Can the output voltage of a circuit be larger than its input

In summary: L = (2*pi*L/V)^2If the voltage is greater than the supply voltage (1), then Vr will be greater than V and current will flow through the inductor.
  • #1
Abunada
3
0
Hi guys, this is my first post in this forum so I hope I get some help.

My question is: can an output voltage of ANY circuit be larger than its input/source voltage. I realize that in DC circuits, this is not possible because of KVL. Since the total voltage around a loop has to be zero and the fact that voltage across resistors is always positive.

However, can this (Output swinging higher than the input) happen in AC circuits. This really pisses me of, because sometimes out college professors speak about "voltage drops" in AC circuits and yet they say that the output voltage CAN go larger than its input. For example in power systems, they say that the voltage at the load can be larger than that at the generator side. Can somebody please help me understand this point.
 
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  • #2
Yes, even in passive circuits the output voltage can be greater than the input voltage. There are many examples, resonant circuits and transformers are two key example.
 
  • #3
Can you explain just a little bit why this is so, an example would be great. Since there are many arguments we made in electronics that the output cannot swing above the source voltage and thus it will clip, which happens in the MOSFET amplifier for example.

If this is true why do we need an transistor amplifier at all. Can't we build passive circuits that amplify the signals for us.

Thanks
 
  • #4
For passive circuits, it's only possible using resonance and alternating voltage. Example: http://en.wikipedia.org/wiki/Transformer" [Broken]
For active circuits, it's entirely possible to have a DC-DC converter which steps up the voltage. Example: http://en.wikipedia.org/wiki/Boost_converter" [Broken]

Another example that I can speak to is that in Medtronic implantable ICDs, the internal battery voltage is much less than that required to stimulate a fibrillated heart. To gain enough voltage to shock the heart, a bank of capacitors are first charged in parallel, then rapidly switched to a series connection, building up enough voltage momentarily to defibrillate the heart tissue.

Abunada said:
Since there are many arguments we made in electronics that the output cannot swing above the source voltage and thus it will clip, which happens in the MOSFET amplifier for example. If this is true why do we need an transistor amplifier at all. Can't we build passive circuits that amplify the signals for us.

The reason it's so common to hear this is because for most circuits, it is true. DC voltage step-up tends not to happen on its own when not intended, as least not on a steady state scale. Note how every voltage-increasing circuit I've mentioned contains energy storage devices, either capacitors or inductors. This is the kernel of their operation; that's how the "magic" happens.

There are several reasons we don't use such a circuit in place of an amplifier, for starters: they are imprecise, they are slow, and have poor frequency response. These are all very undesirable characteristics for passing signals!

Fun fact: If you measure a circuit such as an op-amp circuit or logic gate, when the output switches it has a short "spike", and that voltage can exceed the supply voltage. This is actually due to the parasitic capacitances inside the transistors and other circuit elements!
 
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  • #5
KingNothing said:
Fun fact: If you measure a circuit such as an op-amp circuit or logic gate, when the output switches it has a short "spike", and that voltage can exceed the supply voltage. This is actually due to the parasitic capacitances inside the transistors and other circuit elements!
Back in the 1960's I had one of those Heathkit experimenter/educational circuit boards. It had a number of parts on the board and spring connectors to let you wire different parts in different ways. It was powered by 4 C-cells.

The instruction manual had a number of different circuit projects to try. But I ended up just trying my own stuff, too. I found two way to make the mechanical SPDT relay oscillate. One way was for the energize-to-close contact short out the coil circuit. The other way was for the energize-to-open contact interrupt the coil circuit. The latter one had a higher oscillating frequency. One day I touched the relay to try to mechanically dampen the vibration. But I got a really strong electrical jolt. Given that the contact was arcing, it was obvious to me later when I learned about inductors, that the inductance of the coil was creating a rather high voltage as it was being interrupted. Unfortunately, I never had an opportunity to measure the voltage.
 
  • #6
An example is here
ferranti.jpg


In the above diagram take Vs = 1<0 (complex) and solve for Vr. You should understand why this happens mathematically.
 
  • #7
The voltage V through an inductor with inductance L is given by the following equation, which depends upon the rate of change of current through the inductor (di/dt):
[tex]V=L\int{\frac{di}{dt}}[/tex]

By forcing current through an inductor, you can get higher (or even negative) voltage. This is the basis of many switching regulators--the transient response of an inductor.

If you think of a charged capacitor as a battery, you can see how you can easily create a voltage inverter / doubler / tripler / etc. just by moving the connections around once you've charged up the capacitors. This concept is the basis of many simple and low-current voltage references.
 
  • #8
Okay so let's get this straight, if the circuit is DC powered then the voltage will only rise above the input during the transient response phase which will eventually die out. So in the steady state case, it will not exceed the output voltage.

However, in the AC case the output voltage (At least in magnitude) can easily exceed the input voltage as thecritic showed. But their are a few points I still did not quite get:

- For the MOSFET case I don't quite understand what KingNothing means when he says: "DC voltage step-up tends not to happen on its own when not intended".

