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Approach needed: Charge/Discharge circuit for big capacitor

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Odysseas

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Hello everybody,

I am designing a circuit used for safe charging and discharging of a large capacitor (1F, 12V => automotive). An LED should indicate whether the cap is charged or discharged, i.e. I want it to stay on when the cap is being charged/discharged, and turn off once the cap is fully charged/discharged.

I've attached a basic idea, but I have no idea if it will work like that, this one is only for discharging. The idea is, when there is less than 0.7V applied to the +Cap terminal, the NPN should turn off, otherwise, allow current to flow through the LED (please excuse the missing pre-resistor :) ).

Please let me know your thoughts on this one.

Thanks!!
 

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The circuit as drawn will not work.
You will need about 2.8 to 3 volts at the base of the transistor before it conducts, and when it does the transistoe and LED will be destroyed as there is no current limiting resistor.
The 1N4001 will clamp the base voltage at 0.6 - 0.7 volts, this will stop this circuit from doing anything at all.

Move the LED to the collector of the transistor, and add a 1K resistor in series with the LED.

Remove the 1N4001, it is not necessary.

JimB
 
Okay, I've reworked the schematic. But as I understand it, the transistor will turn off once the base voltage drops below 3V approx, right? Because I'd like it to stay on for as long as possible, would using two transistors in a row make sense here?
 

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Odysseas said:
Okay, I've reworked the schematic.
Your new schematic is OK.
Odysseas said:
But as I understand it, the transistor will turn off once the base voltage drops below 3V approx, right?
Wrong.
The circuit as it is now drawn, the transistor will turn on when the capacitor voltage is greater than 0.6-0.7 volts.

JimB
 
Ok, thanks a lot! Will it allow the LED to emit a good, visible amount of light when the transistor is at about 0.7 volts?
 
The light output of the LED will reduce as the voltage drops below about 1v and will be zero at about 0.6 volt.

I suggest that you build the circuit and test it.

JimB
 
Nigel Goodwin said:
Non-English posts deleted!.

Please note, only English is allowed here.

Moderator.

Sorry, forgot about that.
 
Discharging a Cap

Hi Oddyseas,

OK, let me try to polish my English a bit. I can understand that posts in German are being deleted. 80% of Germans speak English, but only a hand full of English and Americans speak German.

Well, the circuit looks alright so far. However the whole thing is an estimating circuit. If the base voltage goes below 0.7V the LED will fade and finally extinguish. At that time the cap is not fully discharged.

**broken link removed**

A more elegant method is using an OpAmp wired as a comparator. You can select any voltage just above zero using an unsymmetric power supply. As long as the threshold is above the set value the LED will be lit and extinguish when the voltage of the cap is almost at zero.
 
Discharging a Cap

This circuit has a trip point of 24mV (12/1000*2).

The charging voltage is selected 200mV (any voltage in excess would be of an endlessly long duration) Note that the 100Ohm resistor has to be a 2W type since it has to dissipate 1.44W (12/100*12). A 1W resistor would do it because the max current flows for a short time only decreasing rapidly.

**broken link removed**
 
Discharging a Cap

**broken link removed**


Finally the reference voltage is higher than the input voltage and the comparator changed its output from high to low. Note R6(470K). It prevents both LEDs to light up simultaneously with the positive feedback to the non-inverting input when the voltages are almost equal. The simulation was paused at a capacitor voltage of 22.3mV. Actually the change of the output was at a higher voltage (about the preset value).

As a cheap extra you might add the green LED as well. It indicates a fully discharged cap.

Do not connect both LEDs in series. You must use separate current limiting resistors like in the schematic.

How is it going in Kaufbeuren? Does Jantsch electronic (Neugablonz) still exist? I was there a few years as a navigator instructor.

Regards

Hans
 
Boncuk said:
80% of Germans speak English, but only a hand full of English and Americans speak German.
What is even worse is that only 80% of English people can speak and write English as well as you and Odysseas. (Just look at some of the posts on this board!).

JimB
 
Boncuk said:
OK, let me try to polish my English a bit. I can understand that posts in German are being deleted. 80% of Germans speak English, but only a hand full of English and Americans speak German.

My daughters boyfriend speaks German - but then he has a German mother! :D
 
I got another question, If you can`t manualy press/connect the "discharge" switch, you only have access to switch on/off the power supply, Vcc. What is the easiest way to discharge the cap then? som kind of electro swich?

**broken link removed**
 
jdahl said:
I got another question, If you can`t manualy press/connect the "discharge" switch, you only have access to switch on/off the power supply, Vcc. What is the easiest way to discharge the cap then? som kind of electro swich?

Don't like to mention this, but there's no 'charging' circuit in that diagram, it's just banged directly across the battery - so it will charge up pretty instantly with a huge current surge.

What's the reasoning behind wanting an LED anyway?.
 
@Boncuk: Oh, jeez :) Amazing how small earth is (wie klein die Welt doch ist *g*) Yes, Jantsch still exists, and in fact, I intend to buy all the parts for this circuit there :) You want me to give anybody there your regards?

Thanks a lot for the circuit you elaborated, I really appreciate! Just one question regarding the circuit, I think I should charge the cap from the 12 V supply (why 0.2 volts?) through a current-limiting resistor. If so, I could use a simple alternating switch to select charging or discharging operation, is that correct?
 
JimB said:
What is even worse is that only 80% of English people can speak and write English as well as you and Odysseas. (Just look at some of the posts on this board!).

JimB

Hi JimB,

thanks for the compliments. The problem concerned with the English language is that most people write the same way they talk. That applies to the German language only (which is correct) - as far as I know. You are right talking of some posts. They really make me scratch my head for a while thinking what the author is getting at, especially being confronted words like: "grzmmlhyly". Can you translate that for me? :D
 
Discharging a Cap

Hi Odysseas,

hopefully I got your login name correct this time.

No! No! Just use the 12V charging voltage as you intended to do. May be you overlooked that part in my describtion. I just didn't want to sit around and wait for the cap to discharge all the way from 12 to 0. (It takes about 10 minutes). I selected 200mV to illustrate the interesting part of the circuit to shorten the process. This simulation is a real time simulation and looking at the time bar at the bottom of the picture you can see how much time it took from 200mV to the trip point.

If there is still somebody of my old friends around I guess he could be "Puemmer" Bergmann. Sorry, forgot his real first name. Do they still have that excellent service at Jantsch: "If you want to buy a part, you must know the part number! I don't know of any functions." :confused:

The service in my shop was slightly better. I helped a student with his schematic, while his pal stole the soldering irons hanging at the wall. :mad:

Regards

Hans-Juergen

P.S. Schrech mer ueber Obschd, ned obschd Obschd magschd, obschd mi magschd?

P.P.S. Hi Nigel, please don't delete that post for German language! :D
 
Discharging a Cap

Hi Odysseas,

here I am again.

This is the wiring you should use selecting one switch for charging and discharging. I omitted a current limiting charge resistor. Select one according to the power supply's capacity. A high charge current will just flow during the initial charging phase. For R6 you better use 1MOhm. I observed a slight fading of the red and simultaneously slightly increasing brightness of the green LED. This measure will influence the hysteresis slightly but LEDs will jerk from red to green.

As you can see, the simulation with the cap fully charged to 12V took exactly 10minutes, 47 and 6/10 seconds.

A useful application for the circuit could be an alarm circuit to activate after a time period (to leave the house in time not triggering false alarm) which you might select calculating another trip point (voltage divider at the inverting input).

This is the final circuit using one switch.

**broken link removed**

Again kind regards

Hans-Juergen
 
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