# Load Resistor Problem!

Discussion in 'The Projects Forum' started by NM2008, Feb 16, 2008.

1. ### NM2008 Thread Starter Senior Member

Feb 9, 2008
135
0
Hi there,
I have a 12v 17AH battery, from this I want to run a receiver and servo set-up.
The receiver operates at 6-8v 300mA max.
What I did was set up 3 25ohm 25w load resistors. Two in parallel and one in series, to divide the voltage into 4, 8 and supply 12v.
I connected the receiver to the 8v output but once I turn on the transmitter the servo turns to max left and max right continuously,I turn the transmitter off and it stops. The transmitter has no control over the servo at all.

On the power supply at 8v the receiver and servo work fine with transmitter, when I set it to 12v and put the load resistors on it, the same thing happens as with the battery.

So it must be something that I am missing with the resistor set-up?
Could someone please advise me as to what I am doing wrong?
Thanks in advance NM

2. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
Sounds like the servo is drawing peak currents that are causing the 8V from your voltage divider to drop below the 6V minimum rating on your servo.

Have you tried adding a high valued filter capacitor across the 25 Ohm leg of your voltage divider?

hgmjr

3. ### SgtWookie Expert

Jul 17, 2007
22,201
1,809
You really need to use a voltage regulator instead of those power-guzzling load resistors.

Your receiver may use 300mA maximum, but what's the minimum? It's going to use a variable amount of current, so you need to supply it with a constant voltage. 300mA across 7V means that the load is 23 Ohms at that point. However, it may draw far less current, depending on a number of factors.

An LM317 voltage regulator with input & output capacitors will give you a nice steady supply voltage.

Connect the +12v battery terminal to the LM317's input, and the negative terminal to ground.
Connect a 120 Ohm resistor from the LM317's output terminal to the ADJ terminal, and a 1K Ohm potentiometer set to about 530 Ohms from the ADJ terminal to ground. This will give you about 7V out.

Use a 0.1uF ceramic capacitor and a 10uF electrolytic capactor on the output pin to ground to keep the output stable.
See the attached schematic.

The LM317 is going to need a heat sink. A discarded PC CPU heat sink works quite well for this. Discard the fan and any other hardware attached, drill a small hole, and fasten the LM317 to the heatsink using a self-tapping sheet metal screw.

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4. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
Take a look at the datasheet for the power resistors you are using. They are probably wire-wound resistors which means they have an inductance associated with them. This can cause there resistance to be a function of the di/dt or current slew rate.

hgmjr

5. ### jpanhalt Expert

Jan 18, 2008
7,533
1,808
I would go with SgtWookie's suggestion. I have experience with model receivers and servos. The description of the problem sounds like a voltage problem. First confirm that receiver and servos are rated at 6 to 8 volts, not something like 4.8 to 6 V.

At voltages outside of range (high or low) servos can do a variety of things. In this case, it looks like they drive to the stops, which causes their drain to increase a lot more. That in turn causes more voltage drop in your divider. How many servos are you trying to run from the 12V supply? If it more than 3, be sure to consider the wattage needed for the voltage regulator, if the servos are put under stress. As a general rule, modelers do not use linear regulators at greater than 10V with more than 3 (4?) servos. Six servos at 11.1 volts requires a switching regulator to handle the high drain. A good heat sink might get you by, but keep in mind the high current requirements of a servo under stress. John

6. ### NM2008 Thread Starter Senior Member

Feb 9, 2008
135
0
Thanks for the help and diagram of regulator circuit.

I tried the load resistor circuit again with an older model receiver and servo of the same values. The servo responded to the transmitter as normal.
The older receiver dates back to early 90,s model and is about twice the size of the 2008 model, therefore could this problem also be down to tolerances within the modern components and ic's?
Anyway the load resistors are a big drain and generate alot of heat, so as you were saying it is probably best if I eliminate them !

