Hi Everyone,

I am a newbie to this forum and to Electrical technology, I am not sure if this is the right place to post the following. I was a welder earlier(and thanks to recession) and am now studying to become an Electrician. I joined the electricians course a little late and am now befuddled with some terms that i am not able to negotiate very well. I have googled/binged but have not found any satisfactory answers for the following terms. It would be great if someone can in layman terms(analogies may be?) explain the following

1. Impedance matching
2. line-to-line, line-to-phase voltage
3. balanced load and unbalanced load in a 3-phase system
4. commutation components in a circuit.

TIA,
Bob

 

Welcome to the AAC (All-About-Circuits) forum, we hope you enjoy your stay!

I will expain impedance mathing. Impedance is measured in ohms, denoted by the symbol "Ω". It is in regards to AC circuits only, not DC circuits. Suppose we have a speaker with a 8Ω impedance, and an amplifier that also has an 8Ω impedance. We can now say that the impedances are matched giving a full transfer of power. Since they are both 8Ω, they cancel out delivering the highest potential of power distributed from the amplifier to the speaker (8/8 x 100 = 100%). What if we had 4Ω for the speaker and 8Ω for the amplifier? We can say that we would have a 50% loss of power, which is quite a bit (4/8 x 100 = 50%). So it is important to match impedances between circuits, you will deal a lot with this since electrical home systems are based entirely upon AC circuits.

 

Thanks Electronerd for your prompt response. I have some queries though, hope it is ok(since they are very basic). Consider the following, a single phase system delivering power to a motor. When you say impedance of the supply system, do you mean the impedance caused by the supply wire as an inductor ? also given
P=vi = i^2*r
doesnt it mean that decrease in destination impedance would lead to increase in current and therefore the power delivered remains constant ? I didnt understand the 50% logic that you mentioned in your post

 

Thanks Electronerd for your prompt response. I have some queries though, hope it is ok(since they are very basic). Consider the following, a single phase system delivering power to a motor. When you say impedance of the supply system, do you mean the impedance caused by the supply wire as an inductor ? also given
P=vi = i^2*r
doesnt it mean that decrease in destination impedance would lead to increase in current and therefore the power delivered remains constant ? I didnt understand the 50% logic that you mentioned in your post

No, there are two impedances. The impedance of the supply system, and the impedance of the motor. If you want to be really accurate you would add the impedance of the wire that feeds to the motor. They both need to match for maximum power transfer. The P = V x I isn't significant for impedance, that just lets you find out how much power you are using. Impedance is made up of resistance and reactance. Find the reactance of the motor (as in inductance) and also the power phase system using the following equations:

XL= 2∏fL - Where ∏ = Pi or 3.14, f = frequency, and L is the inductor value.

XC = 1 / 2∏fC - Where C = capacitor value.

Once you find the reactance, you can add it to all the resistance values forming the impedance.

As for the 50% query, imagine that the power phase system has an impedance of 8Ω and the motor has 4Ω (they will be more than that, but this is just an example). To find the power ratio we do (4Ω / 8Ω) x 100 = 50%. That means that only 50% of the power you put in is going to the motor. You want 100%, and you would need a motor of 8Ω to accomplish that. Another example: Suppose the power phase system has an impedance of 8Ω and the motor 6Ω. Well, (6Ω / 8Ω) x 100 = 75% You don't need the "Ω" symbols just do 4/8 or 6/8.

 

Note that in our examples we are using low impedance values (4Ω, 6Ω, 8Ω). Devices that need to be matched should have a relatively high impedance for high efficiency. Many MP3 players have high impedance and therefore their headphones also for high efficiency.

 

I used to be quite confused about what the meaning of impedance matching is all about and I dont know if i am jumping the gun here if you are learning about impedance then you will be soon introduced to impedance matching with the use of a transformer. As already mentioned, the idea of impedance matching is to achieve maximum power delivery to your load and the only way this is possible is if the load has the same impedance as the source. The idea of an impedance matching transformer is for the energy source to "see" or deliver energy to a different impedance that what it actually is. A convenient analogue of this is the bicycle. Try and contemplate the purpose of the gears on the bike in terms of impedance matching. Your legs are most efficient at delivering power at certain RPMs and changing the gears on your bike faciliate this.

take a look at the calculus based proof of this in this wiki

 

Note that in our examples we are using low impedance values (4Ω, 6Ω, 8Ω). Devices that need to be matched should have a relatively high impedance for high efficiency. Many MP3 players have high impedance and therefore their headphones also for high efficiency.

