Call Search
     

New to Ham Radio?
My Profile

Community
Articles
Forums
News
Reviews
Friends Remembered
Strays
Survey Question

Operating
Contesting
DX Cluster Spots
Propagation

Resources
Calendar
Classifieds
Ham Exams
Ham Links
List Archives
News Articles
Product Reviews
QSL Managers

Site Info
eHam Help (FAQ)
Support the site
The eHam Team
Advertising Info
Vision Statement
About eHam.net

   Home   Help Search  
Pages: [1]   Go Down
  Print  
Author Topic: DC voltage regulation  (Read 465 times)
KG6IBW
Member

Posts: 13




Ignore
« on: February 28, 2002, 09:52:14 PM »

I want to put in a linear to run on lead acid cells (part of my desire for disaster communications).  The Ameritron ALS-500M rates the voltages like this:   18 damage, 17 max, 16-14 good, 13 compomised, 12 weak, 11 damage.  A bank of 3x6v batteries in series works about like this:  idle 21v, low drain 20v, moderate drain 17v, high drain (needed by the 500M at full power) 15v.  

How can I build a regulator to fit between the batteries and the amp to reduce very high voltages slightly under low load conditions but deliver the whole enchilada when the current draw is high (80A)?
Logged
NB6Z
Member

Posts: 550


WWW

Ignore
« Reply #1 on: February 28, 2002, 10:55:14 PM »

It does not seem like High Power Amps were designed for battery operation... If you must use a regulator and if you find a device to operate at 80 amps, I would recommend a switching type supply and use it with a very large 12 volt lead acid or several in parallel. 12 V batteries are easier to charge and more available. (Always start with fresh NEW batteries if you run them in Parallel, so they will "age" together for longer cycle life.) A switching supply will boost the voltage from the battery wich could be more efficiant than reducing... In any case don't expect to find a linear type regulator to simply drop the voltage for that much power, as that is not efficiant.
Maybe contact Ameritron and see what they recommend.
Logged
LA1SJA
Member

Posts: 55




Ignore
« Reply #2 on: March 01, 2002, 04:26:09 AM »

Fully agree with NB6Z.
Consider this: If you were lucky and found a switcher capable of handling 84 Amps peak (witch is the spec. of Ameritron) it would at best have efficiency in the range 70 - 80 %. This means a peak current from the battery between 105 and 120 Amps at 13.7 Volts. That is more than 1600 Watts worst case. Even two 110 Amp batteries in paralell would not last long. It would require "Deep Discharge" type batteries that can live with both long standby time and deep discharge cycles.
They cost a fortune and still need to be replaced regularly.
You might consider the "Field Day solution".
Buy a 2.5 KW petrol generator to deliver emergency mains supply. Operate at less than 500 watts, say 400. Set your mains-to-DC power supply to 13.76 Volts nominal (fairly close to the 14 Volts that Ameritron describes as OK), and float a cheap 110 Amp battery over the supply to keep you going the few minutes it takes to fill up the generator every 1 to 2 hours depending on how much RF output you actually run.
Total cost might even end up lower.
Logged
KG6IBW
Member

Posts: 13




Ignore
« Reply #3 on: March 01, 2002, 12:23:21 PM »

Thanks to both of you for your thoughtful input.  Here is how far I have gotten on the problem.

Imagine a power supply of 3 6v lead acids in series.  18v nominal, 21v no-load, 15v at 100A.
Imagine a regulator which is a "magic resistor" connected in series with the linear.

Here are the specs:

Draw  Voltage   %  Rtot   Rext   Pext  Pint
   .1   21     75   200   150      .6     1.5
  1     20     85    20     3     3      16
  5     18     85     3.5    .5   2.5    80
 10     17.5   94     1.7    .1  10     160
 20     16     98      .8    .03  6     300
100     15    100      .1   0     0    1400    

Draw is the total draw of the regulator and linear
Voltage is the battery voltage under that draw
% is the percentage of the supply to be delivered to the linear
Rtot is the total resistance, Rext is the regulator resistance
Pext is the power dissipated by the regulator
Pint is the power used by the linear

How does one build such a "magic resistor"?   Does this look like the reverse path on a zenar?

Thanks again for the help.   Dave  KG6IBW
Logged
LA1SJA
Member

Posts: 55




Ignore
« Reply #4 on: March 02, 2002, 06:28:56 PM »

The resistor you look for might only be available in heaven.

