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capacitor
- Thread starter domnic
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Isn't C sized to achieve a desired power factor based on winding R?
for starting you may want a pf of 0.7, a larger phase angle for positive start
running of 0.9 lower losses but a ph angle for smoother less pulsed operation
assume R = 0.01
cos(arctan X/R) = pf
or X = R tan(arccos (pf))
X = 0.01 tan(arccos 0.7) = 0.01 Ohm
X = 2 Pi f C or C = X/(2 Pi f) = 0.001/(2 Pi 60) = 270 mfd
similar for pf = 0.9
X = 0.0048
C = 128 mfd
for starting you may want a pf of 0.7, a larger phase angle for positive start
running of 0.9 lower losses but a ph angle for smoother less pulsed operation
assume R = 0.01
cos(arctan X/R) = pf
or X = R tan(arccos (pf))
X = 0.01 tan(arccos 0.7) = 0.01 Ohm
X = 2 Pi f C or C = X/(2 Pi f) = 0.001/(2 Pi 60) = 270 mfd
similar for pf = 0.9
X = 0.0048
C = 128 mfd
David Goodman
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- Pahrump, NV, USA
I'd say the results of your calculations look good. Hard start kits with relays for hermetic compressors are likely in the mfd range you describe. The manufacturers generally don't disclosed the rating. The hard starts are in the circuit for such a short amount of time that my test equipment won't pick up accurate voltage or current readings on these devices.
thanks
it was for illustrative purposes only
but generally run = 1/2 x start, give or take
still thinking about the voltage ratings lol
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As I was taught years ago:
In what we call a 240VAC system, that is 240V RMS, the peak voltage seen by the capacitor is going to be 338V. Then if the AC RMS goes 5% high, as is allowed for utilities, the peak voltage might be be as high as 355V. Add to that the stresses golddigger mentioned, 370V is going to be minimal.
In what we call a 240VAC system, that is 240V RMS, the peak voltage seen by the capacitor is going to be 338V. Then if the AC RMS goes 5% high, as is allowed for utilities, the peak voltage might be be as high as 355V. Add to that the stresses golddigger mentioned, 370V is going to be minimal.
kwired
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- NE Nebraska
running capacitor is in the circuit all the time, it still provides a phase shift while running, just that depending on design more shift is needed for high torque starting ability and they put a start capacitor in parallel with the run capacitor until the motor has reached a certain speed, then they switch the start capacitor out of the circuit. The capacitor may limit the current of the aux winding but it's capacitance is selected by considering characteristics of the winding it is matched to as well as the desired performance of the motor/expected load to be put on the winding.
Difference would mostly be design of insulation between the capacitor plates would need to be able to withstand a higher voltage between the plates.
Let me add that I am not well refreshed on capacitor and inductor calculations, but is likely possible they could build single phase motors without capacitors (actually they do but typically only in about 1/6 HP or less) but probably is less costly, less bulky and more efficient to use capacitors instead of just different impedance combinations of aux and main windings to try to accomplish the task.
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gar
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160911-2255 EDT
Something for you to think about:
Suppose you consider a 50 ufd capacitor at 60 Hz is in series with a 10 ohm resistor and we have an adjustable inductor also in series. The capacitive reactance is 50 ohms. Adjust the inductor to also have 50 ohms of inductive reactance, an inductance of 0.13 H. then we have a series resonant ciccuit resonant at 60 Hz. In series apply a sine wave source voltage of 100 V RMS.
What is the current flow? 10 A.
What is the voltage across the resistor? 100 V.
What is the voltage across the capacitor or inductor? 500 V.
What is the phase angle of the current relative to the source voltage?
Detune from resonance and what happens to the voltages and current?
.
Something for you to think about:
Suppose you consider a 50 ufd capacitor at 60 Hz is in series with a 10 ohm resistor and we have an adjustable inductor also in series. The capacitive reactance is 50 ohms. Adjust the inductor to also have 50 ohms of inductive reactance, an inductance of 0.13 H. then we have a series resonant ciccuit resonant at 60 Hz. In series apply a sine wave source voltage of 100 V RMS.
What is the current flow? 10 A.
What is the voltage across the resistor? 100 V.
What is the voltage across the capacitor or inductor? 500 V.
What is the phase angle of the current relative to the source voltage?
Detune from resonance and what happens to the voltages and current?
.
David Goodman
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- Pahrump, NV, USA
FionaZuppa
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i thought i answered this. yes, caps need to handle the max peak voltage, not the peak-to-peak voltage. it needs to have a dielectric strength to prevent that EMF (volts) from pushing through the insulator and shorting.
