Considering Inrush Current in Voltage Drop Calculations

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GoldDigger

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Right. The inrush is not because the coil is an inductor, the inrush is a result the moving core (in the case of a solenoid) or armature (in the case of a contactor) and the time it takes to complete its motion, lowering the inductance in the circuit.
Independent of the actual armature motion, the magnetic circuit is tighter once the pole faces are in contact. That means higher inductance and therefore lower current once the contactor has finished closing.
This does not apply to DC except for the final time constant.

mobile
 

gar

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170207-2411 EST

Ingenieur:

Following is the plot of your request. Source voltage 24 V. Added series resistance to get 18 V across the relay coil is 161 ohms. This added resistance is what is used to measure current. Scope common reference for both voltage and current is the negative end of the relay coil. Two 680 k resistors in series are across the relay coil with the voltage probe connected to the midpoint. The normally open and normally closed contacts of one relay pole are in parallel and shunt the more positive 680 k resistor.

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DS2_QuickPrint53E.JPG
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The normally closed contact opens 17 mS after application of current, and initial closure of the normally open contact occurs 10 mS later.

The initial time constant is about 2.5 mS, and the second one around 7 mS.

I had tried to use a bridge to measure open and closed inductance, but results were incorrect. I don't know a reason.


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Ingenieur

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170207-2411 EST

Ingenieur:

Following is the plot of your request. Source voltage 24 V. Added series resistance to get 18 V across the relay coil is 161 ohms. This added resistance is what is used to measure current. Scope common reference for both voltage and current is the negative end of the relay coil. Two 680 k resistors in series are across the relay coil with the voltage probe connected to the midpoint. The normally open and normally closed contacts of one relay pole are in parallel and shunt the more positive 680 k resistor.

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View attachment 16746
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The normally closed contact opens 17 mS after application of current, and initial closure of the normally open contact occurs 10 mS later.

The initial time constant is about 2.5 mS, and the second one around 7 mS.

I had tried to use a bridge to measure open and closed inductance, but results were incorrect. I don't know a reason.


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excellent!
thank you very much

so it looks like if the V drop at operating current is held less than the mfgs spec'ed pick-up voltage it should function reliably
and since drop-out is so low (<5 vdc) it is not an issue

can't say how impressed I am with your effort and results
 

gar

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Ann Arbor, Michigan
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EE
Summary points on a DC relay

Summary points on a DC relay

170108-0820 EST

Some summary notes about a DC solenoid, contactor, or relay.

1. If the device actuates there is a change in the inductance with respect to time, but this is still a series resistance-inductance circuit.

2. There is no inrush current. If the initial current in the circuit is zero at time t = 0 when a voltage is applied to the circuit, then at time t = 0+a small increment of time the current is still zero. This is a characteristic of an inductance. The current thru an inductor can not be changed instantaneously.

3. At no time after t = 0 will the current exceed V/R where V is the source voltage and R is the total series resistance.

4. If at any time the series current exceeds the pull-in current of the device, then the device will pull-in. It will stay pulled-in until the current drops below the drop-out current.

5. What the current is at any time is a function of the plunger or armature position and past current history.


Next I will discuss an AC device.

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Response to GAR as to why not local power supplies (which would make a lot of sense).

Our highly esteemed customer has a spec and an approved component list. They have these 24 VDC power supply panels with two power supplies each supplying two branches rated at 15 amps each. They will only allow 14 AWG trunk feeders to the field devices. Our problem is that the type of conveyor system their "tooling automation" scheme is being applied to is not concentrated in one place like a tooling cell, but rather is over 3000 feet long. The ArmorStarts are located at 25 foot centers along the conveyor path. We are running out of voltage way before we run out of 15 amps. This is going to require somewhere around 24 of these power supply panels, just becuase of the distances involved. You can see why voltage drop plays an important part in minimizing our cost.
 

Phil Corso

Senior Member
RichWaskowitz...

Thread # 5452312, "Series Voltage-Drop All Inclusive" started by SprayMax on 10-22-'15, may be helpful It provides a solution to the problem of calculating wire-size covering series-positioned (evenly-spaced or not) loads DC or AC)!

Regards, Phil Corso
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
170208-1211 EST

richwaskowitz:

Assuming you really have to live with these somewhat centralized power sources, then can you use an Edison distribution system? In other words three wires.

With an Edison system you only have voltage drop on one wire. Also feed the loads from the center.

I will use some simple values. Per wire maximum current is 7.5 A (1/2 of the power supply because of center feeding). Approximate wire resistance of #14 copper at 2.5 ohms per 1000 ft, or 0.07 ohms per 25 ft by rounding up. This somewhat compensates for wire temperature rise.

Assume 1 A per ArmorStart. Put one ArmorStart at the power supply. Thus, we only start with 7 A to the distribution line.

At 1st 25 ft location from power supply the voltage is Vps - 0.49, the drop to the next is 0.42, and as follows 0.35, 0.28, 0.21, 0.14, 0.07 and we are done. Total drop is 1.96 V, or using 24 V as the source, then 24 - 2 = 22 V at the last ArmorStart. Use your own figures.

If we assume that the only time that power is switched on to ArmorStarts is when the 24 V power supply is powered, then you can probably ignore any initial inrush current. Thus, the real important current to know is the peak current used when the ArmorStarts are working and doing their intended job.

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