Ohh, so the equals sign is a capacitor. I understand the upside-down triangle is ground.

What is the benefit of adding the C1 and C4 capacitors? Is D1 simply for reverse voltage protection?

Is there a difference between ceramic, electrolytic and metalized capacitors? Google search says ceramics are better?

C2 and C3 (in my schematic) are to prevent the LM7806 from oscillating.
D1 is to prevent connecting the input with reversed polarity.
Ceramic capacitors are better in this application for C2 and C3.

 

C1 and C4 are additional filtering for the AC source that you are not using.
D1 is reverse polarity protection. My drawing does the reverse polarity protection, but it doesn't just quit if you're wrong, it corrects the connection and runs anyway.

 

Try this approach.

 

Thanks for the clarification #12.

And tracecom you beat me to it (the diagram).

Just confirming:

D1: 1amp diode
C1: 0.33 uf ceramic capacitor
C2: 0.10 uf ceramic capacitor

And the ONLY benefit of adding a 1amp diode is for reverse voltage protection? Even if I did reverse the polarity, I don't see how any damage would be done to the battery, circuit, or scanner? I'm thinking skip the diode, unless theres a reason to keep it?

Also, whats the difference between negative and ground? Why would I want to connect the negative terminal of my 12V to my car chassis?

I just found draw.io and drew this.

 

Even if I did reverse the polarity, I don't see how any damage would be done to the battery, circuit, or scanner?

You are dead wrong.

 

You are dead wrong.

Fair enough. But I don't understand why. If the only power source is the 12V battery, then how could there by any reverse voltage heading back towards the battery? And reversed polarity on the battery side... oh.. well I guess it would fry the voltage regulator? I suppose that's a good enough reason for a diode.

EDIT: Perhaps I don't properly understand what a diode does. It's a one-way gate, right? So if I want to prevent reverse polarity, wouldn't I put it on the ground/negative line, so if I were to connect the battery backwards, it doesnt go through the voltage regulator backwards (counterclockwise on my diagram)? I must be missing a fundamental point here.. but I'm trying.

Would using a larger capacitor be of any benefit? I assume not, especially since this is an isolated battery and is not being used to start my car.

 

You will fry the regulator if you connect it backwards.

 

I just hooked up a 7805 from Radio Shack to a 12V battery and to my scanner with some wires plugged in to a bread board. No capacitors or diodes. It works, but the voltage regulator is so hot that when I put some spit on my finger and touched it, it sizzled and was painful. I thought this wouldn't generate that much heat? Is this normal or should I be wondering what I did wrong? I will find my infrared temperature gauge and get a reading, but I'd guess at 100 degrees Farenheit or more. Feels as hot as a pan out of an oven, or more. My multimeter reads 4.95 volts on the output.

Given that my test on my bread board is working, what actual difference would the capacitors make? I could only find a 0.10uf at RS. They say they don't stock 0.33uf.

 

Yes, this is normal. The dissipation is (12-5)x0.5 = 3.5 Watts. That's a mini soldering iron .

Very little, you won't notice.

 

You don't need to wonder what you did wrong. You already know. You failed to connect the capacitors that stop the regulator from oscillating. Lucky you that it works at all.

.33uf ceramic means, ".33uf or more" in this case. Try to buy a larger size than .33uf or use a 10uf tantalum capacitor with a .1 ceramic in parallel.

Now, measure the current through the circuit. The power heating the regulator is the voltage across it times the current through it, unless it is oscillating. After you know the power that is being created, use the datasheet of the 7805 chip to find what temperature that would cause. Power times theta JA = temperature rise above ambient air temperature. If the measured temperature is close to the calculated temperature, add a heat sink. If the temperatures do not match at all, install the capacitors to stop the oscillating.

ps, sizzling is 100 centigrade, not Fahrenheit.

 

You don't need to wonder what you did wrong. You already know. You failed to connect the capacitors that stop the regulator from oscillating. Lucky you that it works at all.

.33uf ceramic means, ".33uf or more" in this case. Try to buy a larger size than .33uf or use a 10uf tantalum capacitor with a .1 ceramic in parallel.

Now, measure the current through the circuit. The power heating the regulator is the voltage across it times the current through it, unless it is oscillating. After you know the power that is being created, use the datasheet of the 7805 chip to find what temperature that would cause. Power times theta JA = temperature rise above ambient air temperature. If the measured temperature is close to the calculated temperature, add a heat sink. If the temperatures do not match at all, install the capacitors to stop the oscillating.

ps, sizzling is 100 centigrade, not Fahrenheit.

I don't know what oscillating means in regards to the circuit? Nor do I understand how a capacitor (temporary storage of electricity) has any affect on it. I won't argue though, as you know more about this stuff than I. I'm just saying I don't understand it.

In an effort to keep the circuit simple, I'll just buy a higher number uf capacitor. As long as the capacitor is 12V, couldn't I just go for as big as a capacitor as I can find without any negative consequences?

I will try again to read the 7805 data sheet, but it's mostly greek to me.

