically improve the Common Mode noise rejection needed from line noise and ground noise differences.

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If a signal is 5V and 1uA (or maybe even 1nA), would you rather be inspecting the large 5V? Or the miniscule 1uA or 1nA? You probably want to look at the voltage since it is larger and easier to read, so we call it a voltage signal since that's the component you are treating as carrying the information.

But if your signal is 10mA being pushed down wire by the transmitter applying 2.5V, which would you rather inspect for data? They both seem to be magnitudes that are reasonably easy to inspect. But wait...that 10mA is a pretty high current that would have caused a voltage drop along the wires so if you inspect the voltage it won't be accurate to what is on the other end of the wire. The 10mA current, on the other hand, remains unchanged down the line and all the extra power was spent to push it down the wire so it would be a waste if you didn't take advantage of the current somehow (you could easily push 2.5V down the wire into a high impedance input so the current is much lower and therefore consumes much less power with much less voltage drop for a more accurate voltage). Why wouldn't you inspect the current for the data? So we call it a current signal.

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  • \$\begingroup\$ Thank you so much for reaching out.What causes a voltage drop across the wire when you have 10mA current. \$\endgroup\$ – rdeep Jun 21 at 0:02
  • \$\begingroup\$ Wires have resistance. More current means more voltage drop so you dont want to send more than is necessary if all you want is to transmit a voltage which is why you want high impedance inputs for voltage. \$\endgroup\$ – DKNguyen Jun 21 at 0:23
  • \$\begingroup\$ Could you also please explain how (high current ,low voltage) and (low current,high voltage) effects power consumption in circuits? \$\endgroup\$ – rdeep Jun 21 at 2:09
  • \$\begingroup\$ I'm not sure what exactly what it is you want to know there. It's a really broad and vague question. \$\endgroup\$ – DKNguyen Jun 21 at 3:01
  • \$\begingroup\$ As you said, there will be voltage drop along the wires when we pass 10mA current. According to Ohm's law, current directly proportional to voltage.Because of voltage drop here, current should also reduce right? How can we measure 10mA current down the line? \$\endgroup\$ – rdeep Jun 21 at 17:29

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