[lacrossanator1]

So, here my problem: I'm 16 and have self taught myself a he'll of a lot about electronics and so on, I have been a main attribution to the building of a Cnc wood lathe that has 2 cutting heads which cut simultaneously and I have also wired a switchboard consisting of 2 transformers, 4 contactors, 2 pneumatic solenoids, 2 three phase motors, a VFD, a user interface of 14 buttons/switches, 28 relays, multiple limit switches, and a sophisticated emergency system! So everyone tells me how they would love to employ me! Enough about me and now to my latest little project which is a simple taper charger with current limiting using lm317 regulators the problem is that I don't know much about transistors and apparently I can somehow incorporate an led with a transistor and resistor that will light up once the charge cycle is finished! That is when the input current to the charger drops below about 20mA! Anybody like to help me? Preferably not tell me how to but teach me how to?

[Al_The_Man]

It would be ideal to know exactly what circuit you have right now, in order to incorporate the LED.
BTW, here is a nice little Gellcell charger if charging low energy batteries this may be usefull.
12 Volt Gel Cell Charger
http://www.rason.org/Projects/projects.html
Al.

[SteveWill]

Where to start. Transistor that covers a lot of area. I will hit some high points.

I assume that you know what this means and can use it: Volts = Amp times Ohm. What a diode is and can use it.

Let start with the old PNP and NPN transistor and that is as far as I am going in the area of transistor.

1: PNP = Arrow Points N (in). That is our Diode. I will not explain the PNP but it is the same but the opposite.
NPN = Arrow Not Pointing N (in).

2: There are 3 leads. Named B= Base, C=Collector, and E= Emitter.

3: Think of the Arrow in the transistor symbol as an Diode. Put an line at the point of the arrow so it looks like a diode.

4: When a Diode is forward bias (Turn On) it will read a voltage of 0.2 to 1 Volt (V), most will say 0.7 and that is what I will use and also for the diode in the transistor.

5: I will talk about Voltage when talking about the Base Emitter junction, and Current (I) when talking Collector to Emitter.
We get a voltage drop from Base to Emitter and we have current flowing in Emitter to Collector.
OK let see if we can make the NPN transistor work.

1: Circuit We will input (In) to a 100K resistor to B, Tie E to Ground (G) and C to a 1K to +5V.
Draw it out and label.
0 to 0.6 V at In: Measure; 0 to 0.6V at B, 0V at E (it will always be 0V in this example, last time for it), C at 5V. ) 0I C to E.
The base to emitter is said to be off and the transistor is said to be off.
0.7V at In: With E tie to ground the transistor is switching and It is anybody guess what the V and I are.
0.8 to 5V at In: Measure; 0.7V at B, C at 0.2V ( less then 1V). The base to emitter is said to be on and the transistor is said to be on.
Remember we said 0.7V is the B to E voltage, if we apply 1V directly to the B it will blow up, just like a diode will. There is just enough Current flowing in the 100k resistor to drop the voltage to the B so it does not blow up. Remember Volts = Amp times Ohm. I use it a lot with transistor. Also the current is limited by the 5K resistor so transistor does not burn up.
Now if you place a LED and a resistor sized to the LED current in the Collector side circuit you could turn if on and off.



2: Circuit We will input (In) to a 100K to B, Tie E to a 100ohm resistor to G and C to a 1K to +5V.
Draw it out and label. This Circuit is said to have a gain of 10, 1K /100 Ohm=10
0 to 0.6 V at In: Measure; 0 to 0.6V at B, 0V at E , C at 5V. The base to emitter is off and the transistor is off. No current in E to C
Now this is the part you can use.
Remember we said this circuit has a gain of 10 so the useful input voltage is a 0 to 0.5voltage change since the supply voltage is 5V. 0.5V time 10 = 5V. Called hitting a voltage rail. Or from 0.7 to 1.3 input voltage.
So 0.8 In Volt at the B give us .1 volt at E (0.8-0.7) across the 100 ohm resistor and that is how much current? 0.001. That current is the current that is also flowing in the EC part of the transistor and it will drop how much volt across the 1K resistor, now minus that from the 5 V supply and you get 4V on the C. See you have just got a gain of 10. 0.1 volt change in, 1V change out.
Now try it for 0.9 in. You should get 3 volt out. If I have explain it correctly.
Remember at the start I said the useful range was 0.7 to 1.3 let see if that is true. We saw that .7 would give use 0 v across the 100 ohm so we would have 5 V at the collector now lets try 1.3v in we get 0.5V across the 100 ohm that would give us 0.005 Amp. Now time it against 1K = 5V. Now that from supply 5V =0V but wait a minutes, I can not have 0.5V on the 100 and 5V on the 1K that would be 5.5V and that is what is meant running into the rail. So the max input useable voltage is between 1.2 and 1.3.
If you run your current thru a resistor to give you a voltage and now with the 2 circuits, you should be able to work something out.

