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Please never add text on what I have written. Try to reread my post. I never infer that the inductor will produce larger inrush of current. I said sudden change of current. Inrush current and change of current are two different things. Inrush is a spike of current, normally seen in a capacitive load or lamps (due to high demand of current for the filament to energize). change of current is more on discrete, from low to high (50mA to 250mA). Both of these currents are function of time.
Change of current is the di of V=L*di/dt and the Inrush current can be approximated to thru this Iinrush=C*dv/dt.
I agree with you that the inductive load should have low inrush current but it should have higher transient voltage.
BTW, we have no choice... we should consider the 25ohm in the design. It is the resistance of the inductor coil (not in series nor parallel). It is called ESR.
Please never add text on what I have written. Try to reread my post. I never infer that the inductor will produce larger inrush of current. I said sudden change of current. Inrush current and change of current are two different things. Inrush is a spike of current, normally seen in a capacitive load or lamps (due to high demand of current for the filament to energize). change of current is more on discrete, from low to high (50mA to 250mA). Both of these currents are function of time.
Change of current is the di of V=L*di/dt and the Inrush current can be approximated to thru this Iinrush=C*dv/dt.
I agree with you that the inductive load should have low inrush current but it should have higher transient voltage.
BTW, we have no choice... we should consider the 25ohm in the design. It is the resistance of the inductor coil (not in series nor parallel). It is called ESR.
Thanks for all of you.
I think that i still didnt understand how come the capacitor decreases the input impedance?
Please explain it to me, its sitting on my mind and i cant get it right.
The impedance of a power supply is a measure of its "quality" - its ability to deliver a current.
The impedance of a power supply is measured in ohms but we don't say the resistance of a power supply because we are not actually measuring any resistance values but determining the value of resistance by measuring the voltage and the current it is capable of delivering. From these values we get a value we call impedance.
If we connect a load to a power supply, the output voltage will drop.
If we connect a load to a supply for a very short period of time, the voltage will drop then rise again. This dip in voltage can upset some of the other components in the circuit and this is something we can prevent by adding a capacitor across the output of the power supply. The capacitor is called an electrolytic – due to the way it is made and has a high capacitance value.
The electrolytic charges and sits in the circuit in a fully charged state. It is like a small battery connected to the circuit.
When the voltage drops suddenly, it delivers energy to the system and “tops up” the supply voltage so that the voltage does not dip as much.
The power supply is now classified as being “tighter” and because it is capable of delivering a higher current (for a short period of time) it is said to have a lower impedance.
Now i understand! i really thank you!
You said in previous posts that the voltgae drops when PSU cant deliver the demanded current.
But why when the PSU can deliver the demanded current, the voltage still goes low?
For example, many times i measured the AC mains voltage (~230V) in open circuit (no load) and when 3KW load connected, and the AC mains voltage was higher in the first case.
I truely know that it can deliver more than 3KW/230V= 13A current, so Why does it happen?
You have to introduce some terms into your discussion.
It we take a power supply such as the mains (240v) and say: I want regulation of 10v for a current of 20 amps.
If you now draw 20 amps and the supply drops 8v, the power supply is within limits.
If it drops 16v, the supply is not within your requirements.
It’s all to do with your requirements.
If you have a 5v supply and require it to drop less than 5mV for 1A delivery. It will not be accepted if it drops 17mV when delivering 1 amp.
No power supply can deliver a high current without some sort of reduction in voltage.
Even a car battery drops in voltage when delivering 20 amps to the headlights. It might be only 0.2v, but it will be detectable. And when it is delivering 450amps to the starter motor, the voltage can drop to less than 10v.
The mains also has tolerance values. In other words, the voltage will be within certain limits.
The power companies try to deliver a few volts above 240v as any increase in voltage causes a larger increase in current and the electricity bill is higher. A 5v increase (2%) increases the bill 4%. Our supply voltage is 249v. That’s why I am paying 13.6% more. Cunning isn’t it. I bet you didn't even realise the consequences.
These are the sort of things no-one thinks about or knows about.
Well, you gave me lots of important examples of PSU's voltage dropping down, and i thank you for that (and for your patiance).
but you didnt tell me why it happens.
I have a theory (a friend of mine just told me).
In open-circuit, no current flows through the wires, therefore no voltage drops on the wires, and therefore the voltage that you measure (after the wires) is maximal.
Bigger load you connect, gives larger current flows through wires and therefore larger voltage drops on wires, therefore I measured less voltage on AC mains.
But this theory has nothing to do with what you said about impedance of PSU being V_PSU/I_PSU, becaues V_PSU doesnt drop, it just that V_PSU - r*I (where r is the wires resistance) drops, and that is what i measure, V_PSU - r*I.
That's right.
When a current flows, a voltage drop appears across any of the wiring that is carrying the current. The voltage drop is proportional to the current flow. In other words it corresponds to the equation (formula)
V = IR where V is the voltage in volts, I is the current in Amps and R is the resistance in Ohms. If the voltage drop is 3v when 1 amp flows, the drop will be 6v when 2 amp flows and 15v when 5 amp flows etc.
This also applies to smaller values of mV when mA currents flow.
But we've also got power losses in the wiring and this is equal to I x I x R so that when the current doubles (increases 200%), the heating increases 4 times (that is 400%)
Since the inductive load (such as the relay) has a resistive characteristics. As viewed from the output of the power supply, there is a sudden change of current (not inrush) because the load impedance decreases due to the low resistance of the coil. Note this is viewed from the output of the power supply, therefore, the load is the whole circuit. Let say initially, the circuit draws about 50mA (relay off), and rises up to 250mA when relay is activated. As seen by the power supply, there is a change of current for 200mA. We can say that it is caused by the relay.
With this change of current, if the power supply can not tolerate this change, there is a chance that the voltage supplied will be fluctuated. In order to resolve this, we can add a bulk capacitance (temporary energy storage) so that during the this sudden change of current, the voltage output is still stable and no fluctuation occur.
The problem by adding the capacitor is the inrush current (not change of current) during power up. As seen from the power supply, the capacitor is parallel with respect to the total load (the whole circuit). Therefore,
Ctotal=Cbulk+Ccircuit
The higher the capacitance, the higher the inrush current. This is because
I=C*dv/dt
So to resolve the inrush, we can add a resistor in series of the supply line. The trace impedance is not realiable to consider as your resistance. But if you design is part of a system (such as automotive) which would have long cables before the supply reaches your ciruit, this impedance is significant. This would render at least 1Ω of impedance and can reduce the inrush significantly.
As viewed from the relay coil, it can see a change of current from maybe 0A or a leakage of few µA to 200mA. With the characteristics of the inductor,
V=L*di/dt
it will have will rapid increase its voltage across itself and it will stop only when there is no more change of current (di). This phenomenon is called inductive kick off voltage. The problem here lies on the energy it gives off during the kick off (specially when you're using semicon in driving the relay). Since this energy is given, something should absord. If you design has no protection, it will destroy your relay driver or power supply or both. To resolve this, there are few solutions.
1. free-wheel diode that will absord the energy.
2. Transient suppressor such as transil, MOV, etc. that will clamp the kick off voltage so that the other circuits can tolerate the kick off voltage. But the suppressors partly absorb the energy too.
3. snubber, this is more on changing the waveform of the voltage ringing into lower frequency. From fast, high peak and destructive ringing into a more friendlier waveform where your components can tolerate.
These are all what I'm trying to infer here. I have simulated, calculated and design all of these situation. Maybe I stated something wrong but these are all I've seen during the past few years. If someone here who is knowledgable or an expert in design for EMC compatibility(specially the ISO pulses) please do correct me.
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