What does the 555 IC do here?

Thread Starter

Sid2001

Joined Dec 7, 2016
29
Hello, for my GCSE electronics I am making a solar powered phone charger. I came across this circuit on the internet. I recreated it on circuit wizard.upload_2017-5-20_16-22-20.png
The 12V DC power supply on the left represents the solar cell and the 5v battery on the right represents the 5v USB output. I can't seem to understand what the 555 does or what the circuit does in general. The website I got it from has this as an explanation, but I can't understand it. Layman's terms of some sort would be helpful.

Here is how the website explains it:
"Charging of the cellphone battery is a big problem while travelling as power supply source is not generally accessible. If you keep your cellphone switched on continuously, its battery will go flat within five to six hours, making the cellphone useless. A fully charged battery becomes necessary especially when your distance from the nearest relay station increases. Here’s a simple charger that replenishes the cellphone battery within two to three hours. Basically, the charger is a current-limited voltage source. Generally, cellphone battery packs require 3.6-6V DC and 180-200mA current for charging. These usually contain three NiCd cells, each having 1.2V rating. Current of 100mA is sufficient for charging the cellphone battery at a slow rate. A 12V battery containing eight pen gives sufficient current (1.8A) to charge the battery connected
across the output terminals. The circuit also monitors the voltage level of the battery. It automatically cuts off the charging process when its output terminal voltage increases above the
predetermined voltage level.

Working
Timer IC NE555 is used to charge and monitor the voltage level in the battery. Control voltage pin 5 of IC1 is provided with a reference voltage of 5.6V by zener
diode ZD1. Threshold pin 6 is supplied with a voltage set by VR1 and trigger pin 2 is supplied with a voltage set by VR2. When the discharged cellphone battery is connected to the circuit, the voltage given to trigger pin 2 of IC1 is below 1/3Vcc and hence the flip-flop in the IC is switched on to take output pin 3 high. When the battery is fully charged, the output terminal voltage
increases the voltage at pin 2 of IC1 above the trigger point threshold. This switches off
the flip-flop and the output goes low to terminate the charging process. Threshold pin 6 of IC1 is referenced at 2/3Vcc set by VR1. Transistor T1 is used to enhance the charging current. Value of R3 is critical in providing the required current for charging. With the given value of 39-ohm the charging current is around 180 mA. The circuit can be constructed on a small general-purpose PCB. For calibration of cut-off voltage level, use a variable DC power source. Connect the output terminals of the circuit to the variable power supply set at 7V. Adjust VR1 in the middle
position and slowly adjust VR2 until LED1 goes off, indicating low output. LED1 should turn on when the voltage of the variable power supply reduces below 5V. Enclose the circuit in a small plastic case and use suitable connector for connecting to the cellphone battery."



What I am wondering about the most is how the 555 creates pulsing for the LED. The website says that when no battery is connected, the LED flashes at 765Kh (unnoticeable to eye). When there is a battery connected it flashes at 4.5Hz and when the battery is fully charged it turns off. I don't see how it does this as it doesn't utilize pin 2,6 or 7 for monostable/astable function. I can't actually see the flashing since Circuit Wizard doesn't really simulate charging batteries.


Thank you for helping. If you'd like to see the website I got the circuit from, here it is.
http://electronicseveryday.blogspot.co.uk/2009/07/mobile-cellphone-charger.html
 

zijo

Joined May 20, 2017
8
Replace battery with V source. and add resistor.

It could be that each time it is different voltage level.
Why would it be 4.5HZ is hard to know.
 

Thread Starter

Sid2001

Joined Dec 7, 2016
29
It's not homework as such. It's coursework. I did originally put this in homework help but someone moved it here.
 

ronv

Joined Nov 12, 2008
3,770
Hello, for my GCSE electronics I am making a solar powered phone charger. I came across this circuit on the internet. I recreated it on circuit wizard.View attachment 127218
The 12V DC power supply on the left represents the solar cell and the 5v battery on the right represents the 5v USB output. I can't seem to understand what the 555 does or what the circuit does in general. The website I got it from has this as an explanation, but I can't understand it. Layman's terms of some sort would be helpful.

Here is how the website explains it:
"Charging of the cellphone battery is a big problem while travelling as power supply source is not generally accessible. If you keep your cellphone switched on continuously, its battery will go flat within five to six hours, making the cellphone useless. A fully charged battery becomes necessary especially when your distance from the nearest relay station increases. Here’s a simple charger that replenishes the cellphone battery within two to three hours. Basically, the charger is a current-limited voltage source. Generally, cellphone battery packs require 3.6-6V DC and 180-200mA current for charging. These usually contain three NiCd cells, each having 1.2V rating. Current of 100mA is sufficient for charging the cellphone battery at a slow rate. A 12V battery containing eight pen gives sufficient current (1.8A) to charge the battery connected
across the output terminals. The circuit also monitors the voltage level of the battery. It automatically cuts off the charging process when its output terminal voltage increases above the
predetermined voltage level.

