0
\$\begingroup\$

Its said that if you get more electric current, the electronic device in a circuit will just take what it needs and its not dangerous, so if we raise the voltage and therefore the electric current increases but the device will take just the electric current it needs, where is the parameter I am missing to undertand why increasing the voltage is dangerous for the device?

Thanks

\$\endgroup\$
  • 1
    \$\begingroup\$ "if you get more electric current" is meaningless. Current can only flow. You are probably confusing a situation where a supply is rated for more current than a load will draw at a given voltage - hence you don't "get" more, rather you have a supply which under other circumstances could supply more. \$\endgroup\$ – Chris Stratton Jan 2 at 22:24
  • \$\begingroup\$ if we raise the voltage. All components have a maximum voltage that can be applied across any two pins. \$\endgroup\$ – CharlieHanson Jan 2 at 22:24
  • \$\begingroup\$ Well... I would I say either... This is a vague question though. For voltage, if you supply too much voltage, you could really fry something. For current, if you don't calculate Ohm's Law properly, you may have too much current draw and could fry something. Which one applies to you? (I don't think devices "need" current. They need voltage, however and depending on the material and wiring of that circuit will you get your current draw.) \$\endgroup\$ – KingDuken Jan 2 at 22:24
  • \$\begingroup\$ If you drink a load of water you will need to urinate. If you keep drinking more you will urinate more. If you drink too much you could drown \$\endgroup\$ – JonRB Jan 2 at 22:27
  • \$\begingroup\$ If I replace the battery of a circuit from 5V to 12V,and inside the circuit there is a device which works at 5V, the higher voltage is going to burn the device. But flow more current? I dont think because the device just take the current it needs, so, why it burns? \$\endgroup\$ – user3254515 Jan 2 at 23:04
6
\$\begingroup\$

A device draws the correct current only at the correct voltage (usually specified as a range). If you change the voltage (go outside the specified range), then the current that it draws might change too, to an unsafe value.

Here are a few examples:

  • With resistors, current rises linearly with voltage. Power dissipation rises with the square of the voltage. Usually not too much of a problem, at least until you reach the dissipation limits.
  • With capacitors — especially electrolytics — there is a voltage above which the current rises dramatically when the internal insulation breaks down.
  • With transformers, raising the voltage raises the current which increases the magnetic field within the core. When the core reaches saturation, the current rises dramatically, leading to excessive dissipation in the windings.
\$\endgroup\$
  • \$\begingroup\$ I think you covered all the bases. You answered the OP's question about both voltage and current.+1 \$\endgroup\$ – Sparky256 Jan 2 at 22:33
1
\$\begingroup\$

Electronic devices don't "take only as much current as they need" by magic: they're designed to work that way.

The simple answer is that by attempting to power a device with a higher voltage, you are violating the design parameters that make it work. As with any device, operating outside the design parameters often leads to failure.

\$\endgroup\$
  • \$\begingroup\$ ok, but why? what´s happening underhood when we increase the voltage? The more current is flowing and the device is just getting what it needs or no more current is flowing through the circuit? \$\endgroup\$ – user3254515 Jan 3 at 11:35
  • \$\begingroup\$ @user3254515 There's no simple answer: it's about as broad as "what happens when too much force is applied to things?" The simple answer is "they tend to break", the more complicated answer is the whole of mechanical engineering. If you had asked about a specific circuit I could give specific failure modes, otherwise there's no concise answer possible. \$\endgroup\$ – Phil Frost Jan 3 at 13:49
  • \$\begingroup\$ But what devices, wires, and other electronic components see is just electrons moving through them? if its like this, the electrons is what would cause the failure in the circuit because of the overheating \$\endgroup\$ – user3254515 Jan 3 at 14:06
  • \$\begingroup\$ @user3254515 I don't think you can say "electrons cause circuit failure" any more than you can say "steel causes car crashes". Yes, electrons are involved -- only because they are present in all the matter comprising the circuit. \$\endgroup\$ – Phil Frost Jan 3 at 14:14
  • \$\begingroup\$ Ok, but according to Ohm´s law, if you increase the voltage in a certain circuit(changing battery from 5V to 12V), would it increase the current through the circuit? I think that if I change the battery for other one with more amperes per hour, its safe but if I change it with more voltage is dangerous. Any reference(website...) to know whats happening underhood? \$\endgroup\$ – user3254515 Jan 3 at 14:21
1
\$\begingroup\$

There are many failure modes in semiconductors. These are prevented by staying well within the Absolute Maximum Specs.