- For the transformer case it is not considered an AC circuit per say is it right, since it is coupled together using a magnetic field.

- For MATLABdude, I did not understand just how can we force a current through an inductor. Since the voltage is proportional to di/dt we should force the current to change right. I know that switching regulators switch the voltage both on and off, which is like introducing AC into the circuit, since a pulse train is like a series of AC voltages of different frequencies right? So this is what is actually causing the voltage rise right?
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To be honest I always think of capacitors as charged batteries, however, I still can't see how inverters/doublers/tripler etc can be constructed. Can you explain just a little bit more.

Thanks for the help guys and sorry if my questions are many and idiotic. But this has to happen when you have a bunch of worthless no lifer professors that don't understand a crap in EE.
 
  • #9
Abunada said:
Okay so let's get this straight, if the circuit is DC powered then the voltage will only rise above the input during the transient response phase which will eventually die out. So in the steady state case, it will not exceed the output voltage.

For MATLABdude, I did not understand just how can we force a current through an inductor. Since the voltage is proportional to di/dt we should force the current to change right. I know that switching regulators switch the voltage both on and off, which is like introducing AC into the circuit, since a pulse train is like a series of AC voltages of different frequencies right? So this is what is actually causing the voltage rise right?

The transient response is harvested to provide a continuous voltage (be it higher, lower, or inverted--boost, buck, inverting switchers). This is the reason most switchers use square waves of higher frequencies (usually in the tens or hundreds of kHz)--the on-off of the pulse train provides lots of transient response. You can find a quick intro here at:
http://en.wikipedia.org/wiki/Switched-mode_power_supply

The square wave should have only one frequency--however, the Fourier series should show lots of frequency components. This really isn't applicable to voltage conversion (though you can use this fact to double frequencies).

Abunada said:
To be honest I always think of capacitors as charged batteries, however, I still can't see how inverters/doublers/tripler etc can be constructed. Can you explain just a little bit more.

Consider two capacitors: if you have a DC supply giving out 5V, you can charge them in parallel so that both have 5V. If you then disconnect the DC supply, and put the capacitors in series (with the proper orientation), and then measured the total voltage across the two capacitors, you'd have 10V--voltage doubled. Change your multimeter probes around (i.e. switch the capacitors around), and you'd have -10V--voltage inverted.

Abunada said:
Thanks for the help guys and sorry if my questions are many and idiotic. But this has to happen when you have a bunch of worthless no lifer professors that don't understand a crap in EE.

That's pretty harsh. No need to denigrate yourself or your professors. Besides which, what year (or program, for that matter) are you in--if you're in first or second year, you'll probably only have seen bits and pieces of the above. However, with no "find current through this resistor array", the techniques, confidence, and just plain practice required to understand / analyze more complicated circuits just isn't there.

Most of this post above comes from playing around with and trying to design electronics rather than in-class material. Emphasis on the trying, with lots of failing, failing some more (with, or without burning, or release of the magic smoke genies from circuits), giving in and finally properly reading the datasheet, and (sometimes) eventually succeeding. However, all of the understanding and insight was laid down by those classes.
 
  • #10
Abunada said:
Okay so let's get this straight, if the circuit is DC powered then the voltage will only rise above the input during the transient response phase which will eventually die out. So in the steady state case, it will not exceed the output voltage.

In passive circuits, yes, and only due to energy storage elements.

Abunada said:
- For the MOSFET case I don't quite understand what KingNothing means when he says: "DC voltage step-up tends not to happen on its own when not intended".

In a practical environment, if you make a circuit that does some unrelated task, its extremely unlikely to step-up DC voltages on its own. As a designer it's not something that you have to worry about happening in a DC circuit.
 

1. Can the output voltage be larger than the input voltage in a circuit?

Yes, it is possible for the output voltage of a circuit to be larger than the input voltage. This can occur when the circuit includes a voltage amplifier or a transformer, which can increase the voltage level.

2. Is it safe for the output voltage to be higher than the input voltage?

In most cases, it is not safe for the output voltage to be higher than the input voltage. This can cause damage to the circuit components or even pose a safety hazard. It is important to design circuits with proper voltage regulation and protection measures in place.

3. What factors can cause the output voltage to exceed the input voltage in a circuit?

As mentioned before, the use of a voltage amplifier or transformer can cause the output voltage to be larger than the input voltage. Additionally, fluctuations in the power supply or faulty circuit design can also result in a higher output voltage.

4. Can a circuit be designed to always have an output voltage that is larger than the input voltage?

Yes, a circuit can be designed to have a constant output voltage that is larger than the input voltage. This is commonly seen in voltage boost converters, which are used to step up the voltage level for various applications.

5. What are the potential risks of having a higher output voltage than input voltage in a circuit?

The main risk of having a higher output voltage than input voltage in a circuit is the potential damage to the circuit components or external devices connected to the circuit. This can also pose a safety hazard if the output voltage is high enough to cause electric shock or fire.

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