Almost any power source [generator] can be modeled by a voltage source and a series resistance. The series resistance is the internal resistance of the [generator]. For a battery, the most power you can deliver to an external load occurs when the load resistance is exactly equal to the internal resistance of the battery. NOTE! Though this condition results in the maximum POWER TRANSFER, it is by no means the most EFFICIENT condition...in fact it will be only 50% efficient because half the power will be dissipated in the internal resistance of the battery!

An AC generator can be more accurately modeled by an ac voltage source, a series inductor, and a series resistor. For an AC circuit, maximum power transfer occurs when the load impedance is the COMPLEX CONJUGATE of the generator impedance. This is true whether you're dealing with a simple AC generator/ motor circuit, or a radio frequency system with a complex system of transmissio lines and matching networks.


If a generator has a source impedance of 50 +j100 ohms, the load impedance has to be 50-j100 ohms for maximum power transfer. Again, this condition results in a mere 50% efficiency.

Hope this helps.

eric

 

Almost any power source [generator] can be modeled by a voltage source and a series resistance. The series resistance is the internal resistance of the [generator]. For a battery, the most power you can deliver to an external load occurs when the load resistance is exactly equal to the internal resistance of the battery. NOTE! Though this condition results in the maximum POWER TRANSFER, it is by no means the most EFFICIENT condition...in fact it will be only 50% efficient because half the power will be dissipated in the internal resistance of the battery!

An AC generator can be more accurately modeled by an ac voltage source, a series inductor, and a series resistor. For an AC circuit, maximum power transfer occurs when the load impedance is the COMPLEX CONJUGATE of the generator impedance. This is true whether you're dealing with a simple AC generator/ motor circuit, or a radio frequency system with a complex system of transmissio lines and matching networks.


If a generator has a source impedance of 50 +j100 ohms, the load impedance has to be 50-j100 ohms for maximum power transfer. Again, this condition results in a mere 50% efficiency.

Hope this helps.

eric

Right, thanks for mentioning all of that. Although, higher impedance matching (compared to lower impedance matching) would be better since it would be less subtle to noise, and filter it better, correct?

 

Thanks Electronerd, I am now clearer than ever with regards to impedance matching. One final question in this regard, how do power stations manage impedance matching, given they dont know who the user is at the other end.

Could you please explain the rest

2. line-to-line, line-to-phase voltage
3. balanced load and unbalanced load in a 3-phase system
4. commutation components in a circuit.

succinctly as you have done for Impedance Matching ?

 

Thanks Electronerd, I am now clearer than ever with regards to impedance matching. One final question in this regard, how do power stations manage impedance matching, given they dont know who the user is at the other end.

Could you please explain the rest

2. line-to-line, line-to-phase voltage
3. balanced load and unbalanced load in a 3-phase system
4. commutation components in a circuit.

succinctly as you have done for Impedance Matching ?

Power station (or rather system) operators don't concern themselves with maximum power transfer matching - they simply try to ensure there is sufficient generating & transmission capacity within the power network to maintain supply integrity, stability and economic dispatch.

2. I think you meant line-to-neutral not line-to-phase. The terms line-to-line and line-to-neutral voltage refer to the voltage potentials which exist between any two of the 3-phase active line conductors (line-to-line) or any of those conductors and the neutral conductor (line-to-neutral) in a 3-phase power system.

3. If equal currents (in both magnitude and phase) flow in the active lines of a 3-phase power system then the system is balanced. If the line currents aren't equal the system is unbalanced.

4. I'm not sure in reference to what your question about commutation components relates. One can have natural or forced commutation for instance - such as in power electronic systems. In a simple bridge rectifier the diodes are naturally commutated by the AC supply voltage. The commutator in a DC machine is an example of a commutation component.

 

I forgot to mention the terms "line" and "phase" are occasionally used interchangeably.

You might therefore see the term "phase-to-neutral" or "phase voltage". This can sometimes cause confusion. Although most electrical 'bods' would discern a difference between the terms "line voltage" and "phase voltage".

 

ELECTRONERD,

Impedance is measured in ohms, denoted by the symbol "Ω".