240 rms = 339.46V
p-to-p= p x 2 = 678.92v
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As shown by example in post #27 both the RMS and the peak voltage across the capacitor can and will be greater than the RMS and peak voltages on the line terminals. Series LCR circuits do that when anywhere close to resonance.
In the limit as R approaches zero and Xl approaches -Xc the voltage across the cap and across the inductor individually increases without limit. (Or as some are comfortable saying approaches infinity.)
Phil Corso
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- Boca Raton, Fl, USA
Wrong again!
With the cap in series with the winding (an inductor) an RLC circuit is formed! The resultant V-drop across each component vary! Typically, the V-drop across the cap will be higher than the motor's nominal voltage!
Phil Corso
Last edited:
gar
Senior Member
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- Ann Arbor, Michigan
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- EE
160911-2414 EDT
ActionDave:
Draw the series circuit of an ideal voltage source (zero internal impedance), a resistor, capacitor, and inductor. When the capacitive and inductive reactances are equal in value, then the circuit is said to be at series resonance. When this occurs the series circuit impedance is the value of the resistance. In my example at resonance the current is 100/10 = 10 A. This 10 A is flowing thru both the capacitor and inductor. Treating the capacitor alone the voltage across the capacitor is its reactance times its current or 50 * 10 = 500 V, but the capacitor voltage lags its current by 90 deg. Doing the same for the inductor we get 500 V and the voltage leads the current by 90 deg. With respect to lag and lead note that I am talking about voltage relative to current. Whereas you usually think in terms of current vs voltage. So lag and lead are reversed.
Since the capacitor and inductor voltages are equal in magnitude, but are 180 degrees apart their sum is zero in the series circuit. 90 - (-90) = 180.
You can run a bench test with a low source voltage, possibly 10 V, a capacitor, and inductor all in series. Later I can possibly suggest some components you might have that would be suitable values.
.
ActionDave:
Draw the series circuit of an ideal voltage source (zero internal impedance), a resistor, capacitor, and inductor. When the capacitive and inductive reactances are equal in value, then the circuit is said to be at series resonance. When this occurs the series circuit impedance is the value of the resistance. In my example at resonance the current is 100/10 = 10 A. This 10 A is flowing thru both the capacitor and inductor. Treating the capacitor alone the voltage across the capacitor is its reactance times its current or 50 * 10 = 500 V, but the capacitor voltage lags its current by 90 deg. Doing the same for the inductor we get 500 V and the voltage leads the current by 90 deg. With respect to lag and lead note that I am talking about voltage relative to current. Whereas you usually think in terms of current vs voltage. So lag and lead are reversed.
Since the capacitor and inductor voltages are equal in magnitude, but are 180 degrees apart their sum is zero in the series circuit. 90 - (-90) = 180.
You can run a bench test with a low source voltage, possibly 10 V, a capacitor, and inductor all in series. Later I can possibly suggest some components you might have that would be suitable values.
.
Mgraw
Senior Member
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- Opelousas, Louisiana
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- Electrician
As you said this is common mostly in refrigeration. Compressor manufacturers love to use OEM potential relays. Large manufacturers can have hundreds of OEM part numbers. No service guy is going to carry all of them. Most of those OEM relays can be replaced with 8-10 third party relays.
gar
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- Ann Arbor, Michigan
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- EE
160913-1045 EDT
Back to the original post.
Voltage ratings on anything basically are based on experimental data and some desired expected life of the anything.
Capacitors are composed of an insultating material between two electrodes. Dielectric breakdown of the insulator is the first major factor in determining the voltage rating. For a reasonably good discussion see https://en.wikipedia.org/wiki/Film_capacitor .
The DC and AC voltage rating of a particular capacitor can be quite different. AC imposes greater stresses on the dielectric, but in either case it is the peak voltage that is important.
How a particular capacitor is rated may depend upon its application.
In a series circuit consisting of a low impedance voltage source, an inductor, a capacitor, and resistance the voltage across the inductor or capacitor terminals can be greater than the source voltage. More on this in a subsequent post.
.
Back to the original post.
Voltage ratings on anything basically are based on experimental data and some desired expected life of the anything.
Capacitors are composed of an insultating material between two electrodes. Dielectric breakdown of the insulator is the first major factor in determining the voltage rating. For a reasonably good discussion see https://en.wikipedia.org/wiki/Film_capacitor .
The DC and AC voltage rating of a particular capacitor can be quite different. AC imposes greater stresses on the dielectric, but in either case it is the peak voltage that is important.
How a particular capacitor is rated may depend upon its application.
In a series circuit consisting of a low impedance voltage source, an inductor, a capacitor, and resistance the voltage across the inductor or capacitor terminals can be greater than the source voltage. More on this in a subsequent post.
.