Hope this helps.

[wildwestpat]

Before you get too deep into this project you need the characteristics of the battery. There are several types and each has its own demands for charging. For example

Lead acid chemistry is basically 2 to 2.2 volts per cell has a defined safe trickle charge that can be left in that state for a long time. The maximum charge current is defined by the cell physical size and plate design. These cells have a temperature dependent voltage variation which needs to be factored into the charge. These cells usually resent being left in the discharged state and can be difficult to recover from a deep discharge unless of special construction. This chemistry reacts well to a tapered charge starting with a maximum safe charging current with the current being reduced in stages as the cell voltage (also temperature rises) and this is often referred to as tapered charging.

NiCd chemistry is around 1.2 volts per cell - these cells resent being left discharged and also resent being cycled between partial discharge and full charge. This is the oft quoted memory effect of NiCad. Again these cells can be left on trickle charge for long periods. Again the manufacturer will have maximum charge current and trickle current specifications. There are several types of physical construction and this means that rapid charging needs to take this into account.

Lithium rechargeable cells have much tighter requirements on charging current control and temperature during charging. Many of these cells have temperature sensors built in and these are used to control the current. The voltage of these cells is around 3 volts per cell. They will not recover from a deep discharge and many multi cell packs contain electronics designed to cut off the current to the load when the voltage falls to the minimum safe level on load. Examples of this type are the lithium cells used for portable power tools. These cells store charge well compared to lead acid or NiCd and strorage time of 12 months would typically give only 10% power loss and no trickle charge. Due to the problems of overcharging resulting in fire or explosion of lithium chemistry cells I am going to assume these are off limits for your project.

Now for your LM317 you need to determine what sort of charger your project is to be. Is it a rapid charger or one of the sort that takes hour if not days to fully charge your battery. Yes the aim can be achieved but please post the requirement - maximum cell charging current - type of chemistry used in the cell - the power source (transformer secondary voltage with full or half wave rectification) or is it a DC source as in another battery? Is your design for minimal number of components?

There several good basic tutorial paper on the web and this one is as good as any: Light Emitting Diode (LED) Tutorial

Good luck - post more details if you need help - regards - Pat

[lacrossanator1]

Thanks everyone for your willingness to help!
The charger is a taper charger with current limiting! The author of the circuit designed it for charging rc lithium polymer batteries but I intend to use it for charging sealed lead acid batteries of 12v and 7Ah. As regard to my charge current, I was thinking of capping it at 500 mA. My powersupply will be a Meanwell switchmode 24V@6A which I scraped from and old machine! Below is the website from which I found the circuit diagram! Although he says it is for lipo batteries, I cannot see why I couldn't use it for SLA charging since it is a taper charger anyway!
Cheap Lithium Charger for Electronics Tinkerers - RC Groups
Once again thanks alot, Thomas

[lacrossanator1]

Forgot to add that the transformer outputs DC

[Al_The_Man]

Forgot to add that the transformer outputs DC

(Most) transformer do not out put DC unless the rotary kind!
At least without added components.
You might be better off to look at the link I mentioned as this is for lower energy gell cells etc.
The link did not show the schematic?
If you look at the Maxim site for the IC it shows a few applications.
Al.

