It is well known that the voltage divider (two resistors connected in series) is a unidirectional device where we apply the input voltage across the whole network and take the partial voltage across one of the resistors. Thus the output voltage is always less than the input voltage and the transfer ratio is less than unity.
My idea is to make just the opposite - to apply the input voltage across one of the resistors and to take the voltage drop across the whole network as an output. Thus the "output" voltage should be higher than the "input" voltage and the transfer ratio should be more than unity. Is it possible to do such a magic?
Try to answer this question on your own. If you experience a difficulty, you can look at the correspondence and the "circuit riddle" below:
http://en.wikibooks.org/wiki/Talk:Circuit_Idea/How_Do_We_Swap_Circuit_Inputs_and_Outputs%3F
(Below the pictures, I have written in red "normal divider"; above the right picture, I have written in black "reversed divider")
Dear, Colleague Cyril,
From the name the conventional voltage divider outputs a fraction of the input voltage.
It consists normally of two resistors. This divider can not be made reversible since if the input is applied across R2 and the output voltage is taken across the two resistors., the output voltage will be always = the input voltage and not = (R1+R2)/R2 as required. Therefore , the circuit in its conventional form is unidirectional.
To make the circuit reversible we have to use an amplifying device with feed back such that the gain is equal to the inverse of the feedback ratio.From our experience with the op amp amplifier configuration, this function can be implemented by the the noninverting op amp amplifier configuration. For this circuit the gain= (R1+R2) /R1 =1/( (R1+R2) /R1) the inverse of the feedback ratio.
Therefore, the noninverting op amp configuration is said a potinetiometric configuration.
Hi Cyril,
I think your question really boils down to energy conservation. If we think of the conventional voltage divider circuit, there are only passive elements which cannot add energy into the system. Thus, without any increase in current flowing through the network through some energy source, one cannot get a higher voltage than the input voltage source. The same goes for other passives such as inductors, capacitors, and even memristors.
However, the introduction of an op amp changes things since an op amp can be used as an energy source. Almost every circuit element that has "gain" (as defined by Abdelhalim) is a source of energy in the circuit. Any two-port network that increases the voltage without an energy source (even in voltage regulators) are only be able to supply lower current at its output than its input takes in. Otherwise, you will have solved the world's energy crisis with a new device that creates energy out of nothing!
No, we cannot!
But, reactive circuit elements could allow such transient phenomena in a proper switching circuit - this principle is used to making energy converters which are so spread noawadays.
The energy conservation is a basic physical law and devices constructed violating it can never work in the real world. The voltage gain in the reversible resistive divider introduced by Cyril is on the cost of the DC power supply for the op amp.
In the respect of the energy conservation we are all in agreement. You are right.
However concerning the voltage division of an ac voltage, one can use tapped inductors to step up and step down an input voltage. This type of magnetic induction devices as we all know is the auto transformer. The conventional electrical transformers are also reversible. They can step down the voltage in one direction and
step it up in the reverse direction.
As POP hinted in his comment, generally, one can even use switches + inductors or capacitors to step up the voltage. The step down realization is straight forward.
Thank you.
You can Amplify voltages using passive devices and with no power dissipation ...
Use the principle of "Resonance" ...
Apply your Voltage signal to a series RLC and set 1/sqrt(LC) to be equal to your signal frequency (w_input = wo) , the output will be:
Vout(t) = j Q Vinput
where Q is the RLC quality factor .
Amplify Currents by parallel RLC too, Iout = -jQ Iin
But you are not amplifying "Power" in both cases, you are just changing voltage into current and vice versa, something which is like a transformer...
If I understand you question correctly, you are asking can you take a battery and place a resistor across it. Then attach another resistor to one lead, the the voltage across the pair should now be twice the battery voltage? Kirchoff says NO. Vbattery plus zero equals Vbattery.
I like this last explanation based on a law in it's turn based on an energy conservation principle. But the explanation holds.
Oh, may I also add a third variant you forgot in this question Cyril ?
you presented,
a) passive voltage divider, where k 1 (you presented it)
c) why not, an ASTIVE one like (a), with the OPAMP,
What good this would do ? Well, it is IMPEDANCE MATCHING ... In other words, it will create a voltage divider which is immune to load ...