FionaZuppa
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gar
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- Ann Arbor, Michigan
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- EE
160913-1204 EDT
ActionDave:
I have been working on an experiment that uses components that you or some other electricians may have on hand that will allow you to observe some of the effects of a series resonant circuit.
Assumptions:
A high impedance (10 megohms or greater) DVM is available.
A 500 VA 60 Hz transformer with at least one 120 winding can be found. A 1 kVA or 250 VA would work, but just a somewhat different capacitor range for resonance. This is to get an inductor for the experiment that you may have sitting on the shelf. Only one 120 winding of the transformer is used as the inductor. All other wires or terminals are left unconnected.
A small, but sufficiently large, stepdown transformer to provide about 6 V at 60 Hz at close to constant voltage output. I used a Stancor P-8668 transformer with a 120 primary and 28 V 2 A center tapped secondary. Rating is ballpark 50 VA. Only 1/2 of the secondary was used for the source, and a Variac adjusted the primary voltage. Between loaded and unloaded in the test the voltage changed from 6.09 to 6.12 V. This was a sufficiently low impedance that it is approximately a constant voltage source.
An assorment of moderately good quality capacitors ( meaning paper, oil filled, or plastic film, and NOT electrolytic ) in the range of 1 to 30 mfd. Electrolytic capacitors are quite lossy ( substantial internal shunt resistance ). Most good capacitors are of relatively low loss. Mica, polystrene, and polypropylene are very good capacitors, current is very close to 90 degrees from voltage.
Most practical inductors are quite lossy, hard to get zero resistance. Losses are primarily in two places, wire resistance, and magnetic core.
The test circuit:
This consists of the series circuit of the voltage source (set to around 5 to 10 V RMS sine wave), a combination of capactors connected in parallel to form one capacitor that is adjustable by changing the capacitors in parallel, and the one winding of the transformer that is the inductor. The voltmeter is connected to the particular component as desired.
Note: the resistance in the series circuit is mostly that which results from the wire resistance in the inductor, and its core losses. A small part of the resistance comes from the voltage source. Virtually none of the resistance is from the capacitors if they are high quality. Also a small part of the circuit inductance is from the voltage source. At 60 Hz no inductance of significance in the capacitors.
Voltage drop across the capacitor is all reactive. Inductor terminal voltage drop is of a series combination of of inductive and resistive components.
My test circuit results:
Resonance occurs when the capacitor is adjusted to produce maximum voltage drop across the capacitor. At this point the capacitor voltage was 17.58 V from a source voltage of 6.09 V. The capacitor terminal voltage is 2.89 times the source voltage. The capacitance was 6.22 mfd. 13.7 V is the terminal voltage across the inductor and its internal equivalent series resistance from wire rsistance and core losses.
From this capacitance we can calculate the series inductance. f = 1/2*Pi (L*C)^0.5. Rearranging and L*C = 1 / ( 60*2*3.1416)^2 = 1 / 142,123. For C = 6.22*10^-6 the rsult is L = 10^6 / 142,123*6.22 = 1,000,000 / 884,005 = 1.13 H (Henrys).
If you pick a random transformer, but similar to the Sq-D, its inductance should be in the ballpark of my value.
As you move away from resonance the capacitor voltage drops. If the capacitance is made very large, then its voltage approaches zero. If the capacitor is made very small, then its voltage approaches the source voltage.
Conclusion:
In a series resonant circuit the capacitor voltage at resonance can be large compared to the source voltage. As you move off resonance the capacitor voltage drops. So does the inductor voltage drop. Loop current is a maximum at resonance.
A side observation:
I like to check experiments by an alternative means if possible. A measurement with my General Radio impedance bridge produced an inductance measurement around 0.15 H, a 10 X difference. The bridge normally operates at about 1 kHz. The difference between 1000 Hz and 60 Hz should not account for this huge change in inductance. In the bridge circuit the voltage across the inductor was only a fraction of a volt, instead of the measured almost 14 V in the series resonant circuit. Why the apparent difference in inductance? I believe the reason is that when the flux density in a ferromagnetic material (in this case transformer iron of some sort) is very low the permeability drops and therefore the inductance.
So another experiment. In the series resonant circuit I lowered the excitation level so as to have about 0.2 V across the inductor at resonance. Now my calculated inductance was close to that of the bridge circuit.
ActionDave: this is probably still confusing, but if you can setup some similar experiment it may help you see what happens in series resonance. In the motor application of capacitors you are probably way on the side of the resonance curve. Thus, not as much voltage multiplication. http://www.electronics-tutorials.ws/accircuits/series-resonance.html is quite nice.
Have to leave, haven't proofread.
.
ActionDave:
I have been working on an experiment that uses components that you or some other electricians may have on hand that will allow you to observe some of the effects of a series resonant circuit.
