Electronics Repair (sci.electronics.repair) Discussion of repairing electronic equipment. Topics include requests for assistance, where to obtain servicing information and parts, techniques for diagnosis and repair, and annecdotes about success, failures and problems.

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Lenny
 
Posts: n/a
Default Battery charger design for security system

I have a security system in my home that I built around a car alarm
design which I modified. Since the original had none, I designed the
120.0 VAC power supply to be used with this unit. It is a basic full
wave bridge, feeding a 2000 uf electrolytic, and a simple two
transistor series regulator which uses a 15 volt zener diode as a
reference from base to ground of the driver transistor. The voltage
across the filter is about 16.0 VDC. The voltage as measured at the
emitter of the output pass transistor is about 13.80 volts, just about
what I expected it to be. In my first trial design I connected this
output through a 5.0 ohm resistor which was in series with a 12.0 volt
7.0 AH sealed rechargeable lead acid battery. I placed a known good
battery which had been sitting on the shelf for two months into the
cicuit and powered up the system. The initial charge rate was about
180 ma. After charging for a day the charge rate dropped to 70.0 ma
with a battery terminal voltge of 13.6 volts. Noting that this trickle
charge, as well as my battery voltage was much too high, I ran the
battery down to 9.50 volts and this time instead of the 5.0 ohm
resistor, I used a small # 354 incandescent light bulb which is in
series with the battery. I then powered up the system again. My theory
is that the bulb will now act as the current limiter, and now for
example with the battery in this serious discharged state, the initial
charge current will be high, and drop to a trickle of 10.0 ma as the
terminal voltage rises. So now, with the battery at 9.50 volts, my
charge rate started out at around 20.0 ma, and after two days the
battery has come up to 11.20 volts and the charge rate has dropped to
about 15.0 ma. I should back up here at this point and mention that
across each side of the bulb are connected the anodes of two diodes.
The cathodes are connected together and this is where my 12.60 volt
output feeds the alarm circuit. I have also placed a 250.0 ma load at
this point to simulate the alarm equipment when it will be connected.
So my question is: with the charge rate running so low now, will this
battery eventually fully charge up, and if so does this sound like a
good "float" type situation. I would like the eventual float voltage
to be 12.60 volts so as not to overheat the battery thus shortening
its life. I know that this charge circuit can be better regulated
electronically but I really didn't want to get that involved. I've
seen this simple bulb scheme used before with success and thought that
I'd try it for myself. If anyone has any further thoughts or ideas on
this I would sincerely appreciate any and all comments. Thanks, Lenny
Stein, Barlen Electronics.
  #2   Report Post  
Jerry G.
 
Posts: n/a
Default Battery charger design for security system

You wrote a lot in one big paragraph. Very basically, you cannot get more
out than you are putting in. The charge rate will have to be equal to, or
higher than the rate of the load.

You need a situation where when the battery is weak, it is getting charged
at a rate that is safe for the battery. Once charged, then the charge should
become unity with the load in order to not discharge the battery, or
overcharge it.

If the alarm requires more than its standby current, then the battery and
charger should be able to handle it for that period of time.

One method would be to have the battery on a standby charge all the time,
and the alarm system be working from an adequate main supply. Only in the
event of a power failure, the alarm system would be switched over to the
battery supply. When the power returns, the alarm system would then be
switched back to the main power supply.

In the meantime there is always a battery charger feeding the battery that
uses a battery condition sensing type monitor. If the battery goes below a
certain state of condition, the charger will increase its charge current
accordingly. When the battery is back to specs, the charger would then go
back to a low trickle charge. You must take care to never over charge
batteries or let them go too far in to depletion, or they will be destroyed.

Many alarm systems use a main AC supply, and are driven with a UPS. This
way the UPS is always supplying AC to the alarm system's main supply. All
the sophistication would be in the UPS. This is a very dependable and
efficient way to work this out. There are many alarm systems using UPS's
for their operation. The only important maintenance is that every 2 to 3
years, the UPS battery is replaced.

--

Greetings,

Jerry Greenberg GLG Technologies GLG
=========================================
WebPage http://www.zoom-one.com
Electronics http://www.zoom-one.com/electron.htm
=========================================


