Capacitor charging using low voltage supply

[hubble]

Ok, I understood it. I have one more situation. How do I connect this capacitor to the LTC3639 buck converter. Does it need to be directly connected to Vin? Also, how to select components to get the desired output from this converter apart from 3.3 V, 5 V?

Thanks.

 

[Hero999]

You didn't say anything about a buck converter before.

How much current do you need to draw?

What's the point of increasing the voltage to 50V then reducing it to 5V?

According to the datasheet, the LTC3639 can work up to 150V so as long as the capacitor is suitably rating you don't need to limit it to 50V.

The data sheet shows you how to calculate component vales for a given output voltage,
http://cds.linear.com/docs/en/datasheet/3639fd.pdf

 

[hubble]

Earlier, I planned to use super capacitor of 2- 5 F 2.7 V. However, these capacitors are very expensive here, if available at all. Instead, high voltage mF caps are readily available. It is because of this reason the generator voltage is needed to be raised to high voltage of around 50 V (safe limit for a 63 V cap) to store considerable amount of energy and then the cap voltage is brought down to sub 5 V level using a high efficiency buck converter. This cap is intended to supply 20 mA max current.

Thanks.

 

[Hero999]

How long does it need to last for?

I hope you're aware that you will not be able to charge the capacitor and draw 5V at 20mA at the same time.

Before you said the transformer charges the capacitor to 90V when nothing is connected to it, then it drops to around 50V when a 22k resistor is connected to it. Using  Thévenin's theorem the internal resistance of the transformer plus the stepper motor can be calculated as 17.6k. Maximum power transfer occurs when the load resistance equals the source which is when the output voltage is half the open circuit voltage so it can be calculated to be 452/17.6k = 115mW.

The datasheet for the says the LTC3639 says it's 85% efficient so the power required to give 5*0.02 = 100mW out is 0.1*0.85 = 117.6mW.

This means you can charge the capacitor as long as the buck regulator is very lightly loaded.

 

[hubble]

How long does it need to last for?

I would have wanted it to last for as long as it can. But for this prototype, the requirement is under 120 sec.

I hope you're aware that you will not be able to charge the capacitor and draw 5V at 20mA at the same time.

Yes, I am. Once the capacitor is fully charged then only the load is going to be activated.

Shall I directly connect the capacitor to the input of buck regulator?

Thanks.

 

[Hero999]

Why not use a 100V capacitor, then you can utilise the power more efficiently and don't need any over-voltage protection?

 

[hubble]

A 100 V capacitor.......sounds good. But then, it should be available here in the milifarad capacity range. I shall try to find it. If I am able to get this, shall I need to connect this capacitor directly to the input pin of the converter? In the datasheet, there is a small 100 V cap across the input. In my case, there is already going to be a big capacitor. Do I need add something more?

Thanks.

 

[Hero999]

It's the energy stored, rather than the capacitance which is important.

E = 0.5CV2

Doubling the voltage quadruples the energy storage, given the same capacitance. A 5mF capacitor charged to 100V will be storing twice the energy as a 10mF capacitor charged to 50V and it's likely it will be physically twice the size.

Using higher voltages also means when the voltage drops too low for the buck converter, more energy will be extracted from the capacitor so you're able to use the storage more effectively.

It's worth keeping the capacitor shown on the datasheet, even if you're connecting a large capacitor across it because smaller capacitors have superior high frequency characteristics.

 

[hubble]

I finished putting together all the components. One component I am not sure about is the inductor. The datasheet mentions to use ferrite pot-cored type which is not available here. Instead, I managed to get the one which looks like as it is shown in the extreme left in this image:http://www.p-wholesale.com/upimg/5/119a1/leaded-power-inductors-698.jpg . Nevertheless; to test the circuit, I connected at the o/p one 3.5 V LED with 100 ohm resistor. The LED lit up only for a few (may be 10 secs) seconds. This was after charging the 10 mF capacitor to 51 V. Noticed the current going way above (twice or thrice) the calculated 15 mA current. I really don't know what is happening here. How do I find the issues with this circuit. Also noted that when there is nothing connected at the o/p, the o/p shows the same voltage as that of the charged capacitor. Is this normal?   

Thanks.

 

[Hero999]

That inductor doesn't seem to be what you want.

Can you post a schematic of the whole circuit.

 

[hubble]

Yes, the inductor is not exactly the same. But, does it have so much affect on the circuit? What other type of inductor can be suitable for this type of application? I have followed datasheet to implement the schematic. I have posted the schematic.

Thanks.

PS: Can it be because of inductor saturation? If yes, can I use high value inductor?

View attachment 42541

 

[Hero999]

How is the circuit physically constructed? Is it small and compact or are there lots of long leads and wires?

Did you read pages 10 & 11 on the data sheet on inductor selection?
http://cds.linear.com/docs/en/datasheet/3639fd.pdf

 

[hubble]

The circuit is designed on a perforated board. However, it is compact in component placement. LTC3639 is very small, so very thin magnet wires (enameled copper wire with almost zero resistance) needed to be connected to its leads in order to be able to have them soldered to other components.

And, yes I have already read the datasheet for inductor selection. This is where it is mentioned to use torroid or shielded pot core in ferrite type inductor. I am having one torroid type inductor but I don't know its inductance value nor its core type.

PS: Looked at the circuit connections again and found a few stray metal pieces stuck between unconnected points. Removed them and now I notice a completely different behavior. When the cap voltage is 50 V and discharged through the LED and 82 ohm resistor, the current is limited to around 18 mA which is what I am looking for until the voltage drops to around 34 - 35 V when the current shoots up to 150 mA and stays that way till the cap is completely discharged. When it is again charged to around 30 - 31 V, it exhibits the same behavior when voltage reaches to 26 - 27 V. The similar behavior can be seen at different voltage levels in a different charge-discharge cycle. What is this strange behavior?

Thanks.

 

[hubble]

Someone around to help me............please?

 

[Hero999]

It seems like your circuit is working.

The current drawn from the capacitor by the switched mode should increase as the voltage drops. This is because the regulator tries to keep its output voltage irrespective of the input.

 

[hubble]

Here are some more observations:

Condition: Capacitor is charged to about 30 V.

Case 1: Nothing is connected at the output. Almost same voltage as that of the capacitor.

Case 2: Only resistor at the output. I see only 1.5 V at the output with the current remaining unchanged at around 18 mA. When the voltage drops below 26 - 27 V, the output current rises to more than 150 mA and the voltage rises to more than 15 V.

Case 3: Resistor with LED at the output. Around 18 mA at 5 V. Similar behavior when voltage drops below 26 - 27 V.

Even when the capacitor is charged to, say, 20 V and allowed to discharge, the output current is over 150 mA. So, it has just started discharging but the current is already over the maximum value.

 

[Hero999]

1) That doesn't sound right but some regulators do need a minimum load current to regulate properly. What does it say on the datasheet?

2) What resistor value are you using? That sounds like the expected short circuit current. Is the meter in series with the resistor or just connected across the meter?

3) That sounds right.

 

[hubble]

Thanks all for the guidance; especially you, Hero999. I replaced the converter ic with the new one and now everything is fine. I think it had something to do with voltage feedback pin of the regulator. Nevertheless, it is working fine now with slightly less efficiency due to not so efficient inductor.

Thanks.

 

[Hero999]

It works, congratulations!