Charging of capacitor through rc and with pulse

[chrisggg]

Hi quick question, I have an rc circuit which is to be charged with an input voltage of 3.6 V but what happens if a pulsing current source is connected to the capacitor as well?

So I have a rc circuit but also a pulsing current source connected directly to the resistor.

I have attached the circuit incase I have not explained it properly. My question is how does one work out the vc voltage, its not so straight forward and believe is involves laplace but am very confused.

Thanks for any hints.

 

[gneill]

Hi chrisggg. Welcome to PF.

When you describe the current source as pulsing, what precisely do you mean? Is it a squarewave? A unit step? something else? You've indicated on your diagram ton = 50μs. Is that the time after t0 that it turns on, or is it the length of time that it is turned on?

 

[chrisggg]

Hi, many thanks for the reply.

It seems simple but infact is quite complex, I think of it as a load that requires a certain amount of current. I think of this load being pulsed where the time it is connected to the rc circuit is the ton time. 50us is just a figure I plucked out the air, I am searching for some sort of expression that can allow the capacitor voltage to be calculated taking into consideration the current being drawn from the RC circuit from this pulsed load.

I know the general expression for an rc circuit is straight forward but I am finding it difficult to express in equation form how the capacitor voltage is affected by this switching load.

 

[gneill]

Well, I think you'll have to specify your conditions a bit more precisely. Is the capacitor allowed to fully charge from the voltage supply before the first time that the load is applied? How long is the interval between pulses ("recovery" time for the capacitor voltage)?

 

[chrisggg]

Well what I am trying to figure out is a general expression that can be applied therefore the conditions should not be of huge importance but for example say the load is pulsing with a period of 1ms and the on time 5us. And for simplicity sake let's say the capacitor is allowed to fully charge to the 3.6V before the load starts to pulse.

 

[chrisggg]

That would be enough time for the capacitor to recover.

 

[chrisggg]

Ton( time the load is on ) as 5us by the way, not 50us.

 

[gneill]

I see. The long term behavior of the circuit will depend upon whether or not the capacitor voltage can fully recover between load pulses. If it can, then you need only analyze one cycle to characterize the behavior. If it doesn't, more work will be required to find the "steady state" max and min capacitor voltage.

With the component values shown the time constant for the RC circuit is 10μs, which for a 1ms time between pules, allows ample time for recovery. [STRIKE]You can then adequately analyze the circuit by finding the capacitor voltage "a long time" after the 50mA load is applied.[/STRIKE]

EDIT: For a 5μs on-time, you will have to determine a function for the capacitor discharging over that time, then the recover function. Find the discharge curve as though the load was permanent, but cut it off after 5μs, then switch to the charging function.

 

[chrisggg]

Hi, thanks for your reply.

yep, I am more than aware of how to do it all pratically, I can even simulate it and find my result this way but I am particularly hoping to understand how to express this effect on vc through an equation of some sort. I am under the impression it becomes very comples and needs the use of laplace a couple different circuit theorems to arrive at a final conlusion.

 

[chrisggg]

Also, I am not so worried about the long term effect as I am aware generally the capacitor voltage will remain as 3.6V as the recovery time is long compared to the time where the load is applied, I just want a way of working out the effect the applied load will have on vc, how much vc is likely to 'droop' and from this kind of expression i could then calculate the actual load current :)

 

[gneill]

Also, I am not so worried about the long term effect as I am aware generally the capacitor voltage will remain as 3.6V as the recovery time is long compared to the time where the load is applied, I just want a way of working out the effect the applied load will have on vc, how much vc is likely to 'droop' and from this kind of expression i could then calculate the actual load current :)

Rather than a current load, can the load be expressed as a resistance? That would allow the load current to vary according to the available voltage.

 

[chrisggg]

Yes, I believe the solution would be to use thevenins equivalent?

 

[gneill]

Yes, I believe the solution would be to use thevenins equivalent?

You could use Thevenin, but the circuit is simple enough that you will probably not require it.

In the figure, V1 is 3.6V, R1 = 10Ω, C1 = 1μF, RL = 72Ω (Giving your 50mA current when Vc is 3.6V).

When the switch closes at t=0 the time constant of the circuit changes (what is it?) and the capacitor voltage "heads" for the new target determined by the voltage divider R1:RL. It follows this curve until t=5μs when the switch opens. The voltage will then head back towards V1 (3.6V) with the time constant determined by R1 and C1.