Working
Timer IC NE555 is used to charge and monitor the voltage level in the battery. Control voltage pin 5 of IC1 is provided with a reference voltage of 5.6V by zener
diode ZD1. Threshold pin 6 is supplied with a voltage set by VR1 and trigger pin 2 is supplied with a voltage set by VR2. When the discharged cellphone battery is connected to the circuit, the voltage given to trigger pin 2 of IC1 is below 1/3Vcc and hence the flip-flop in the IC is switched on to take output pin 3 high. When the battery is fully charged, the output terminal voltage
increases the voltage at pin 2 of IC1 above the trigger point threshold. This switches off
the flip-flop and the output goes low to terminate the charging process. Threshold pin 6 of IC1 is referenced at 2/3Vcc set by VR1. Transistor T1 is used to enhance the charging current. Value of R3 is critical in providing the required current for charging. With the given value of 39-ohm the charging current is around 180 mA. The circuit can be constructed on a small general-purpose PCB. For calibration of cut-off voltage level, use a variable DC power source. Connect the output terminals of the circuit to the variable power supply set at 7V. Adjust VR1 in the middle
position and slowly adjust VR2 until LED1 goes off, indicating low output. LED1 should turn on when the voltage of the variable power supply reduces below 5V. Enclose the circuit in a small plastic case and use suitable connector for connecting to the cellphone battery."



What I am wondering about the most is how the 555 creates pulsing for the LED. The website says that when no battery is connected, the LED flashes at 765Kh (unnoticeable to eye). When there is a battery connected it flashes at 4.5Hz and when the battery is fully charged it turns off. I don't see how it does this as it doesn't utilize pin 2,6 or 7 for monostable/astable function. I can't actually see the flashing since Circuit Wizard doesn't really simulate charging batteries.


Thank you for helping. If you'd like to see the website I got the circuit from, here it is.
http://electronicseveryday.blogspot.co.uk/2009/07/mobile-cellphone-charger.html
I haven't modeled it but this is how I think it works:
The 555 is used as a comparator. It turns on the charger when the battery voltage is low. The battery acts like a big capacitor so it's voltage rises and the 555 turns off. Apparently this happens at a 4.5 Hz rate.
With no battery the voltage rises very quickly and the 555 runs at it's maximum speed looking like it is on.
When the battery is finally above the threshold while being charged or at rest it turns off.
It is made to charge batteries directly, not thru a USB port.
 

AnalogKid

Joined Aug 1, 2013
11,037
The circuit uses a 555 as a flipflop, which is not what most people would use, but it is ok. The *serious* problem is R3. He is right, the value is critical, but there is no way to determine the correct value without building the circuit and measuring the output current. He is using R3 to "dangle bias" Q1 by starving it of base current until it's max output current is 100 mA. But the gain of Q1 varies with current, temperature, age, and from one part to the next. This is a very poor way to limit output current, and will almost certainly lead to problems down the road. Also, when no load is connected the output rises to almost 12 V, a very bad starting point when a dead battery is connected. Finally, at 100 mA output current there will be a 10 V drop across R2, dropping the output voltage to less than 2 V. Overall, this is not a good circuit.

Also, most cell phone batteries are Lithium ion, not Ni-Cad, and this changes the charging current profile. Search for Lithium battery charger schematics and USB power supply schematics to get a feel for how this game is played. This can be done with an LM317 instead of a 555. The 317 datasheet has a schematic for an output current limiter. Also, there are many products on the market that take some regular batteries and produce a 5 V USB-compliant power source.

ak
 

Thread Starter

Sid2001

Joined Dec 7, 2016
29
The circuit uses a 555 as a flipflop, which is not what most people would use, but it is ok. The *serious* problem is R3. He is right, the value is critical, but there is no way to determine the correct value without building the circuit and measuring the output current. He is using R3 to "dangle bias" Q1 by starving it of base current until it's max output current is 100 mA. But the gain of Q1 varies with current, temperature, age, and from one part to the next. This is a very poor way to limit output current, and will almost certainly lead to problems down the road. Also, when no load is connected the output rises to almost 12 V, a very bad starting point when a dead battery is connected. Finally, at 100 mA output current there will be a 10 V drop across R2, dropping the output voltage to less than 2 V. Overall, this is not a good circuit.

Also, most cell phone batteries are Lithium ion, not Ni-Cad, and this changes the charging current profile. Search for Lithium battery charger schematics and USB power supply schematics to get a feel for how this game is played. This can be done with an LM317 instead of a 555. The 317 datasheet has a schematic for an output current limiter. Also, there are many products on the market that take some regular batteries and produce a 5 V USB-compliant power source.

ak

I've just done some research and found out Li-ion cells require very specific values for the voltage and current variables. I also found out that modern smartphones have a protection circuit. If this circuit was incompatible with one of these modern phones do you think the phone would just not allow itself to charge? I have ordered the components and a breadboard (they should arrive tommorow). I shall breadboard it and see the results.

Thanks for the help guys.
 

wayneh

Joined Sep 9, 2010
17,498
If this circuit was incompatible with one of these modern phones do you think the phone would just not allow itself to charge?
Yes, many phones reject charging if they don't see what they want. For example I have a cheap car USB charger that is rated to 0.5A. My iPhone will not charge at all on that device even though 5V and a few hundred milliamps would do a nice job while in the car. A similar adapter I have (rated to 3A) works great.
 
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