160° Farenheit, whoa! Hopefully once it's in a project box with a heatsink it'll dissipate better.

 

Yes, this is normal. The dissipation is (12-5)x0.5 = 3.5 Watts. That's a mini soldering iron .

Very little, you won't notice.

I'm getting conflicting answers.. one says I won't notice a difference, the other says it's because I don't have the capacitors in that is causing it to "oscillate" and thus the high temp.

With it "on" for about 10 minutes now, with a 0.10uf capacitor on both sides (see the pic in the above post), it's now 180° Farenheit.

 

it's now 180° Farenheit.

If you go on like that, it gets cold again in the near future only to remain cold for the rest of your life .

 

Well these will be on for an average of 8 hours every night. I seem to be stabilizing at about 185°.

I read somewhere that a diode drops the voltage by approximately 0.7 volts, depending on the diode. Would dropping the voltage down a little bit before it reaches the voltage regulator help, and if so how could I accomplish this?

It works great.. but the temperature makes me uncomfortable, so I'd like to improve upon this.

EDIT: I have 2 of these diodes on hand. It says power dissipation is 1 watt. If my voltage regulator is dissipating 3.5 watts of power, could 2 of these inline just after the battery drop the power by 2 watts (more than 50% less heat), or am I misunderstanding this?

 

Every diode "steals" 0.35 Watts from the 7805. The total loss remains the same of course, only the heat is spread over multiple components.

EDIT. I just noticed that these are Zener diodes with a current rating that is way too low and a voltage that is too high. I thought you had an 1N400x or something similar at hand? My mistake.

 

Those were the only 12 volt diodes I could find at Radio Shack. Everything else was outrageous, some even 600 volts. This is why I despise Radio Shack. They never carry what you need, and when they do, it's 10x more expensive. But I was too lazy/impatient to drive 20 miles to a "real" electronics store.

Is there a way to reduce the heat (aside from a heatsink) without distributing it across multiple components? To literally reduce the heat?

Maybe once I mount it to a project box heatsink it will be much better, but I have an inquisitive mind and have to ask.

 

Is there a way to reduce the heat (aside from a heatsink) without distributing it across multiple components? To literally reduce the heat?

Yes. Liquid cooling. You have a can of soda there

 

That's not soda, thats solder flux.

Oh well, I guess I'll have to put up with the heat. Thanks for the help everyone!

 

Here is a DC-DC converter that is about 80% efficient. For 2.5 watts out, it will only dissipate about a 600mW. The bad news is, it costs about US $14.

You might be able to find one that is less expensive.

EDIT: This might suit you better. Same pinout as 7805, less money.

 

Some basic electronics will really help. I will give you the two minute version so don't feel stressed.

Resistors are simple. Capacitors are a little more complicated. Induction which means any wiring and coils adds a bit more complication.

The Oscillation you see with the 7805 without the capacitors is caused by induction in the wiring resisting the regulators attempts to maintain the voltage. Every change in current will cause a voltage change across the wiring because of induction. That is why the capacitors are needed. That changing voltage is easily drowned out by the fact that a capacitor requires a sum capacity of current; some amount of current over some period of time before its voltage will change. AC components or oscillations will be filtered down towards ground by the capacitor. The 78XX regulators will not oscillate if they are properly filtered with input an output capacitors.

I will come back to inductors and oscillations to explain why an SMPS is much more efficient.

First I want to discuss using linear regulators for this application. If you were using linear regulators, (with a 12V battery) definitely use the 7806 instead of the 7805 and definitely use a bridge rectifier like this one.

The rectifier will save you the grief of a reversed battery connection. The bridge rectifier will also drop about 1.5 volts. That is a good thing. With that drop and the 7806 instead of the 7805 you only drop from about 10 to 11 volts down to 6 volts. Car batteries are actually 12.6Volts and often end up overcharged which will quickly ruin them. About 4 volts across the regulator is much easier to handle. You can also reduce the stress on the regulator by placing a power resistor in the regulators input circuit path. A 1 ohm power resistor with 1.5 Amps across it will drop another 1.5 Volts. When there is very little current the regulator won't be stressed anyway so the fact that the resistor won't be dropping any voltage in that situation is fine.

You might enjoy the choice of linear regulators on cold nights. A few watts of heat might be enough to keep you from turning on the engine just to run the heater.

More likely you want to save battery power. On the very busy night when all your radios are humming with exciting news worthy events, your battery will be drawing down by whatever current your radios need - if you use linear regulators.

If you use SMPS regulation your 12Volt battery will draw less current than your radios actually use. I mean that if your radios are humming at 2.5 amps 6 volts, then your 12 volt battery will only need to provide about 1.5 amps at 12Volts with an 85% efficient regulator.

Remember that I said that I would get back to inductors and oscillations to explain why a SMPS is a better choice? Where do you think those magic extra amps for the radios come from? They come from inductors in the SMPS. These work like a transformer allowing you to reduce voltage and increase current at the same time to try and equal output power to the input power, instead of wasting half your power on heat.