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Maximum Voltage in passive capacitive circuit

fsonnichsen

Oct 23, 2013
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This question is academic insofar as I am assuming a circuit with no inductance.
In a circuit having only passive capacitive elements, when driven with a square wave--is it possible to have a voltage greater than the drive voltage at any node? I am assuming linear components--e.g. no diodes or active elements--just a capacitor and resistor network.
From an energy perspective I think not. Overshoot from "ringing" for example, always (in my experience) is associated with inductances and a 2nd order circuit.

Fritz
 

duke37

Jan 9, 2011
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I have seen an oscillator circuit many years ago which had a string of capacitors, base to earth. The junctions between the capacitors were driven with resistors from the emitter. A very low increase in voltage was obtained, sufficient to promote oscillation. The frequency was low and was used for vibrato modulation in an electronic organ.
 

AnalogKid

Jun 10, 2015
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I have seen an oscillator circuit many years ago which had a string of capacitors, base to earth. The junctions between the capacitors were driven with resistors from the emitter.
Sounds like a phase-shift oscillator.

ak
 

AnalogKid

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In a circuit having only passive capacitive elements, when driven with a square wave--is it possible to have a voltage greater than the drive voltage at any node?
If I understand the question correctly, then yes, it is.

5 V square wave, 50/50 duty cycle, 0 ohm source impedance, 1 kHz. positive pulse width = 500 us
0.1 uF capacitor and 1.667K resistor in series; time constant = 166 us
Cap connected to square wave source, resistor tied to +5 V.

The pulse width is three time constants, so when the square wave is low for 500 us, the cap is charged to 95% of Vcc. When the squarewave goes high, the R-C node jumps up to 9.75 V and then starts decaying down to 5 V. As the C gets larger, the R-C node looks more and more like the input squarewave, now centered about 5.0 V (2.5 V to 7.5 V) by simple AC coupling.

ak
 

Ratch

Mar 10, 2013
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This question is academic insofar as I am assuming a circuit with no inductance.
In a circuit having only passive capacitive elements, when driven with a square wave--is it possible to have a voltage greater than the drive voltage at any node? I am assuming linear components--e.g. no diodes or active elements--just a capacitor and resistor network.
From an energy perspective I think not. Overshoot from "ringing" for example, always (in my experience) is associated with inductances and a 2nd order circuit.

Fritz

The following is a passive circuit that contains only capacitance and resistance. No overshoot and no ringing. The output is at zero phase with respect to the input at a particular frequency, and the voltage gain is enough to insure enough feedback to make an oscillator. The transfer function is

fsonnichsen2.JPG
and the frequency at which the output phase with respect to the input phase is zero is
fsonnichsen3.JPG
The magnitude increase at the zero phase frequency is 30/29 = 1.03%.

The following shows the circuit with C = 1uF, R = 1k, and AC voltage at 390 Hz. Comparing the output probe with the input probe displays shows a very slight increase in peak to peak amplitude.

Ratch

fsonnichsen.JPG
 

AnalogKid

Jun 10, 2015
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Not quite.
Not sure how you can say that. I was winging a guess based on an incomplete, two-sentence description with no schematic. Based on only that, a string of resistors and capacitors and one transistor still sounds like a phase shift oscillator to me.

ak
1470647057.PNG
 

WHONOES

May 20, 2017
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Being a little pedantic, a circuit without inductance does not exist. That is also true of capacitance.
 

Ratch

Mar 10, 2013
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Not sure how you can say that. I was winging a guess based on an incomplete, two-sentence description with no schematic. Based on only that, a string of resistors and capacitors and one transistor still sounds like a phase shift oscillator to me.

ak
1470647057.PNG
The circuit above is really a phase shift oscillator. The 3 caps and resistors shift the phase 180°. The loss of voltage across the 3 RC network is compensated by the voltage gain of the common emitter transistor.

The circuit presented by me has no phase shift and actually amplifies the input voltage slightly. That means it can be used in a common collector transistor configuration which, as you know, never exceeds unity voltage gain and has no phase shift. Therefore, I consider it wrong to call an oscillator incorporating my network a "phase shifter" because it does not shift phase.

Ratch
 

Ratch

Mar 10, 2013
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Being a little pedantic, a circuit without inductance does not exist. That is also true of capacitance.

The difference is that the stray inductance present is mostly compensated for when necessary and rarely used as a primary component in the circuit design.

Ratch
 

Ratch

Mar 10, 2013
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Ah, but in #3 I was not commenting on your circuit, which had not yet been posted. That is why I quoted the OP.

ak
OK, I see your point. I should not have commented on your observation. I think we both agree my circuit is not a phase shifter.

Ratch
 

Audioguru

Sep 24, 2016
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A voltage multiplier circuit is passive and boosts the voltage. The diodes are passive.
 

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Ratch

Mar 10, 2013
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Discussion about this topic:

Fialkow, A.D. & Gerst, Irving. (1953). The Maximum Gain of an RC Network. Proceedings of the IRE. 41. 392 - 395. 10.1109/JRPROC.1953.274389.

Also: https://ieeexplore.ieee.org/document/4050644

Also: https://electronics.stackexchange.c...-network-with-voltage-gain-greater-than-unity

The IEEE references are useless to anyone who does not live close to a library that racks that 1951 journal, or is not a member of IEEE.

The circuit in stack exchange does not have a phase shift of zero degrees when its gain is greater than 1. That makes it less useful. Tell me if I am wrong.

Ratch
 

Ratch

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AnalogKid

Jun 10, 2015
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The circuit in stack exchange does not have a phase shift of zero degrees when its gain is greater than 1. That makes it less useful. Tell me if I am wrong.
I'll tell you I don't understand why that matters. There is nothing in post #1 about phase shift being part of the question or answer. also, *any* circuit with nothing but Rs and Cs will have phase shift. That's kind of a thing.
The StackEx circuit and the single-stage version I described in #4 are lossy charge pumps. Seems simple enough.

Note, I think the OP has abandoned this thread.

ak
 

WHONOES

May 20, 2017
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The difference is that the stray inductance present is mostly compensated for when necessary and rarely used as a primary component in the circuit design.

Ratch
Agreed. But stray capacitance can make itself felt and is not so easy to ignore.
 
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