# Can't understand concept 'stiff source'

What does it mean in itself and in context?. Bolded phrase in section 10.5.2 IEEE Std 519.

Methods for compensating for existing or potential flicker are much the same as those used to compensate for subtransient disturbances, such as those evidenced by notching or harmonic currents. The simplest and generally most effective technique is to provide a sufficiently stiff source of power so that the effect is negligible at the point where the flicker source is tapped off from the rest of the power distribution system. Compensatory methods are used to emulate the stiff source. Series capacitors, thyristor switching of inductors with shunt capacitors (static var control), saturating shunt inductors, and thyristor switched shunt capacitors may be used to maintain a relatively steady voltage at the tie point. As in cases in which such schemes are used to provide subtransient compensation, the possibility of overall distribution system instability must be thoroughly investigated before one can confidently apply the technique.

Thanks!

The other answers are good, but you might find a different approach better. Start with a voltage source and a load. The voltage source will have an output impedance. The circuit will look like

simulate this circuit – Schematic created using CircuitLab

Let's say that Rout is large. Then changes to the load resistor will cause changes to VLoad. The voltage (measured at the load resistor) is "elastic". It will vary with changes to the load.

Now let's say that Rout is very small. Then changes to RLoad will have little effect on VLoad. Using the image of the load on the end of a beam, the source is now said to be "stiff". It does not change with changes in load.

• Thanks for your answer, it's much clear. Now, what would it be, for example, that Rout in power systems? It's located at distribution or generation level? Commented Apr 7, 2017 at 3:15
• @Diegov - For power systems, Rout is determined by every step from the generating plant to the power outlet. Every transformer, every power line will add resistance. Of course, every added resistance wastes power and reduces the efficiency of the system, so such losses are carefully considered and their effects counteracted to the degree economically feasible. Commented Apr 7, 2017 at 14:28
• I was perplexed with the term stiff until you mention "beam". Thanks.
– KMC
Commented Apr 15, 2020 at 8:49

It means a relatively low impedance source, compared to the disturbances in the system.

In the context above, it seems the power source needs to have a low impedance in order to minimize the effect of the transient disturbances. By compensatory methods, they mean the the power supply is designed using a control loop or other schemes which "further stiffen" the source: regulation methods lower the effective impedance of the power supply.

% "Load regulation" is like a voltage divider with Vout= (Zout/Zout+Zsource)Vin and typically we just say R ratio of load/source for high ratios. Thus load-dependent Voltage regulation depends on low Source Z so the output "resists" change in voltage hence "stiff source".

In this case, active power factor using thyristor switches and passive LC components are used to regulate distribution voltage within 5% typically for generation and 5% for distribution with variations in load to stay within tolerance. Hence stiff source with compensation. So when transient inductive or capacitance loads are dominant, grid controls may change to accommodate best transformer source impedance and tap voltages to improve regulation.

In India it may be not be so well regulated yet for any reasons, but here in Toronto area my voltage is well regulated within a few % year round with rare exceptions. They use a complex network of telemetry at all tie points and neighbourhoods to regulate this and isolate detected faults.

My brother-in-law who is a Prof at U of T in EE Energy networks, recently visited our PowerStream network control and said it was possibly the best in the world ( he has seen or heard) for power regulation and control. But then our peak demand energy costs were also very high (\$175/MWh) but recently rates were rolled back by Premier due to consumer complaints.

We know that Op Amps reduce Output impedance with negative feedback, but this is quasi static and not real-time high BW like an Op Amp, although the compensation has some effect of lowering effective load regulation by occasional changes to these series shunt reactors and caps.

• +1 for the voltage divider viewpoint! Commented Apr 26 at 19:41

"Stiff source" means low source impedance, high ratio of prospective short circuit current / rated current, low effect of source voltage by changing load current.

In the context to IEEE 519, notching is essentially a microsecond range short circuit caused by reverse recovery current of a thyristor. Harmonics are caused by load current with high peaks occurring every cycle.

# The power of step-by-step presentations

The most natural way to intuitively understand circuit ideas is to follow step-by-step the evolution of the circuit solution. CircuitLab gives us a great way to do it in a working way through a sequence of "live" simulations instead of the usual "dead" verbal explanations through banal verbal clichés. This is how we will get closer to the natural way of understanding by doing familiar to us from life.

# The role of figurative names

Of course we can simply say that "a perfect voltage source does not change its voltage when the load increases". But how much stronger visual idea of ​​this phenomenon we get when we say that it is "stiff" ("hard") and does not yield to the "pressure" of the load.

Conversely, instead of saying that an imperfect voltage source changes its voltage when the load decreases, we can metaphorically say that it is "soft" and easily yields to the "pressure" of the load.