The answers are shown by the units below. {V,mA,'C}

enter image description here

\$\endgroup\$
0
\$\begingroup\$

If a device fails because of heat, that's power, which is a combination of voltage and current. If it fails by punching through a gate, that's voltage.

\$\endgroup\$
0
\$\begingroup\$

Its said that if you get more electric current, the electronic device in a circuit will just take what it needs and its not dangerous,

It depends on what the circuit is. most consumer appliances are voltage fed and there that statement is correct.

so if we raise the voltage and therefore the electric current increases but the device will take just the electric current it needs, where is the parameter I am missing to undertand why increasing the voltage is dangerous for the device?

The device may be designed to expect a certain voltage and if the wrong voltage is applied the wrong current will be taken.

Above a certain level the device cannot withstand the voltage, resulting in permanent damage.

Above a different level the device may overheat causing thermal shutdown and/or permanent damage.

\$\endgroup\$
  • \$\begingroup\$ But when we increase the voltage, acording to Ohm´s law, the current will increase, so I assume the current will increase through the circuit but the device will just take what it needs or no more current is flowing through the circuit when we increase the voltage? \$\endgroup\$ – user3254515 Jan 3 at 11:37
0
\$\begingroup\$

Both are dangerous. For example, consider the rate of change of both voltage and current. That is, \$ \frac{\text{dv}}{\text{dt}}\$ and \$ \frac{\text{di}}{\text{dt}}\$ resepectively. Now, if \$ \frac{\text{dv}}{\text{dt}}\$ is high i.e. which is called a voltage overshoot or voltage spike, the electronic devices may run into a problem of incompatibility with the voltage and may fail. This can be prevented by using snubbers and a simple snubber will have a series connected \$RC\$ in parallel with the device to be protected. The capacitor's discharge current in the snubber circuit will be limited by the resistor \$R\$. If \$ \frac{\text{di}}{\text{dt}}\$ is high, it means that the increase in current is higher than spreading velocity of the current hence leading to the concentration of current (i.e. high current density) at a particular point called the hotspot and it heats up rapidly and the internal components will get damage due to the high heat produced as a consequence of the Joule's heating effect. This high \$ \frac{\text{di}}{\text{dt}}\$ can be prevented by using a diode in parallel with a series \$RC\$ in parallel with the load and finally an inductor in series with the supply. An example is shown below for an SCR based half wave rectifier.

enter image description here Various other implementations of \$ \frac{\text{dv}}{\text{dt}}\$ protection is shown below. enter image description here

Capacitor blocks sudden voltage change and can be proved mathematically below as: \begin{equation} C = \frac{dq}{dV} = \frac{idt}{dV} \end{equation} Thus, for a constant current, \begin{equation} C \propto \Bigl(\frac{dV}{dt}\Bigr)^{-1} \end{equation}

Similarly for an inductor, \begin{equation} V = \frac{d}{dt} (LI) \end{equation} Assuming inductance as constant, \begin{equation} V = L \frac{di}{dt} \end{equation} Thus, for a constant voltage, \begin{equation} L \propto \Bigl(\frac{di}{dt}\Bigr)^{-1} \end{equation} Thus, inductor opposes sudden change in current.

Image credit for all the above images: Rashid, M. H. (2018). Power Electronics(4th ed.). Noida, UP: Pearson India Education Services Pvt. CIN: U72200TN2005PTC057129 ISBN : 978-93-325-8458-7

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.