Charging 35V 470uf with pic 12F683 board.

[LukeP]

This project derived from Bobledoux project.
I want to charge a 470uF capacitor to 35V, then discharge it into a solenoid. The two actions will be separated by PIC control.
I'm using this board by now: 9V battery, .14 Period and .3 Duty give me 14mA draining during recharge, 9mA when fully charged, about 14seconds to full charge.
Battery life and dimension is the problem: it's too short the first one, too big to fit project.
I tested the board discharging 80times during 2hours, first load battery is 9,20V, after the test is at 8.60V, doesn't sound great to me, i need at least 3000 cycles with one battery.

I made some calc:
Energy in cap = 0.5 x C x V^2 = 0.5 x 4.7x 10^-4 x 35^2 = 0.2879 Joule
solenoid request is 130W x 0,002seconds discharge = 260mW assuming 60% efficency = 433mW
If powered by 3V battery, 20mA x 3V = 60mW ... 433/60 = 7 seconds to recharge, near my goal: 24mA = 6sec recharge time.

As for battery i was thinking about cr2450, but datasheet refer "continuous standard load" as low to 0,20mA, others indicate as not too much life reduction with 10mA, both value are far away my calc (20-30mA).

By now i tested the second board attached, lot of problems: mainly i must double VRCON value to obtain the same level, and i don't know why; plus the battery pack (2xAA) is too big to fit inside real project but i cannot use coin cell because of too high current draw.
Wich is the right way to make the board functioning with 2xAAAA, at least 3000click?
Thanks,
best regards,
LukeP.

 

[ronsimpson]

I do not like the idea of 9V to 4V to 35V. You loose almost 1/2 your power in the 9V to 5V. If you use this keep the PIC on the 5V and move the boost up circuit to +9V.

To keep the current low, use a large cap at the input of the boost up circuit, keep the maximum duty cycle low. It will take some time to charge 470uF up to 35 volts with low current. Start out with a very low duty cycle and you will see the current is low.

 

[LukeP]

Ok, just connected 9V to L1, recharge very fast a pair of seconds or so.
.42 Period and .3 Duty give me 12mA draining during recharge, 8,5mA when fully charged, about 7 seconds to full charge..
.2 Duty .32 Period recharge time 10seconds, 7-9mA.
.1 Duty .5 Period, time 10sec, 11-9mA
So 9V battery it's connected best way possible i think.

My goal is 6seconds, with a much smaller battery. like coin cell 3V or AA 1,5V.

 

[ronsimpson]

You said "8,5mA when fully charged". Why is there that much current when fully charged. Do you turn off Q1 or keep it on but at a very small duty cycle? The LM79L05 will use some current. That will go away with a coin cell. Is the 12F683 running at full speed? It could run much slower when nothing is happening. R6, R8 has voltage across them for a short time, but they could be 470k. That will save a small amount of power. Some power is lost in R9.

 

[LukeP]

It's for maintaining that level of charge. When nothing is recharging, ex when switch is held down there is the discharge into che solenoid, but until you don't relase it doesn't start recharging. I measured less than 5ma during this time: i assume it's the draining of 12F683, LM78L05 and resistor. Led is off during this time.

 

[colin55]

Use 6v battery and diode to PIC chip.
Use sleep and the circuit will shut down to a few microamps.

 

[colin55]

D1 is not needed as pin 6 will never go above 4v.

4 other parts can be removed.

 

[LukeP]

6V PX28L https://www.electro-tech-online.com/custompdfs/2012/03/0900766b801115c3.pdf

Solenoid need 130W 0,002sec; assuming 60% efficency = 433mW.
mW from the battery = 433mw/ 6s (recharge time) = 72mW = 5,2V (battery draining curve) x 14mA. From schematic that battery suffer from anything upper than 5,2mA.

This is only for capacitor charge, excluding board draining.

The chip read every 30 microseconds the imput from S2 switch, every minute or so, i can put the PIC in sleep mode, but i must introduce another switch on Pin3 to determine when to go on sleep mode and when to return fully On because it's the user that decide when i need charge and when not.
To be easy i left the pic always checking and charging.

 

[colin55]

Does the 470u or 1,000u activate the solenoid successfully?

 

[LukeP]

Yes of course, 470uF at 35V = 288mJ energy
It's a the best choice between Capacitor dimension and Volt.
Function also with 25V 1000uf, or 16V 2200uF, but those capacity require a bigger capacitor and i have not too much space in the overall board dimension, about 30x40mm max.
Consider that with 35V i'll use 50V rated capacitor, so with 16V a 25V rated,

 

[colin55]

How long does it take to charge the capacitor and how often do you want to discharge the capacitor?

 

[LukeP]

Desidered time to charge capacitor is 6seconds, one discharge every 60s or so.
I cannot extend the capacitor charge time from 6sec to fully 60sec because it's the user that decide exactly when discharge: and can accept only 6seconds delay, not a fixed 60seconds delay.
I can introduce a supplementary switch to pause any recharge during the minute or so che user decide not to discharge: when user is ready, he close supplementary switch, and after 6sec he can discharge at any moment.

 

[colin55]

Have you tried the circuit or is all this just an "airy fairy" montage of the mind.

 

[LukeP]

No i have severals board to test.
I have the 9V one i'm talking about in this thread and follow the schematic i uploaded: all mA value are tested on that board.
I'm searching solution to reduce battery dimension (2xAAA maximum) and to extend battery life.

 

[colin55]

You can't do much with 3v.

Try 4.5v See if circuit will charge 470u or 1,000u in 6 seconds.

Use digital input on "sense line" and change lower value to 39k.

What is the DC resistance of the 330uH

 

[LukeP]

I have not enough space for 4,5V, only 3V by 2xAAA or 2xAAAA.
The inductor DC resistance is 860 mOhm.

 

[colin55]

If the inductor is less than 1 ohm it is putting too much strain on the batteries.

You need to wind your own with at least 10 ohms and gradually decrease the number of turns until you get the capacitor charged in 6-7 seconds.

 

[ronsimpson]

Using a inductor with more resistance at first seems like a way of limiting current...but...I believe this results in more loss of power.
Using a low loss inductor will make a more efficient power supply. To limit current use a smaller duty cycle! You can limit the duty cycle to 1% or 10% maximum and the current will be low.

Most coin cells have high internal resistance and you can't pull much current from them. You may have to get different batteries if you want to charge in 6 seconds.

 

[colin55]

Using a inductor with more resistance at first seems like a way of limiting current...but...I believe this results in more loss of power.

How?

Using a low loss inductor

Oh?

To limit current use a smaller duty cycle

Oh?

This is all new to me. Tell me more.

Most coin cells

We are clearly talking about alkaline cells. Are you on a different tram?

 

[ronsimpson]

Current builds up in the inductor when the MOSFET is on. When the MOSFET is off the current is transferred to the load. The amount of power is (time and voltage). If you turn on the MOSFET for 1uS and do that every 1mS you get X amount of power. If you turn on for 2uS every 1mS you will charge at 2X. Or if you turn on for (1uS twice) in 1mS you will get 2X the power.

Just adding a resistor burns power. It is like driving your car with the breaks on. It is better to get off the gas.

We are clearly talking about alkaline cells. Are you on a different tram?

From post #1. "As for battery i was thinking about cr2450"