# Electrical Circuit Problem

Berzerkfury02
Hello I am currently designing a circuit but have run into a couple dead ends. I was wondering if anyone would be willing to take the time to help me with this problem. I am a little rusty at finding unknowns in a circuit, but
basically I am trying to find the rating of a resistor I would need to
use to prevent a higher voltage battery from damaging a lower voltage
device. I am including a schematic of my problem to better aid you in visualizing my exact problem.

I am using a 9v battery
(rated 8.4v for actual output voltage) to power three strobe lights
which are in parallel to each other. (each strobe light is normally
powered by two 1.5v AAA batteries which means the device is designed to
be powered by a total of 3 volts). The resistor directly before these
devices must therefore lose 5.4 volts of potential difference before
reaching the strobes. I have measured the resistence of each strobe and
have come up with 3.5 ohms for each. Since these strobes are parallel
to each other, I have used that inverse equation to come up with a
total resistance through these three devices as 1.167 Ohms. From here I
have calculated a little bit more but am unsure I have done them
correctly. (Since the three strobes should only have a potential of 3 Volts. I used the equation V=IR and plugged these numbers in ... (3)=I(1.167) and got a current needed to run these devices as 2.57 amps. I then went back up to the main battery (8.4 volts) and used the current I found in the equation again to find the total resistance I needed for the entire circuit. Here is where I am not sure I am doing the calculations correctly. If I'm not mistaken I can use the amperage needed to run the paralleled strobes (2.57) in this equation V=IR to plug in (8.4)=(2.57)R to find that the total resistance of the circuit must be 8.4/2.57 or 3.268 Ohms. Since I have a resistance in the paralleled strobes of 1.167 I subtracted this from the 3.268 Total resistance to get about 2.1 ohms of resistance. I then needed to find the wattage rating on the resistor where I used the P=IV equation to find the wattage for the resistor which is 2.1 ohms and must lose 5.4 volts. From P=IV I have, P=(2.57)(6) to get a wattage of 15.42 which this resistor must dissipate before the circuit reaches the three paralleled devices. In the electronics store I am visiting The resistors come accurate to the tenths decimal place. These resistors come in the following wattages.
1/4, 1, 2, 5, 10, and 25. I was wondering If you could help
me figure out which resistor I should use since I am not sure I had done this problem correctly. The resistor I purchased could not be matched exactly however I had tried to come as close as I can. It came out to be a 10 Watt resistor, rated 2.1 ohms. I think there is enough information to determine the correct resistor I need, but if there is any thing else I may need to measure, please let me know. I apologize for this extremely long explanation of my problem but any help would be greatly appreciated.

Daniel

#### Attachments

• schematic.bmp
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## Answers and Replies

mustafa
The calculations are just fine. However, the problem I think is that the 9 V battery may not be able to supply 2.56 A.

Why can't you use just two 1.5 V AAA batteries?

You may use the batteries in parallel, if required.

Berzerkfury02
Hello,

thank you for the information. I actually am using a 9v battery because in parallel to the strobe configuration is a 9 volt device. My battery with an output of 8.4v has a rating of 170 mah. Maybe if I would just parallel two 9v batteries of 170 mah would I end up with an 8.4v battery that has a rating of 340 mah, and if I am not mistaken, I would be able to generate an amperage of 2.856. Also would paralleling the 9v device to the strobe configuration mean that I need even more current for the strobes or does it mean that the current increases in both sides of the circuit. Parallel circuits can confuse me sometimes.

Thank you again,
Daniel

Berzerkfury02
I am also considering maybe creating a series of seven 1.2v rechargeable AAA batteries in order to create an 8.4v battery with a rating of 750 mah. I think this may be able to generate enough amperage for my devices. I think the calculation for the output amps of a battery is something like:

(voltage)(.001)x(capacity in mah).

anyway the output amperage of this configuration would be something like 6.4 amps. Would this be a better solution to the DC power source of my circuit?

pallidin
If you are attempting to drive a high current, momentary device with a low wattage battery, oftentimes it is best to include a capacitor in the circuit.

Berzerkfury02
Oh yes I forgot about that option,

what rating should a capacitor have in this circuit, and also, would that affect the output voltage of my DC power source? I just need any value for a capacitor that seems reasonable to include in this circuit as have very limited knowledge with numericals or other calculations dealing with capacitors. Also do you just place it in the circuit the same way you would a resistor?

Thanks Again,
Daniel

Gold Member
Your math is all correct, but there is NO WAY that these strobe lights can take the kind of current you are talking about and normally be run off of 2 AA cells. What I suggest you do is hook them up the way that they are designed to be run with the AA cells and measure the current. Then redo your math and let us know.

Berzerkfury02
Do you suspect there is going to be too much current going through, or too little current? I will try measuring the current through the devices and get back with the results. If there is too little current coming out of the 9v battery I have somewhat an idea of how to handle it. But otherwise I am not too sure. I will get back to you on the current of the device.

Berzerkfury02
wow I think I realize how there is "NO WAY" my 9 volt configuration is going to power this thing. Taking a current reading, I got 200 mA. Thank you very much for bringing this to my attention. I will be needing to make some design changes. The only reason why I needed a 9 volt battery was to power a 9 volt device also in the circuit. However I think I will separate the 9 volt device into a second circuit with a second switch, and have this circuit run the three paralleled strobes off two AA batteries. This circuit with the three strobes will already have the correct voltage since these are the batteries designed for it and I will be able to completely lose the need to use a resistor. Does everyone agree this is the best solution? because it seems to theoretically work. Thank you everyone for your input.

Daniel

pallidin
Perhaps I should rephrase my original comment.
If you do not use a capacitor your device cannot work due to the extremely high momentary wattage requirements of the strobe light(s) Capacitors are real angels here, and will serve you well.
Take at look at this site, and peruse it's home page as well:
http://www.tkk.fi/Misc/Electronics/circuits/strobo.html [Broken]

Last edited by a moderator:
Berzerkfury02
Thank you for this info, I can actually use this info for my next project, however for this particular project, I had already known from the start that I was going to need at least two switches, (i.e. a master switch, a power switch for the strobes, and another switch to turn on a device running of the 9v power supply. The option of separating the 9v device from the 3v device into two circuits hadn't cross my mind. I still maintain the same number of switches, however, I will now need two power supplies (one in 9v, and one in 3v, which is actually not a big deal, as space is not an actual factor in this project). But I will still maintain the same number of switches, but I would be able to completely bypass the need to use a resistor and capacitor. This way I would be able to have a good battery life for each circuit as each circuit is now powered by two separate DC sources. I thank you for the information as I may just use that info on my next project which also deals with strobes.

On a side note, would you know where you would be able to buy those camera flash units separately (without first buying the camera and removing the flashcube)? I realize that there are some vintage flashcubes such as those by GE and sylvania, however it seems that each flashcube is only good up to 12 flashes, which would not be a very long-term solution at all.

Anyway, thank you for all your help,
Daniel

Mentor
I think you should should consider your switching power supply options. That's how cameras make the voltage for their flashes -- they chop the low battery voltage up to a higher voltage stored on a capacitor, and then dump that energy into the strobe light. When you chop the voltage up, that's called a "boost" regulator. When you chop the voltage down, as in from 9V down to 3.5V, that is called a "buck" regulator. The National Instruments series of "Simple Switcher" ICs make this easy for a beginner to do. Check out their website for some helpful papers and tutorials on applications of the Simple Switcher ICs. Good luck, and have fun!