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I purchased a baofeng car charger for my little UV5Rv2 HT. It was advertised to work with the radio, so I didn't think much of it. It turns out, all it does is pass my car electrical system voltage through to the base charger! Looking at the base charger, I see it accepts 10 V input DC and outputs 8.4 V DC and 400 mA current.

So, I thought I could solder a little resistor inline the charger to add additional resistance to produce a lower voltage. It seems this would work for a fixed load, understanding that my car would produce approximately 14.5 V output while running.

So $$ 10 = 0.4 \cdot R $$ produces $$ {10 \over .4} = 25 $$

So the resistance of the charger is approximately 25 $\Omega$, which is maybe where I'm going wrong somewhere, because if I measure resistance across the + and - terminal of the base station while plugged in, I get 10.56 M$\Omega$. Ignoring that and moving on, I need to figure out what the resistance should be for a 400 mA load creating a 4.5 voltage drop.

$ 4.5 = 0.4 \cdot R $ gives me ${4.5 \over .4} = 11.25$. So I figured putting a $11.25\ \Omega$ resistor inline the positive terminal of the car charger might give me the voltage drop I wanted. This is not the case however. So knowing I have a constant load of 14.5 V and a 400 mA load, how do I calculate the size resistor I should be putting inline to drop 4.5 V to hopefully get to approximately 10 V?

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  • $\begingroup$ You should do that with resistors. Use diode. When you insert diode in circuit it provides 0.7V drop. You can set several diodes in series if you need more voltage drop. check how i did it: pedja.supurovic.net/baofeng-liion-charger-modification All You have to pay attention is that diode can survive current passing through. For that charger almost any diode will do. $\endgroup$ – Pedja YT9TP Nov 27 '16 at 9:00
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You can't use a fixed resistor to control the input the voltage for an arbitrary electronic device, only one which is a passive load, such as a resistor, or an LED. (Note that even a simple LED invalidates your Ohm's-law-based reasoning, because it does not have a fixed value of $R$, though it can still be handled with a resistor.)

A battery charger is not passive — it contains a control system providing the appropriate voltage and/or current for each of the stages of charging (which depend on the particular battery chemistry).

Since you say your “base charger” is marked as having an 8.4 V output, this means it is quite possibly not a charger per se at all, but a power supply for a charging circuit built into the radio. And the 400 mA figure is not the expected output but rather the maximum output. (Hint: No device can ever have an output which is both a fixed current and a fixed voltage. It's physically impossible.)

Whether the device is a power supply (power adapter) or a battery charger, in either case it is going to be regulated — it contains circuits which produce the target output voltage (or current) regardless of the input voltage (within the designed limits). The input power will be maintained, determined by the demands of the load on the output; if the input voltage is lower, more current will be drawn to compensate; if it is higher, then less current.

The result of inserting a resistor in series with a too-high voltage input is that the device will see a too-high voltage whenever it is drawing little current (you can calculate this; just look at $V_\text{across resistor} = IR$ when $I=0$) — and this will happen initially before things are fully powered on, and finally when the battery charge is complete. If there's a protection circuit, this could trip it; if not, and the voltage exceeds the rated voltage of the components, it could damage the device or cause it to misbehave temporarily.

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  • $\begingroup$ Thanks. So instead of a resistor I need something more advanced like a voltage regulator that can produce 10v output. $\endgroup$ – mikew Feb 24 '15 at 12:49
  • $\begingroup$ One of my pass through cigarette lighter adapters had a fairly large body and a little pc board in it. I modified it using an LM317 and two resistors to set the ADJ pin and now get ~10.5V constant regardless of the car being on Acc (~12V) or engine running (~14.6V) $\endgroup$ – mikew Mar 15 '15 at 17:39
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The charger, which accepts 10V on its input and produces 8.4V on its output, is a voltage regulator. A voltage regulator is designed to hold its output voltage constant over possibly varying input voltage and almost certainly varying current.

The "400mA" on the charger is most likely not the current it outputs all the time, but the maximum current it could output while still meeting its other specifications. The actual current drawn will depend on the load.

Because the base charger is a voltage regulator (quite possibly with some additional battery charging circuitry), attempting to lower the input voltage by adding a series resistor is unnecessary. Let it function as designed.

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  • $\begingroup$ Sorry, I might have not been clear or I don't understand your answer. The "car charger" is nothing more than a 12V cigarette lighter adapter that has a dc jack. The "base cradle" for the radio that charges the battery accepts 10V input and comes with a wall wart. If I hook the "car charger" up to the base cradle, it will work with the car off. At this point the car is providing 12V. If I start the car, the voltage raises to 14.5 volt. At this point, the base cradle stops working (i assume it has some protection circuit, as it continues to work when putting it back on the 10V wall wart) $\endgroup$ – mikew Feb 24 '15 at 3:32
  • $\begingroup$ The UV-5R cradle charger outputs 10V - it may lower it if it reads the voltage of the battery, but that's not known. Most likely it doesn't. The battery of the UV-5R stops charging a few volts below that value. $\endgroup$ – user400344 Nov 26 '16 at 23:50
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The way one typically charges lithium-ion batteries, which I believe the Baofeng radios use, is via a constant current circuit. The circuit adjusts the input voltage to meet the desired current level in essence. All a resistor will do is make it that much harder to get the correct power in to the system, but it won't actually detract from the current going in to the battery.

In fact, the circuit isn't simply a resistor, but is a transformer, something in practice a good deal more complex.

I would measure the output of the circuit, see if it is reasonably close to the desired output, and if it seems to work, then leave it alone. If you really want to do something more controlled, you'll need to buy a 13 to 10V DC transformer, a simple resistor won't do.

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    $\begingroup$ It might be more accurate to call it a switched-mode power supply, since a transformer neither is a current source nor works with DC. $\endgroup$ – Phil Frost - W8II Feb 27 '15 at 14:22
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Use a step-down (buck) converter like this -> http://m.ebay.com/itm/251066005460?_mwBanner=1 , and match the output voltage of your base charger when connected to AC.

Edit: my UV-5R charger is at ~10V. I have never measured the charge rate of my radio when docked, but if it should exceed the rated max of the buck converter you end up using, you can simply heatsink the LM2596 chip itself.

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