Assumptions:
A high impedance (10 megohms or greater) DVM is available.
A 500 VA 60 Hz transformer with at least one 120 winding can be found. A 1 kVA or 250 VA would work, but just a somewhat different capacitor range for resonance. This is to get an inductor for the experiment that you may have sitting on the shelf. Only one 120 winding of the transformer is used as the inductor. All other wires or terminals are left unconnected.
A small, but sufficiently large, stepdown transformer to provide about 6 V at 60 Hz at close to constant voltage output. I used a Stancor P-8668 transformer with a 120 primary and 28 V 2 A center tapped secondary. Rating is ballpark 50 VA. Only 1/2 of the secondary was used for the source, and a Variac adjusted the primary voltage. Between loaded and unloaded in the test the voltage changed from 6.09 to 6.12 V. This was a sufficiently low impedance that it is approximately a constant voltage source.
An assorment of moderately good quality capacitors ( meaning paper, oil filled, or plastic film, and NOT electrolytic ) in the range of 1 to 30 mfd. Electrolytic capacitors are quite lossy ( substantial internal shunt resistance ). Most good capacitors are of relatively low loss. Mica, polystrene, and polypropylene are very good capacitors, current is very close to 90 degrees from voltage.
Most practical inductors are quite lossy, hard to get zero resistance. Losses are primarily in two places, wire resistance, and magnetic core.
The test circuit:
This consists of the series circuit of the voltage source (set to around 5 to 10 V RMS sine wave), a combination of capactors connected in parallel to form one capacitor that is adjustable by changing the capacitors in parallel, and the one winding of the transformer that is the inductor. The voltmeter is connected to the particular component as desired.
Note: the resistance in the series circuit is mostly that which results from the wire resistance in the inductor, and its core losses. A small part of the resistance comes from the voltage source. Virtually none of the resistance is from the capacitors if they are high quality. Also a small part of the circuit inductance is from the voltage source. At 60 Hz no inductance of significance in the capacitors.
Voltage drop across the capacitor is all reactive. Inductor terminal voltage drop is of a series combination of of inductive and resistive components.
My test circuit results:
Resonance occurs when the capacitor is adjusted to produce maximum voltage drop across the capacitor. At this point the capacitor voltage was 17.58 V from a source voltage of 6.09 V. The capacitor terminal voltage is 2.89 times the source voltage. The capacitance was 6.22 mfd. 13.7 V is the terminal voltage across the inductor and its internal equivalent series resistance from wire rsistance and core losses.
From this capacitance we can calculate the series inductance. f = 1/2*Pi (L*C)^0.5. Rearranging and L*C = 1 / ( 60*2*3.1416)^2 = 1 / 142,123. For C = 6.22*10^-6 the rsult is L = 10^6 / 142,123*6.22 = 1,000,000 / 884,005 = 1.13 H (Henrys).
If you pick a random transformer, but similar to the Sq-D, its inductance should be in the ballpark of my value.
As you move away from resonance the capacitor voltage drops. If the capacitance is made very large, then its voltage approaches zero. If the capacitor is made very small, then its voltage approaches the source voltage.
Conclusion:
In a series resonant circuit the capacitor voltage at resonance can be large compared to the source voltage. As you move off resonance the capacitor voltage drops. So does the inductor voltage drop. Loop current is a maximum at resonance.
A side observation:
I like to check experiments by an alternative means if possible. A measurement with my General Radio impedance bridge produced an inductance measurement around 0.15 H, a 10 X difference. The bridge normally operates at about 1 kHz. The difference between 1000 Hz and 60 Hz should not account for this huge change in inductance. In the bridge circuit the voltage across the inductor was only a fraction of a volt, instead of the measured almost 14 V in the series resonant circuit. Why the apparent difference in inductance? I believe the reason is that when the flux density in a ferromagnetic material (in this case transformer iron of some sort) is very low the permeability drops and therefore the inductance.
So another experiment. In the series resonant circuit I lowered the excitation level so as to have about 0.2 V across the inductor at resonance. Now my calculated inductance was close to that of the bridge circuit.
ActionDave: this is probably still confusing, but if you can setup some similar experiment it may help you see what happens in series resonance. In the motor application of capacitors you are probably way on the side of the resonance curve. Thus, not as much voltage multiplication. http://www.electronics-tutorials.ws/accircuits/series-resonance.html is quite nice.
Have to leave, haven't proofread.
.
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Perhaps a change in the design made the starting capacitor unnecessary, in reality?
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- Placerville, CA, USA
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- Retired PV System Designer
Motors that are able to start unloaded are less likely to require the extra torque that a switched starting capacitor could provide.
As long as the refrigerant pressure has been allowed to bleed down the starting mechanical torque load on the sealed compressor is pretty low.
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