Yes, an op-amp inverting amplifier with R2/R1 < 1 is an "inverting voltage divider"... It is interesting that besides the invertion its transfer ratio is less with 1 than the non-inverting amplifier based on the classic voltage divider... Why?
https://www.researchgate.net/post/Why_the_gain_of_the_non-inverting_amplifier_is_1_R2_R1_while_the_gain_of_the_inverting_amplifier_is_only_R2_R1_Is_this_a_desired_difference
Somehow this "inverting voltage divider" seems to be an "ideal voltage divider"... Why?
Maybe we have to ask a separate question, "What is the relation between the classic voltage divider and the op-amp inverting amplifier with K < 1"? In other words, let's try to see the humble passive voltage divider within the more sophisticated inverting amplifier... or to show the metamorphosis of the passive voltage divider into an inverting active one... So I have not forgotten the third (active inverting) version of the voltage divider since it is not exactly a "reversed voltage divider"; it is a "reversed, inverting and with different transfer ratio active voltage divider":)
Hi Vasile! Thanks for the response; I always enjoy your perfect circuit analysis. Sorry that I hid the secret of the circuit through Bulgarian words:) Here is a clarification - below the pictures, I have written in red "normal divider"; above the right picture, I have written in black "reversed divider".
I want to assure you that my questions are not accidental but are derived from my long experience of a "circuit thinker":) While circuit designers invent and make circuits, I understand and explain them (sometimes to themselves but this was not the case with Gordon Deboo:)
I have posed this question as an example of the amazing negative feedback property to swap unidirectional devices (a voltage divider here) thus converting them into electronic negative feedback circuits (a non-inverting amplifier here). I have generalized it in the question below:
https://www.researchgate.net/post/Can_we_reverse_swap_the_input_and_output_of_any_unidirectional_device_by_applying_a_negative_feedback_If_yes_what_will_be_the_use_of_this_trick
Your explanation of the situation when we directly "attack" the op-amp output (not so crudely that to "kill" the input source or the op-amp:) is another sophisticated idea on which famous negative feedback circuits are based:
https://www.researchgate.net/post/Can_we_apply_the_input_voltage_to_the_output_of_a_voltage_follower_instead_to_its_input_and_to_take_the_current_as_an_output_Is_it_correct (common-base stage)
https://www.researchgate.net/post/Can_we_apply_an_input_current_to_the_collector_of_a_BJT_whose_base_is_held_at_a_constant_voltage_and_to_take_the_collector_voltage_as_an_output (dynamic load)
But if you want to reverse "more civilly" the op-amp:), you should insert it in the feedback loop of the main op-amp... in the place of the voltage divider... (interesting, what would be the overall transfer ratio?)...
I will be grateful if you take part in these discussions... Thank you in advance, Cyril
Vasile, you might have introduced a new term to Electronics : stubbornness :)
No, this doesn't sound technical. Let's call it stubbornitivity :) :) :)
Your explanations sound just like mine. I love it :)
One remark, is stubbornness (i.e., drive capability) is limited in SOME OPAMPs as Short-Circuit-Current .
If you look at the LM324 (this is the all-time-most-popular OPAMP on the planet, second only to 741 :) in the datasheet, it says "Short Circuit Current to Ground" which is "continuous." So, this OPAMP is not stubborn enough to self-destruct.
This might very well be one of the reasons why it became so popular. In summary, electronics engineers do not like stubbornness to cause dead circuit parts, since there is always an OOPS. :)
Hi Cyril - as you will remember, I was "out of service" for a long period.
Now I am back - and I am somewhat confused about the status of your question.
Is it answered already - yes or no? And, more than that, what is the CORE of the question. It is not clear to me (may be because of language problems).
Thank you. Lutz
Dear Lutz,
By this question I want to show one of the most remarkable features of circuits with negative feedback - to reverse (swap the input and output of) the negative feedback network. In this case, we have connected a voltage divider (atenuator) with ratio of R2/(R1 + R2) in the feedback loop. As a result, the whole circuit with negative feedback begins amplifying with a gain of (R1 + R2)/R2... But why? It would be useful to explain it in a human friendly manner... not formally... in the manner used by me in the early 90's...
I think, this shouldn´t be a great problem. If we accept that negative feedback keeps the operational amplifier in an equilibrium leading to a negligible input differential voltage, the output voltage must be greater than the input voltage by a factor K. And this factor K must be identical to the inverse of the feedback factor. Only then the product of both is unity and we have a voltage at the inv. terminal which equals the input voltage (non-inv. terminal). However, I agree that this "explanation" is nothing else than a description of the formula`s content.
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