[lacrossanator1]

here is the schematic!
Lithium Battery Charger Plans
This is basically what I have been using to charge lithium cells up to 1250mAH. It's simple, cheap and the components are all common as. I'm putting it here for others to view for ideas on building their own and take no liability for any inaccuracy or mistakes.

You only need basic electronics skills to make it but it must be checked before use with a good quality multimeter that will allow you to set the output voltage accurately and test the current limiting as well.

One of these on their own is ok but I make a bank of 4 to charge 4 flightpacks at a time. I built one with a switch to select the number of cells but after plugging in a 2 cell pack with the switch set to 3 cells I decided it was best to build dedicated charger that do one size flightpack.


Lithium Charger parts list:

2 x LM317 adj Voltage Regulator IC
2 x Heatsinks for the above
5W Wirewound resistor (1.2 ohm resistor for 1.04A current)
(1 ohm for 1.25A)(1.8 ohm for 700mA)(5 ohm for 250mA)

120 ohm resister 1/4 Watt MF 1%
Voltage regulating resistor 1/4 Watt MF 1 %

Veriboard on which to mount components and other items like cable, vented case, and capacitors if you fell they a necessary.

current regulating resistor ( do not exceed a current greater than 1C):
R = 1.25/Current in amps of current regulator.
e.g. for a 700mAH Flightpack
1.25/0.7A = 1.78 ohm

voltage regulating resistor:
R = (96 x Vout) - 120
eg. for a 2 Cell pack being charged to 8.4 volts
(96 x 8.4v) - 120 = 686 ohms
e.g. for a 3 Cell pack being charged to 12.6 volts
(96 x 12.6v) - 120 = 1090 ohms

With voltage regulating resistors you may have to use 2 to get the value you want.

After you have built it check that the output voltage is no greater than 4.2 volts per cell. Then set your multimeter up to read amps and short out the outputs with the probes to make sure the current is correct.

If you put a 5 Ampmeter on the input side you check the state of charge of packs and watch the amps go down as charging is completed ( For the first hour of charging a flat pack the reading
should be 1CAmp i.e. 0.7amp for a 700mAH pack )

As I have said above do not attempt to build unless you know what you are doing or can be helped by someone that does. Shop around for components if you can as big retailers here charge double the price of elsewhere.

[wildwestpat]

Here is the voltage state of charge graph for lead acid thanks to WiKi. Your circuit will work provided you adjust the run out voltage correctly and set the current for one tenth or less of capacity. Check the ICs for heatsink capacity to dissipate the heat generated. I think a simple LED with current limiting resistor can be added between battery positive and the orange link to indicate charge has entered voltage control but this needs checking out as you might find the action is too soft to be meaning full.

[harryn]

Pb acid batteries are pretty forgiving, but Li and Li polymer are not. Charge balancing each cell is critical, as it absolutely managing the max and min voltage and current.

Just so you know, serious RC guys that run Li cells charge them in a bucket or a special charge bag.

Good luck

[lacrossanator1]

Thanks for that advice, I've seen exploding lipo batteries on YouTube, I see how they can be dangerous! Fortunately for this project, I'm not playing with those...

[lacrossanator1]

Where to start. Transistor that covers a lot of area. I will hit some high points.

I assume that you know what this means and can use it: Volts = Amp times Ohm. What a diode is and can use it.

Let start with the old PNP and NPN transistor and that is as far as I am going in the area of transistor.

1: PNP = Arrow Points N (in). That is our Diode. I will not explain the PNP but it is the same but the opposite.
NPN = Arrow Not Pointing N (in).

2: There are 3 leads. Named B= Base, C=Collector, and E= Emitter.

3: Think of the Arrow in the transistor symbol as an Diode. Put an line at the point of the arrow so it looks like a diode.

4: When a Diode is forward bias (Turn On) it will read a voltage of 0.2 to 1 Volt (V), most will say 0.7 and that is what I will use and also for the diode in the transistor.

5: I will talk about Voltage when talking about the Base Emitter junction, and Current (I) when talking Collector to Emitter.
We get a voltage drop from Base to Emitter and we have current flowing in Emitter to Collector.
OK let see if we can make the NPN transistor work.