1) Series regulation.
Designing regulators with less than 3 volt internal drop at 100 Amps is rather an Olympic sport.
Bipolar transistors drop 0.7 volts nominally, but much more at high currents.
FETs go down to fractions of ohms, but for both types you would need a large number in parallel.
The heatsink might be the size of an average tranceiver.
This is the type of black art beeing practised in the labs of forklift truck factories and similar, and at tremendous cost. (Sometimes you even hear some interesting explosions in there.)

2) Shunt regulation.
For the same reasons as above, and to avoid the voltage losses when sun intensity/ voltage output is low, many solar panel linear voltage regulators simply shunt away some power by sitting as an extra load across the output from the panel in parallell with the intended load to keep the charging voltage regulated.
This means that as the voltage from the panel decreases, the voltage regulator goes Hi Z in the end, and the intended load gets all available energy.
The efficiency and battery life would end up extremely low in this case. The battery would see a continous load, either from the PA or from the regulator (!)

As a general remark; - I have spent some 20 years designing telecom electronics, including switching power supplies and linear current controls.
As I have had my bit of fun (before moving into sales and management), it would be wrong to discourage you from the sheer fun of discovering these problems firsthand. Including resistors burning wildly starting fire alamrs, exploding capasitors and burnt-out expensive transistor arrays, not to forget the whole lab staff on the floor laughing out loud.
Chances are though that if and when you have completed a design that is tested and proven to be stable enough to be trusted in emergency situations, you will have spent a considerable time away from amateur radio.

Still think switchpowers, - and fossil fuel for long term enegy storage seems the best way.

Have fun, whichever solution you choose.
QRU es 73 de Svein, LA1SJA.
Logged
WB6BYU
Member

Posts: 13112




Ignore
« Reply #5 on: March 04, 2002, 11:19:21 AM »

Perhaps it is time to consider some other alternatives.

For example, an amplifier that operates from 48 volts instead of
12 volts.  Or one that will operate better directly from a 12.6 volt
battery.

Or it may make more sense to improve your antenna system so you
don't need so much power in an emergency.
Logged
W7DJM
Member

Posts: 1




Ignore
« Reply #6 on: March 04, 2002, 05:41:09 PM »

Believe it or not, the following is true, I saw it with my own eyes.    Years ago, while working in a large hardware/auto/industrial supply store,   someone (as the story goes) had some sort of battery operated system, and wanted to run their CB rig.

At that time Western Battery offered a sort of industrial battery which still had the old tar top with individual cells.   They made up a special "12v" battery with ONE MORE CELL!!!!!!

Don't ask me how he planned to charge it up, I never got to talk to the customer.
Logged
N8FVJ
Member

Posts: 692




Ignore
« Reply #7 on: March 04, 2002, 08:01:56 PM »

You could use a bank of transistors (called pass transistors in this application) and drive the base (input) of the transistors with a regulator that has an output sense lead such as a SGS Thompson L200. You will need about 17 volts or higher input to get 15 volts out (a little head room). I'll send you a L200 when I get back to Michigan + schematic if you so wish. It is a linear type power supply only you are using a battery to replace the AC transformer and rectifier- no problem.
Logged
KG6IBW
Member

Posts: 13




Ignore
« Reply #8 on: March 12, 2002, 12:02:01 PM »

Well, here is the design I came up with.   A 6v and a 12v lead acid in series.  A relay that selects between just the 12v and both in series.  The relay coil driven by the drop across the 12v, so that when the voltage across the 12v drops, the coil releases and the 6v is added to the supply.   When the current drops, the 12v battery shows a bigger voltage drop across it, the relay snaps, and the 6v is removed from the circuit.   Problems:  Relay needs he-man contacts (150A?), not clear how the linear will like those sudden power changes (from 10v to 15v when the 6v kicks in, or worse 15v to 19v when the draw stops).

It was the only simple solution I could come up with.  

As it was, I gave up on the DC supply and went with a Heathkit SB-220.   Now I need a generator to be able to drive it in an emergency.
Logged
Pages: [1]   Go Up
  Print  
 
Jump to:  

Powered by MySQL Powered by PHP Powered by SMF 1.1.11 | SMF © 2006-2009, Simple Machines LLC Valid XHTML 1.0! Valid CSS!