I am still stumped on this curcuit (Correction repost)

I am trying to make a simple battery float charger for a 12V lead acid battery into a simple power supply. I have a test design that you can veiw at;

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First I would like to thank all you for responding to my problem at the link listed above. I have reposted this again (Above) with the changes and I hope I will not fry the transister is this curcuit.


I have made the following changes do to the spacs of the 12 volt 5 amp/hour battery I will be using for the backup. This battery is sealed and has a standbye charge current of 1.5A maximum. I am now useing a ballest resistor

39 ohm 10W (R2) The transeformer will be now a 15VAC 3A instead of the 28 VAC one. This should reduce charge current even if the battery is flat to about .750 I hope well within the 2N3055 I will be putting Q1 on the outside of my metal case. The hole case should act as a heatsink. I will be fan cooling.


R2 the bias and D2 the zener will be a 13V 1N4743 this will give me a output at the emitter of Q1 of 13.7 volts (well with float specs of the battery). But D4 (There is a mistake in the schematic d4 should be a 1N4743

13v zener) D3 looks like a .6 volt drop. Iahve installed a blue LED (Just looks cool) on the power line side of the curcuit. I have also put a green led in this curcuit to show that the battery is holding a charge and stays on when the 120 volts is lost. I have put a yellow led to show when cutoff/standbye is hit.


I am useing a 741 for U2 because they are rugged and can opperate up to 36V for the comparator. R6 and R7 that bias Q2 are to saturate it it because U1 does not go rail to rail ot floats 2.5v above and below rails. R8 is the hystersis that prevet the relay from chattering when cut is hit bt making the "No Zone" several volts from trigger point.

Please direct replys to the URL below; (Because my e-mail is munged because of spammers)

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Reply to
Scott Wiper
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"Scott Wiper" wrote in news:dFb0c.3523$ snipped-for-privacy@nnrp1.uunet.ca:

Q1's emitter voltage will be 13V MINUS 0.7V! The base voltage must be ~13.7+0.7 = 14.4V. I would replace Q1 with a darlington: the mimimum hFE of a 2N3055 is just

  1. An emitter current of 2A requires a base current of 100mA. The voltage across C1 is ~21V, which leaves 21-14.4 = 6.6V. The current through R1 will be 6.6V/100ohms = 66mA, which is less than the required base current. Add an extra transistor to make Q1 a darlington. If this extra transistor's hFE is 100, the base current will drop to 1mA. D2 must be a 13.7 + 2*0.7 =
15.1V zener. 15V will suffice.

If D2 needs 10mA to be stable, R1=(V[C1]-V[D2])/(I[D2]+I[B,Q1])= (21V-15V)/(10m+1m)=6V/11mA=545ohms. 470ohms is a good choise.

At a 2A output current Q1 will dissipate about 15W. If the max room temperature is 30C and the max temp of Q1 is 150C, the total thermal resistance (junction to ambient) should not exceed (150-30)/15 = 8C/W. The thermal junction-to-case resistance is 1.5C/W, so the resistance of the heatsink must be 8-1.5 = 6.5C/W or less. You don't need fan cooling.

You also may need to mount the 7805 on a heatsink. If the Project draws

0.5A, the 7805 will dissipate ~4.5W. The R[th,J-C] is 5C/W, so the minimum resistance of the heatsink is (150-30)/4.5 - 5 = 21.6C/W.

BTW, you've got the LEDs D7 and D11 reversed!

Gert van den Heuvel
webmaster www.HobbyElectronics.info
Reply to
Gert van den Heuvel







The 15V Zener for D2 is right; the voltage at the emitter of Q1 will then be

15-1.4 = 13.6V, cause the darlington pair has a Vf of two diode drops. Another poster suggested using a beefier transister for Q2; However, 100mA isn't too much for that one. If you are cautious, you could use a 2N2222A with a metal case, which might dissipate heat a little better.


You don't need 36V for the comparator, since your rail to rail voltage is only 12V. Also, I'd say to use a comparator rather than an opamp, just because they are designed for this application. According to Robert Pease of Nat Semi, opamps generally don't like to have their inputs very far away from each other, but its ok for comparators. With your circuit, instead of the resistive divider feeding Q3, you can use a 3.3k resistor to the 12V rail. A comparator will pull Q3's base to within about 0.2V of ground when it the low battery indication happens, and that will open the relay.


Reply to
Robert C Monsen

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