# Perfect voltage sources

The textbook statement that a "constant voltage source has zero output resistance" gives the impression that the source is connected by a piece of wire to the load.

## Zero output resistance

Let's explore this arrangement by changing the resistance of a resistor load RL. As you can see, this is actually the Ohm's circuit.

simulate this circuit – Schematic created using CircuitLab

## Visualized quantities

In CircuitLab, it is possible to set the meter's resistance in the parameters window. So we can replace the load with a voltmeter with the same internal resistance (1 kΩ). It will serve as a "visualized resistor", and the schematic will be simplified.

simulate this circuit

As you can see, the entire voltage V appears across the load because there is no voltage drop (loss) across the "piece of wire". You can think of the wire as a "stiff rod". On one side it is immovably attached to the source, and on the other side, the load pushes and pulls it...but can not move it.

# Imperfect voltage sources

Real voltage sources have (are modelled by) some constant output resistance Rout.

## Inserting output resistance

Let's represent it by a 5 kΩ resistor Rout = RL, and to increase the input voltage to 20 V. The two resistors form a voltage divider, and its output voltage is equal to half (10 V) of the input 20 V.

simulate this circuit

## Visualized quantities

As above, we can visualize the two resistors by meters with internal resistance of 5 kΩ.

simulate this circuit

When sweeping RL from 1 to 10 kΩ, the load voltage VRL significantly varies since there is a voltage (loss) across Rout.

Now you can think of the resistor as a "spring" which under the influence of the load contracts or stretches (the US "zig-zag" symbol is very suitable for the purpose). The greater the resistance, the weaker the spring.

# How to improve imperfect voltage sources

Natural perfect voltage sources have very low output resistance but their voltage cannot be adjusted. Imperfect voltage sources have high resistance but their voltage can be adjusted. Therefore improved imperfect voltage sources are used in electronic circuits. They have the advantages but lack the disadvantages of both.

The problem with imperfect voltage sources is that the voltage V and resistance Rout are static. So to improve them, we need to make either the resistance or the voltage dynamic.

## Dynamic resistance

A possible way is to change the resistance Rout in the same direction as the load resistance RL changes so that the voltage divider gain does not change.

Conceptual circuit: The best way to understand how this happens is to manually set a few typical RL values.

RL = 2.5 kΩ: For example, RL decreases to 2.5 kΩ but we also decrease Rout to 2.5 kΩ, and the voltage VRL across the load does not change (10 V).

simulate this circuit

RL = 10 kΩ: Then, RL increases to 10 kΩ, but we also increase Rout to 10 kΩ, and VRL = 10 V.

simulate this circuit

## Transistor implementation

In electronic circuits, dynamic resistors are implemented by transistors. For example, in the circuit of an emitter follower below, the collector-emitter "resistance" Rce of the transistor serves as the dynamic resistor Rout above.

RL = 5 kΩ: At the initial load resistance RL = 5 kΩ, the static resistance Rce = Vce/Ic becomes 5 kΩ as well; VRL = 10 V.

simulate this circuit

RL = 2.5 kΩ: Then, if RL decreases twice, Rce becomes also 2.5 kΩ; VRL = 10 V.

simulate this circuit

RL = 10 kΩ: Finally, if RL increases twice, Rce becomes also 10 kΩ; VRL = 10 V.

simulate this circuit

When sweeping RL from 1 to 10 kΩ, the load voltage does not vary.

# Dynamic voltage

With the same succes we can change the input voltage in the same direction as the load resistance changes to keep the output voltage constant.

## Conceptual circuit

The best way to understand how this happens is to manually set a few typical RL values.

RL = 5 kΩ

simulate this circuit

RL = 2.5 kΩ: For example, when RL decreases to 2.5 kΩ we increase the input voltage V to 30 V, and the voltage VRL across the load does not change (10 V).

simulate this circuit

RL = 10 kΩ: Conversely, when RL increases to 10 kΩ, we decrease Rout to 15 kΩ, and VRL = 10 V.

simulate this circuit

## Op-amp implementation

Now the op-amp does this routine job of raising and lowering its output voltage to keep the voltage across the load constant. For example, in the circuit of an op-amp follower below, the op-amp output voltage serves as the input voltage V above.

RL = 5 kΩ: At 20 V input voltage, the load voltage is 10 V.

simulate this circuit

RL = 2.5 kΩ: Then, if RL decreases twice, VOA becomes 30 V; VRL = 10 V.

simulate this circuit

RL = 10 kΩ: Finally, if RL increases twice, VOA becomes 30 V; VRL = 10 V.

simulate this circuit

In the graphs below, when we sweep RL from 5 kΩ to 10 kΩ, the op-amp decreases its output voltage from 20 to 15 V to keep the load voltage constant (10 V).