Paul K0ZYV asks, “What is the maximum safe input voltage to µBITx?”
Paul has lithium ion batteries that provide up to 4.2 volt when fully charged, and he hoped to put four in a pack which could max out at 16.8 volts providing about 2300 maH to power the µBITx.
The consensus seems to be around 15v is the maximum voltage that should be applied to the µBITx.
The reason is that the audio amplifier absolute voltage limit is 15V. All the other components can handle the 15V voltage. This assumes of course that the 5V regulator on the raduino has a heatsink and better still has a series resistor to limit power dissipation.
Regulating the voltage when using Lithium Ion Batteries
VE7WQ uses a $1.45 Boost Buck DC adjustable step up down Converter XL6009 Module with a 4 cell 18650 Li-ion Rechargeable Battery pack. This has the following characteristics:
Wide input voltage 5V ~ 32V;
Wide Output Voltage 1.25V ~ 35V
Built- 4A MOSFET switches, efficiency up to 94%.
How about adding a linear voltage regulator between the battery and “+12V”, while applying the full battery voltage to “PA-PWR”? The 470uF caps in the PA may need to be changed to a higher voltage type, but otherwise, wouldn’t that be an idea? The regulator could be a low-dropout linear type such as L4940. The battery voltage still should to be above the regulator dropout, though. This raises the questions: Could the regulator output be a bit below 12V? What is the minimum voltage that could be applied to “+12V”?
I simply use 2 diodes in series to feed the board.
I then use 13.8 volts to power the unit.
13.8 volts goes to the PA and about 12.4 to the board.
KG0BK
Why not using a LM7812 regulator in series to the 13.8V line, to supply the low power circuitry..? With this simply addition it would be possible to use an even higher voltage, get more power, and protect the most delicate components from reverse polarity.
An LM7812 would work fine up to about a 30VDC input voltage. However, linear voltage regulators draw more current. If you want to maximize the discharge time of your battery, using some regulator diodes, at about 0.7VDC drop each, is a very good idea.
I personally have taken advantage of the dual supply inputs to the main board and have installed another powerjack that has a normally closed circuit when no power plug is inserted. In this configuration I have only one 12V battery powering the main circuits and a second 12V battery, added to the main battery, supplying a total of 24 volts to the final PA FETs. Be sure to provide plenty of cooling to the FETs and increase your fuse amperage accordingly.
In this configuration, I can apply the senario thus. Use 2 of the battery packs in series to the main power jack, and 1 or more battery packs to the PA power jack. If anyone is interested, I can provide a copy of the wiring diagram forthis configuration.
Sorry, I did not explain the senario well. Use 2 of the 4.2 volt battery packs, connected in series, plugged into the first, Main, power jack and 2 more 4.2 battery packs, connected in series, plugged into the PA power jack. Providing 8.4 volts to the main circuitry and a total of 16.8 volts supplied for the PA power circuitry.
The normally closed jack, mentioned above, is implemented to allow a single supply to supply both the main and PA power when a single supply is plugged into the Main power jack.
Can you use dual voltage on the ubitx or only the bitx40. If so, what color is the wire feeding the finals, brown?
Thanks, KN4GVV