Thanks again NM

7. ### SgtWookie Expert

Jul 17, 2007
22,201
1,809
Yes, it could be that the tolerances are different.

Jpanhalt's suggestion of a switching regulator is a very good one. They are far more efficient than linear regulators like the LM317, which is really an ancient design. I simply suggested that because they are so ubiquitous, many "tinkerers" would have a few of them lying around.

You can actually build a switching regulator using an LM317 for a base, along with a number of extra components. See National Semiconductor's LM117/LM317 datasheet, in the Applications section. However, there are modern switching regulator ICs that are more efficient and can accomplish the desired result with a far lower parts count and smaller size.

Greater efficiency = less waste heat = extended operation time.

8. ### jpanhalt Expert

Jan 18, 2008
7,533
1,808
Might be be able to give a little better advice, if we knew your application and how many servos you need to drive. Since you mention 12V, I assume that is a lead acid battery, so it is not going to be flying in a model.

Would you please confirm the receiver spec was 6 to 8 volts? if so, it may be designed for a 2S lithium battery with a nominal voltage of 7.2 volts. And further guessing that is the case, it may have a regulator built in to drop to 4.8 to 6 volts for the servos.
John

9. ### NM2008 Thread Starter Senior Member

Feb 9, 2008
135
0
The application is a simple 2ch receiver, on one channel there is the servo in question, which controls a throttle. There will only be one servo.
I originally decided to use load resistors to divide the voltage as the battery is massive.(12v 17ah sealed lead acid). The receiver is rated at 7v, but I was able to push it to 8v on the psu without an abnormalities.

The resistor which the receiver is on has a 7.6v drop across it, therefore I thought that I would have no problems.

I also tried to filter the circuit, it worked, but it took a 1000uf capacitor across each resistor to achieve normal servo operation.

This seems like a long, hit and miss alternative to SgtWoogies diagram of the regulator circuit, but none the less it is interesting.

Could you tell me what are the capacitors I added smoothing in this case, and why they have to be such a high value? And is there any way of eliminating this problem if I were to continue to use the load resistor circuit?

Thanks for your help NM.

10. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
The capacitors are there to supply the short-duration high-current load demands.
The reason they have to be so large is because the resistance of your voltage divider is so low. You may be familiar with the RC time constant. To achieve a given RC time constant, the lower the R, the higher must be the capacitor used.

The linear regulator or better still the switching regulated power supply would serve you better than the resistor divider.

hgmjr

11. ### SgtWookie Expert

Jul 17, 2007
22,201
1,809
What's a Woogie?

In addition to what hgmjr said, I suggest that running your receiver at higher voltages than it is rated for may work for a period of time, but it increases the likelihood that it will fail more quickly than normal.

To ensure the best combination of performance and longevity from electronic devices, always strive to power them within the limits of the recommended voltage and current supplies.

A "normal" tolerance range is 10%. Running your receiver at 8v is 14.3% above it's rating.

12. ### NM2008 Thread Starter Senior Member

Feb 9, 2008
135
0
Sorry about the spelling mistake!

Thanks for info, I had a feeling that I was pushing it a bit with 8v and even 7.6v
I will build the switching regulator, and see how things work out!

Thanks again NM

13. ### jpanhalt Expert

Jan 18, 2008
7,533
1,808
Your battery will probably last longer with the switching regulator because of less heat dissipation (it's more efficient). However, the hook up of the switcher is a little more complex. You will probably have to find an inductor, which some people consider art rather than science.

Since you are only running one servo and a switch, you may want to reconsider the LM317 that was suggested earlier. That hook up is about as simple as they come. It will dissipate a max of 20W with heat sink. Going from 12 V to 6 V at 1.5 A is 9W. With other linear regulators, you can get to 3 or 4 amps (6V drop 4 A = 24W) with a heat sink. Your battery is 17 AH at 12V, so even throwing energy away though heat loss, it will still run your receiver and servo many hours. John