Usually impedance is denoted by "Z", reactance by X, and resistance by R. You will never see a coil or an capactor stamped with a Ω as a resistor sometimes is, because both reactance and impedance are highly frequency dependent.

It is in regards to AC circuits only, not DC circuits.

Impedance is composed of both resistance and reactance. So a DC with resistance only can still be called a impedance with no reactance.

Suppose we have a speaker with a 8Ω impedance, and an amplifier that also has an 8Ω impedance. We can now say that the impedances are matched giving a full transfer of power.

You mean the maximum power available at 8 ohms of source impedance.

Since they are both 8Ω, they cancel out delivering the highest potential of power distributed from the amplifier to the speaker (8/8 x 100 = 100%).

What does canceling out mean? The speaker will receive [8/(8+8)]100 = 50% of the source power available at 8 ohms source impedance.

What if we had 4Ω for the speaker and 8Ω for the amplifier? We can say that we would have a 50% loss of power, which is quite a bit (4/8 x 100 = 50%).

The speaker would receive [(4/(4+8)]100 = 33% of the power available at 8 ohms of source impedance.

So it is important to match impedances between circuits, you will deal a lot with this since electrical home systems are based entirely upon AC circuits.

That is fine and good, but did you know that you can transfer more power from the source to the destination if you can possibly lower the impedance of the source?

As for the 50% query, imagine that the power phase system has an impedance of 8Ω and the motor has 4Ω (they will be more than that, but this is just an example). To find the power ratio we do (4Ω / 8Ω) x 100 = 50%. That means that only 50% of the power you put in is going to the motor.

The correct calculation is (4/(4+8))100 = 33% of the power available at 8 ohms of source impedance.

You want 100%, and you would need a motor of 8Ω to accomplish that.

The motor would have to be of infinite impedance to attain a 100% power ratio at any real world value of source impedance.

Another example: Suppose the power phase system has an impedance of 8Ω and the motor 6Ω. Well, (6Ω / 8Ω) x 100 = 75% You don't need the "Ω" symbols just do 4/8 or 6/8.

(6/(6+8))100 = 43% of available power at 8 ohms of source impedance

Ratch

 

Interesting Ratch, I've never seen those suggestions before. I've always thought that you would want your impedances to match from each circuit, so if the power system was 8Ω you would want the motor to be 8Ω also. Has anyone else believed this too?

 

a balanced system is where current and it's phase angles are equal in all three loads in the configuration.

Phase voltages are just that, voltages across the phases. In Delta, that equals line voltage. In wye, that line voltage is srt(3) times it's phase voltage.

 

ELECTRONERD,

Interesting Ratch, I've never seen those suggestions before. I've always thought that you would want your impedances to match from each circuit, so if the power system was 8Ω you would want the motor to be 8Ω also. Has anyone else believed this too?

It is not a matter of blind faith or belief. It is a matter of proof that everyone will accept. Proof is easy to show, too. Let's say that we have a source voltage of 16 volts and both the source and load impedances are 8 ohms. That gives a current of 1 amp and a load dissipation of 8 watts. Now suppose we could somehow lower the impedance of the source by one-half to 4 ohms. That increases the current from 1 amp to 4/3 amps and the load dissipation from 8 watts to 14.22 watts. There are two things working to make this happen. 1) The total resistance of the circuit is decreasing, and 2) the load is taking a greater share of the voltage.

Now of course that method most often cannot be done because the source impedance is fixed or otherwise unavailable for tinkering. Also the source voltage has to be able to support the higher current caused by lowering its impedance. But it should be kept in mind in case there is a possibility that the source impedance can be lowered.

Ratch

 

Has anyone else believed this too?

I've always thought the maximum power transfer theorem has a limited range of applications. You wouldn't do it unless there were sound reasons.

Take a not too extreme example where you wouldn't intentionally apply it. I live in a state in which the base load power requirement is probably around 10,000 MW - I haven't checked lately but it will be of that magnitude.

Suppose the power authority used the maximum power transfer concept (with 50% efficiency) to distribute power to the community. They would be having to deal with extraordinary losses ~ 10,000MW. That would be intolerable & bad for greenhouse! In fact they would be hoping to achieve very much higher efficiencies. So they build electrical transmission & generation systems in which they intentionally design the 'internal' supply side impedances to be a fraction of the effective load impedances - i.e. as low as is practically possible within the engineering and cost constraints.