Hi All,

Please help me out to calculate the short pulse and voltage of capacitor

I have a inductive load i.e. motor which works under below condition .

V= 5vdc
I=200mA

Now I want to put a capacitor across 5vdc.

So if I know the inductance of of a motor.
Now how will I calculate a

1. short pulse time means DeltaT
2. Peak current during short pulse
2. Voltage dip_ DeltaV

If capacitance is fixed let's say 47uF and let's assume some inductance.

I know some below formula

DeltaV= ((I×deltaT)/C)

But there is 3 unknown quantity i. e DeltaV, deltaT and Ipulsed

I want to do its theoretical calculation.

I am not much sure about above so please help me out. It was just an assumption.

 

What are you pulsing? Can you describe your pulse as a function of time?

More importantly, what are you really trying to do? Why are you trying to model something you are not prepared for? You need to learn the basics before taking on more challenging topics.

 

Your request is sort of a mishmash of statements.
Please clarify exactly what you are doing and what you want to calculate.
A schematic diagram would be a huge help.

 

What are you pulsing? Can you describe your pulse as a function of time?

More importantly, what are you really trying to do? Why are you trying to model something you are not prepared for? You need to learn the basics before taking on more challenging topics.

Probably I was unable to make you people understand.

Let me give it a try once again

I wanted to calculate Turn On voltage Dip of capacitor. For a above specified load how much capacitor voltage fall down when inductive load get turn ON. In other word we can say transient voltage Dip of capacitor when load get Turn ON.

Let’s say a capacitor 22uF/10V is connected at supply voltage of 5vdc so it will store the charge. When a Inductive load turn ON rated 5V@200mA. It will have some Inrush current for a short while and there is possibility of voltage dip at that time. So now I wanted to know how much 5Vdc will fall down below 5V and what is time period.

If inrush current increase there might be voltage dip of device.

This is what I wanted to know !!!

 

Ah, OK, thanks. So you want to calculate the voltage of an RCL circuit, with R, C, and L all in parallel. What is happening with the power supply when the load (R & L) is applied to the capacitor? Any dip in voltage will depend on the output impedance of the power supply. No impedance, no dip.

 

Ah, OK, thanks. So you want to calculate the voltage of an RCL circuit, with R, C, and L all in parallel. What is happening with the power supply when the load (R & L) is applied to the capacitor? Any dip in voltage will depend on the output impedance of the power supply. No impedance, no dip.

Thanks for your reply.

I do not how much I understood your comment but moving ahead of this..

I understand two this -

1. Continuous dip of voltage i.e. at full load means let's say if 5V supply gives 5V at no load and when load is connected it's marginally come down let's 50-200mV.

2. If we turn ON the inductive load let's say motor with 5V so now for instant period of time there will high peak current or surge current. probably it could be 10 time average current. Now what do understand is increase in sudden current definately causes a voltage dip.

Ex.
Let's say 1A rated inductive load has peak current is 10A at turn on time say mS/uS and 5V fall down say 1V before it reaches its final value.

So I wanted to know how to determine that current and time period.

Sorry to again ask you same questions. Could it will be limit that current by using some capacitor or other devices.

I did not understand the same questions you have answered me. Provided me tutorials if possible.

Thank you so much once again.

 

You are correct in your description, but you cannot begin to calculate a number without much more detail describing the power supply. Even the wiring from the supply to your circuit will impact any sharp transients. It is not enough to only know about the load.

Suppose your power supply already has a super-capacitor at its output pins. Adding your little capacitor at the load will have only a tiny effect. But if the load is a small 5V battery, the voltage dip would be severe.