1: Circuit We will input (In) to a 100K resistor to B, Tie E to Ground (G) and C to a 1K to +5V.
Draw it out and label.
0 to 0.6 V at In: Measure; 0 to 0.6V at B, 0V at E (it will always be 0V in this example, last time for it), C at 5V. ) 0I C to E.
The base to emitter is said to be off and the transistor is said to be off.
0.7V at In: With E tie to ground the transistor is switching and It is anybody guess what the V and I are.
0.8 to 5V at In: Measure; 0.7V at B, C at 0.2V ( less then 1V). The base to emitter is said to be on and the transistor is said to be on.
Remember we said 0.7V is the B to E voltage, if we apply 1V directly to the B it will blow up, just like a diode will. There is just enough Current flowing in the 100k resistor to drop the voltage to the B so it does not blow up. Remember Volts = Amp times Ohm. I use it a lot with transistor. Also the current is limited by the 5K resistor so transistor does not burn up.
Now if you place a LED and a resistor sized to the LED current in the Collector side circuit you could turn if on and off.



2: Circuit We will input (In) to a 100K to B, Tie E to a 100ohm resistor to G and C to a 1K to +5V.
Draw it out and label. This Circuit is said to have a gain of 10, 1K /100 Ohm=10
0 to 0.6 V at In: Measure; 0 to 0.6V at B, 0V at E , C at 5V. The base to emitter is off and the transistor is off. No current in E to C
Now this is the part you can use.
Remember we said this circuit has a gain of 10 so the useful input voltage is a 0 to 0.5voltage change since the supply voltage is 5V. 0.5V time 10 = 5V. Called hitting a voltage rail. Or from 0.7 to 1.3 input voltage.
So 0.8 In Volt at the B give us .1 volt at E (0.8-0.7) across the 100 ohm resistor and that is how much current? 0.001. That current is the current that is also flowing in the EC part of the transistor and it will drop how much volt across the 1K resistor, now minus that from the 5 V supply and you get 4V on the C. See you have just got a gain of 10. 0.1 volt change in, 1V change out.
Now try it for 0.9 in. You should get 3 volt out. If I have explain it correctly.
Remember at the start I said the useful range was 0.7 to 1.3 let see if that is true. We saw that .7 would give use 0 v across the 100 ohm so we would have 5 V at the collector now lets try 1.3v in we get 0.5V across the 100 ohm that would give us 0.005 Amp. Now time it against 1K = 5V. Now that from supply 5V =0V but wait a minutes, I can not have 0.5V on the 100 and 5V on the 1K that would be 5.5V and that is what is meant running into the rail. So the max input useable voltage is between 1.2 and 1.3.
If you run your current thru a resistor to give you a voltage and now with the 2 circuits, you should be able to work something out.

Hope this helps.

Thanks a lot for this help; I am starting to gather an understanding as to how I can use these! Where you say in input of for example 1v to the base, what would be the current draw of the transistors base so I can use ohms law to find what resistor I must put in series with the base in order for the led to be switched on? Using the equation r=V/I

Also, what resistor would be suitable for the application I talk of, I was thinking of ordering some bc548 to play with but I wasn't sure if they would be appropriate?
Thanks again, Thomas

[wildwestpat]

The base current is related to the collector current by the current gain of the dipolar transistor when it is normally forward biased. The relationship is called the hFE or small signal current gain. You need to be aware that saturating the base with excess current slows down the turn off time (hole saturation in the semi conductor's crystal lattice) FETS on the other hand are voltage driven. There is a range of voltage that need to be applied that needs to be overcome to get the base to accept current. Think of this as a pedestal and is about 0.7 volts for silicon small switching transistors. I suggest you look at the data sheets and Wikipeadia for more information Bipolar junction transistor - Wikipedia, the free encyclopedia.

I suggest you regard transistors (junction type) as current gain as the DC voltages are coincidental to the way the semiconductor works and are useful for fault finding. For circuit design it is the current that determines how the device will perform. The field effect transistor (FET) on the other hand is a voltage driven device that converts an applied voltage to a current.