"Lenny" wrote in message
m...
I have a security system in my home that I built around a car alarm
design which I modified. Since the original had none, I designed the
120.0 VAC power supply to be used with this unit. It is a basic full
wave bridge, feeding a 2000 uf electrolytic, and a simple two
transistor series regulator which uses a 15 volt zener diode as a
reference from base to ground of the driver transistor. The voltage
across the filter is about 16.0 VDC. The voltage as measured at the
emitter of the output pass transistor is about 13.80 volts, just about
what I expected it to be. In my first trial design I connected this
output through a 5.0 ohm resistor which was in series with a 12.0 volt
7.0 AH sealed rechargeable lead acid battery. I placed a known good
battery which had been sitting on the shelf for two months into the
cicuit and powered up the system. The initial charge rate was about
180 ma. After charging for a day the charge rate dropped to 70.0 ma
with a battery terminal voltge of 13.6 volts. Noting that this trickle
charge, as well as my battery voltage was much too high, I ran the
battery down to 9.50 volts and this time instead of the 5.0 ohm
resistor, I used a small # 354 incandescent light bulb which is in
series with the battery. I then powered up the system again. My theory
is that the bulb will now act as the current limiter, and now for
example with the battery in this serious discharged state, the initial
charge current will be high, and drop to a trickle of 10.0 ma as the
terminal voltage rises. So now, with the battery at 9.50 volts, my
charge rate started out at around 20.0 ma, and after two days the
battery has come up to 11.20 volts and the charge rate has dropped to
about 15.0 ma. I should back up here at this point and mention that
across each side of the bulb are connected the anodes of two diodes.
The cathodes are connected together and this is where my 12.60 volt
output feeds the alarm circuit. I have also placed a 250.0 ma load at
this point to simulate the alarm equipment when it will be connected.
So my question is: with the charge rate running so low now, will this
battery eventually fully charge up, and if so does this sound like a
good "float" type situation. I would like the eventual float voltage
to be 12.60 volts so as not to overheat the battery thus shortening
its life. I know that this charge circuit can be better regulated
electronically but I really didn't want to get that involved. I've
seen this simple bulb scheme used before with success and thought that
I'd try it for myself. If anyone has any further thoughts or ideas on
this I would sincerely appreciate any and all comments. Thanks, Lenny
Stein, Barlen Electronics.


  #3   Report Post  
Jerry G.
 
Posts: n/a
Default Battery charger design for security system

You wrote a lot in one big paragraph. Very basically, you cannot get more
out than you are putting in. The charge rate will have to be equal to, or
higher than the rate of the load.

You need a situation where when the battery is weak, it is getting charged
at a rate that is safe for the battery. Once charged, then the charge should
become unity with the load in order to not discharge the battery, or
overcharge it.

If the alarm requires more than its standby current, then the battery and
charger should be able to handle it for that period of time.

One method would be to have the battery on a standby charge all the time,
and the alarm system be working from an adequate main supply. Only in the
event of a power failure, the alarm system would be switched over to the
battery supply. When the power returns, the alarm system would then be
switched back to the main power supply.

In the meantime there is always a battery charger feeding the battery that
uses a battery condition sensing type monitor. If the battery goes below a
certain state of condition, the charger will increase its charge current
accordingly. When the battery is back to specs, the charger would then go
back to a low trickle charge. You must take care to never over charge
batteries or let them go too far in to depletion, or they will be destroyed.

Many alarm systems use a main AC supply, and are driven with a UPS. This
way the UPS is always supplying AC to the alarm system's main supply. All
the sophistication would be in the UPS. This is a very dependable and
efficient way to work this out. There are many alarm systems using UPS's
for their operation. The only important maintenance is that every 2 to 3
years, the UPS battery is replaced.

--

Greetings,

Jerry Greenberg GLG Technologies GLG
=========================================
WebPage http://www.zoom-one.com
Electronics http://www.zoom-one.com/electron.htm
=========================================


"Lenny" wrote in message
m...
I have a security system in my home that I built around a car alarm
design which I modified. Since the original had none, I designed the
120.0 VAC power supply to be used with this unit. It is a basic full
wave bridge, feeding a 2000 uf electrolytic, and a simple two
transistor series regulator which uses a 15 volt zener diode as a
reference from base to ground of the driver transistor. The voltage
across the filter is about 16.0 VDC. The voltage as measured at the
emitter of the output pass transistor is about 13.80 volts, just about
what I expected it to be. In my first trial design I connected this
output through a 5.0 ohm resistor which was in series with a 12.0 volt
7.0 AH sealed rechargeable lead acid battery. I placed a known good
battery which had been sitting on the shelf for two months into the
cicuit and powered up the system. The initial charge rate was about
180 ma. After charging for a day the charge rate dropped to 70.0 ma
with a battery terminal voltge of 13.6 volts. Noting that this trickle
charge, as well as my battery voltage was much too high, I ran the
battery down to 9.50 volts and this time instead of the 5.0 ohm
resistor, I used a small # 354 incandescent light bulb which is in
series with the battery. I then powered up the system again. My theory
is that the bulb will now act as the current limiter, and now for
example with the battery in this serious discharged state, the initial
charge current will be high, and drop to a trickle of 10.0 ma as the
terminal voltage rises. So now, with the battery at 9.50 volts, my
charge rate started out at around 20.0 ma, and after two days the
battery has come up to 11.20 volts and the charge rate has dropped to
about 15.0 ma. I should back up here at this point and mention that
across each side of the bulb are connected the anodes of two diodes.
The cathodes are connected together and this is where my 12.60 volt
output feeds the alarm circuit. I have also placed a 250.0 ma load at
this point to simulate the alarm equipment when it will be connected.
So my question is: with the charge rate running so low now, will this
battery eventually fully charge up, and if so does this sound like a
good "float" type situation. I would like the eventual float voltage
to be 12.60 volts so as not to overheat the battery thus shortening
its life. I know that this charge circuit can be better regulated
electronically but I really didn't want to get that involved. I've
seen this simple bulb scheme used before with success and thought that
I'd try it for myself. If anyone has any further thoughts or ideas on
this I would sincerely appreciate any and all comments. Thanks, Lenny
Stein, Barlen Electronics.


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