# Step down 6V DC, 5A to 3.7V, 355mA

Discussion in 'General Electronics Chat' started by naveed, Mar 12, 2008.

1. ### naveed Thread Starter Active Member

Mar 12, 2008
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Actually I need a circuit to step down DC voltage from 6V and 5A of current to 3.7V and 355mA. So that I could charge my mobile Nokia 6280 with 6V rechargeable acid battery. As I am beginner in electronics so I need whole circuit diagram.

Dec 27, 2007
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3. ### rwmoekoe Active Member

Mar 1, 2007
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to use that voltage as a source to recharge your cellphone, you don't need to step it down to 3.7v, unless you want to recharge the battery directly, out of the cellphone itself.
you know that we don't just recharge a li-ion battery as is. the cellphone has the complete and reliable procedure of recharging it's batteries.
instead, the cellphone will receive 5 to 7 volts input voltage through it's connector, so you just connect the 6v lead acid battery using the right connector. be careful with the polarity though. (usually it's positive center)

4. ### SgtWookie Expert

Jul 17, 2007
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See the attached.
U1 and R1 thru R3 limit max current to 355mA. (you should use 1/2W for R1 thru R3; 1/4 would work, but it's marginal)
U2 and R4 thru R6 regulate voltage and current during the final stages of charging.
C1 is required. C2 is a lousy simulation of a battery, but you can get an idea of the charge current, and how it decreases as the battery approaches a full charge.

Datasheet for the low-dropout voltage regulator is here:
http://focus.ti.com/docs/prod/folders/print/tlv1117.html

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5. ### naveed Thread Starter Active Member

Mar 12, 2008
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It is not working directly. The phone lights up and displays "Not Charging"

6. ### naveed Thread Starter Active Member

Mar 12, 2008
42
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I did not understand the circuit. Actually I am beginner in electronics. Please tell me what is U1 and U2? And what is the value of IC? Where is the output? resistances are in ohms or Kilo Ohms? Please help me in detail or refer me any other help understanding basic electronics.

7. ### rwmoekoe Active Member

Mar 1, 2007
172
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usually that message comes when the battery is no longer rechargeable (time to replace it with a new one).
is it a new one already? then maybe the problem's in the cellphone.
anyway, recharging a li-ion battery is not so simple as we thought. it requires a fix voltage with limited current charging at first (usually at 4.2v each cell), after some level it needs to be trickle-charge, or things like that. u can learn more in www.batteryuniversity.com

8. ### SgtWookie Expert

Jul 17, 2007
22,182
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Did you see the attached file? You may have to click on it to blow it up to full size.

U1 and U2 are voltage regulator IC's, part# TLC1117C, that Texas Instruments manufactures. "U1" and "U2" are what is known as "reference designators" on a drawing or schematic; a shorthand term representing an item or notation.

The output is taken from where C2 is; C2 is a simulation of the battery and doesn't actually exist. Ignore the "balloon" above C2 that says "CMD1 0V" - that is a Spice command that sets a voltage value upon the start of the simulation.

Resistance values are in Ohms, except for the potentiometer which is 1k Ohms, or 1,000 Ohms.

U1 and U2 will require heat sinks.

The circuit will not be very efficient because the regulators are linear, rather than switching or "buck" type. However, the construction and theory is easier to understand for a beginner.

If you don't understand the circuit, you willneed to do some studying. There is a wealth of information available in E-book form on this website that starts at a very basic level, and builds upon it.

The two pieces of the circuit (U1, U2) were basically excerpted/adapted from National Semiconductor's LM117/LM317 datasheet, available on their site:
http://www.national.com

Using two regulators in series is not as efficient as a single regulator, but it makes the circuit easier to understand and troubleshoot.

9. ### SgtWookie Expert

Jul 17, 2007
22,182
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Attached is a similar circuit, but using just one LM317 voltage regulator IC along with support components.

C1 takes care of transient voltages on the input, in case the power supply is more than a few inches away.
U1 is the LM317 regulator IC.
C2 increases the stability of the regulator.
R4 sets the voltage. With no load, the output should be set to 3.7V exactly.
R2 sets the charging current to approximately 320mA.
R5 causes the charging current to decrease as the battery voltage increases. This is necessary for NiCD, NiMH, and other non-lead-acid batteries; this allows them to "top up" at a slow rate of charge.

One glaring omission is the monitoring of the battery temperature.

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10. ### naveed Thread Starter Active Member

Mar 12, 2008
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This is better circuit. But where is the output of that circuit? And what is the point A which is marked green? And why battery is written on 10mf capacitor?
Also I did not find that IC (TLV111-7C) in my area. It is not available in the market here. I shall see whether LM317 is available or not.
Is it possible to make the same circuit with the help of transister (D313) having heat sink?
I studied DC voltage dividers and parallel circuits. I thought it would be possible to make the same circuit with just 2 resistances in parallel with each other. But it did not worked. Why is that so?

11. ### SgtWookie Expert

Jul 17, 2007
22,182
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I used a simulated 10mF capacitor as a "battery", to show the current flow as charging progressed, as the simulation wouldn't work using the simulation's battery.

Point "A" is a test point measuring the current flow within the "battery" as it charges. Notice that charging current is initally around 320mA, then drops off as the battery charges up. The data for point "A" can be seen on the simulated oscilloscope trace below.

LM317 IC's are available everywhere; they have been in production for many years, and can be obtained very inexpensively. You will need to use a heat sink.

I don't know what your "D313" transistor is, but if is a D313RP, then it is a DC-DC converter that requires between 10.8V and 13.2V for input. You say that you only have 6V, so it would not work for your application.

12. ### naveed Thread Starter Active Member

Mar 12, 2008
42
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Ok then finally please tell me where are the 2 output terminals of the circuit?

13. ### SgtWookie Expert

Jul 17, 2007
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The POSITIVE (+) terminal of the battery gets connected at the junction of R5 and R1.
The NEGATIVE (-) terminal of the battery gets connected at the junction of R2, R3 and R4.

YOUR battery goes where the 10mF capacitor labeled "BATTERY" exists in the schematic.
The 10mF is not necessary in the real circuit. It is only in the simulation to demonstrate how the circuit works.

14. ### naveed Thread Starter Active Member

Mar 12, 2008
42
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Thank you alot. You devoted your precius time for me. Thanks alot.

15. ### &#9794;demaitri New Member

Mar 5, 2009
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btw i have a usb 3.5G modem but the battery is soak. i want to replace the battery with this schematic, i think i will use nokia n70 adaptor (5dcV 380mA)
can you give me idea?
what change can i do? how about i dont want to use 2n2222

thx

Code ( (Unknown Language)):
1. the batt spec is
2. Li polimer battery
3. 3.7VDC 380mA
4. max Voltage 4.2V