Can a resistor cause a voltage drop?

Question

This is my circuit. 7.4V LIPO Battery>100 ohm Resistor>5V>DC-DC step up>12V>4 Outputs

As soon as a device is connected (audio amp 20.5ma) the voltage drops to 6V between the output and the audio amp.

Is it because I have stepped down then back up?

I also tried a LM317 regulator at the output but that doesnt help either.

There are two things I can try:

1) I can take away the resistor and let the battery go directly into the DC-DC convertor and it will give me around 19V which I can regulate at the other end.

2) I can use a 4.8V Nicad battery straight into the DC-DC which fully charged is around 5.5V which will give me just over the 12V I need on the other end and maybe the voltage will not drop.

Any thing else I can try? Also this DC-DC can only do 84mA@12V can anyone recommend me one that can supply at least 200mA+ from this website www.rapidonline.com

Answer 1

If I read you correctly, you are trying to use the resistor to drop the battery voltage to 5v. Not a good idea. The problem with using a resistor for this is that it only works with one current. Change the current and the Voltage changes. That's called regulation and such a circuit has poor regulation.

Next problem is possibly to do with your choice of Voltage regulator. Regulators need a minimum amount of input/output differential. The Input voltage must be greater than the Output voltage by the minimum for the regulator. In the case of the LM317, this value is about 2.3 Volts. A 7.4v LIPO will be very close to this limit. Normal Voltage variations most likely will take it over the limit. You may need to find a low dropout regulator for this purpose.

You would be better taking the Voltage converter directly of the battery. If your converter will work of the battery you have, that might be OK. Your LM317 would work OK there (on the output).

Answer 2

Any time you have current flow through a resistor you will have a voltage drop across the resistor.

It appears to me that the device you are using to increase the DC voltage from 5v to 12v does not have the current furnishing capabilities required to furnish DC input power to the audio amplifier. I think you may be dropping too much voltage (apparently from 12v down to 6v) across the internal impedance of the power supply when the load is applied. This is what occurs when you overload a power supply.  The current furnishing capabilities of any power supply is limited by the internal impedance of the power supply.

If the power supply is a battery then the battery may be in a fully discharged state.  The output voltage of a discharged battery drops considerably when it is connected to a load.

One other possibility is that you may have a partial short circuit in the audio amplifier causing it to draw excessive current and thus overloading the power supply.

Answer 3

yes, of course, that is what a resistor is, resistance to current.

This seems like a complicated way to get 12 volts.  Why not adjust the DC-DC converter to run from 9 volts and deliver 12 volts?

The voltage is dropping because you are drawing too much current for your setup. Without the details, it's difficult to say, but the resistor would be my first suspect.

The LM317 should have solved that, but perhaps you have it configured incorrectly.

You need to make measurements, that is what engineering is all about. Hook it all up. What is the voltage out of the 9v battery? what is the voltage on the other side of the resistor? It is probably much below 5 volts.

Hook up the 317 and repeat. If the voltage into the DC-DC doesn't stay at 5 volts, then the 317 is hooked up incorrectly.

Why not just buy a 12 volt battery?

Let's try a few calculations. You say the amp is drawing 21 ma. That will translate to 51ma on the other side of the converter. Add 20%, 61ma.

61ma thru 100 ohms is 6 volts, so the 9 volts is dropped to 3 volts, and the DC-DC is not seeing the correct voltage. You can try changing the R to 65 ohms.

You should use the 317, and it should work fine under these conditions. As I said above, make measurements.

Answer 4

your dc-dc consumes more current than your battery and serial 100ohm resistor can provide - hence voltage drop.

my suggestion is to eliminate the resistor and use 7805 between the battery ad dc-dc without modifying the rest of the design

Answer 5

Good old Ohms' law: V = I times R.

If current increases, and R is fixed, then the voltage (drop) across the resistor will increase too.

Most DC-DC converters have a range for the input voltage, and a fixed output voltage, with variable current.  As long as your battery is within the range, omit the resistor.

Use 3 converters with outputs in parallel to get more current - read the datasheet carefully, as you'll need to make each converter 'aware' of the others - usually by some particular wiring of the 'sensing' terminals.

Answer 6

A resistor has two functions in a circuit: 1. To drop volage and 2. To limit current.

Answer 7

If you need a quick 'cheap and nasty' fix until you can come up with a more efficient arrangement, you could try replacing the 100 ohm resistor with a string of 4 series connected silicon rectifier diodes.  Each one will produce a forward voltage drop of about 0.6 v.  Four will drop your 7.4v supply to something close to 5 v.

Good luck.