The bias conditions for linear voltage amplification of a waveform i.e. not switching is a bit more delicate when using junction transistors. A form of automatic bias is used with a small resistor in the emitter lead carrying the common base + collector current.




Regards - Pat

[john-100]

Hi Thomas ,

the e-books from here may help - TE Index for website

start with 200 transistor circuits
scroll down the lefthand side and download the two pdf files e-book 1 and e-book 2

John

[SteveWill]

Thanks a lot for this help; I am starting to gather an understanding as to how I can use these! Where you say in input of for example 1v to the base, what would be the current draw of the transistors base so I can use ohms law to find what resistor I must put in series with the base in order for the led to be switched on? Using the equation r=V/I

Also, what resistor would be suitable for the application I talk of, I was thinking of ordering some bc548 to play with but I wasn't sure if they would be appropriate?
Thanks again, Thomas

Lets look at the bc548 Data Sheet first ( I will get to your question later)

Collector–Emitter Voltage VCEO 30 Vdc
1. Is your power supply going to be more then 30 Vdc?

Emitter–Base Voltage VEBO 6.0 Vdc
2. If the emitter is tie to ground the base can be at 6 volt without damage (6.00001 and you will need a new transistor).

Collector Current — Continuous IC 100 mAdc
3. From the supply resistor to the collector to the emitter resistor to ground, There need to be enough resistance not allow more then 100 mAdc

BC548 hFE 110 to 800
4. This is the relationship of the Base current to Collector current.

Now lets see if we can use this new found information.

First the bc548 looks OK to turn on and off a LED.

" Where you say input for example 1v to the base, what would be the current draw of the transistors base so I can use ohms law to find what resistor"

My question to you is how much current do you want/need in the emitter? Sometimes we have to work from the emitter first before we do any thing with the base. Sometime we need to start with the collector to find what we need the emitter to do an then the base. Lets say you have a LED that going need 10mA to be on. That means that you will have 10 mA flowing in the Transistor collector and emitter. So if you have 10 mA in the emitter you will have somewhere between 1/110 to 1/800 of 10mA in the base, that the range of the gain of a BC548. Lets say that you will have an input voltage 0 to 12Vdc, lets tie the emitter to ground, so the Collector lead will have to have the current limiting resistor to stop more then 100 mA. Lets size it using the LED that can only have 10mA. The LED will drop .7 Vdc when "ON" the transistor will drop .2 Vdc when Saturated. Call it a 1 volt total ( I rounded up) 10mA = 11Vdc/resistance. I figure 1100 ohm should give about 10 mA. So LED and 1100 resistor in the collector lead to power supply. We have set the current flow to a max of 10mA which is less then the max of 100mA.
Now that we know how much current we are going to have in the emitter we can figure how much base current and figure the base resistor. We will have .0000125 mA to .0009 ( again using the 800 and 110 gain) in the base in the non Saturation mode of the transistor ( not turned hard on). That would be a little under a 1 Meg resistor in the base lead.
Lets go back to the Data sheet and see if we can find something that called the Collector Saturation Region Chart and if we look at the IC =10 mA Line we see the Base draw is .05mA at the point that the transistor goes into Saturation. So if we use .05mA, 12 Vdc supply, and we know the transistor will blow up if more then 6 Vdc is applied to the base. So we take the 6Vdc and subtract it from the 12Vdc supply and divide that by .05mA and get 120K ohms. We should feel better now that we have done some math.
Remember that Collector Saturation Region Chart, we see the base is capable of a lot more current then .05ma. So just throw a 100K resistor in the base and forget about it. You could put in a 10k and still not get the base over 6 Vdc, do the math it looks like the Collector Saturation Region Chart graft goes to 3mA.
Just thought of another way to check this we want .7Vdc on the base when on so if we take 12Vdc-.7 we get 11.3 divide .05 we get 220K but we want to make sure we get into Saturation so lets double the current and that is 110K.
As you gain experience you will read more and find that there is more going on and you may find way to use it but for now I think this will get you started.
Someone pointed to some circuits to look at and with your new found knowledge should be able to modified them so they do what you need.

Steve