Heat dissipation zener diode and resistor

I have the following circuit

When I am testing the circuit with 5V USB 3.0 power supply, the zener diode and/or resistor does not build heat up. Using the 18650 7.4V battery makes it build heat. The ATMEGA328P only use its TX and RX pins so I am just using minimal current in that part. Why is it that using 18650 7.4V battery makes the resistor/zener diode heat up so much?

• You've chosen the component values to meet the specification exactly. In order to drop the 5v to 3.3v the current through the resistor will be 300mA, and it will dissipate 0.5W. So you're using the resistor at it's max rating. Just because it's rated for 0.5w doesn't mean it won't get hot. And if the ESP8266 is drawing much less than 300mA, the Zener will consume the remainder of the current and get hot also. For example, if the ESP draws only 100mA, the Zener will draw 200mA and dissipate 200mA x 3.3V = 0.7 Watt already – RJR Oct 17 '18 at 4:57
• Using a Zener diode is not usually a great way to produce a regulated output voltage because it leads to excessive current consumption and heating. But if you do, it might be better to use, say, a 3.9V Zener, and then an emitter follower to supply current to the ESP8266. – mkeith Oct 17 '18 at 5:22
• 3.3V Zenar @ ~80mA current - ~0.3W, Resister power (80+300)*(80+300)*5.61 = ~ 0.9W which is above 0.5W resster wattage. As Isdi said, 7805 can't work with USB power to generate 5V. – user19579 Oct 17 '18 at 5:36
• I changed the resistor to 2x 1W parallel resistors equivalent to about 6 ohms and the temperature improves dramatically. Now, I notice the voltage regulator with input 7.4V heats up faster. – Brian Jacob Sanchez Oct 17 '18 at 15:25
• Where is USB power entering your circuit? – marcelm Oct 17 '18 at 20:23

Why area you powering the ATmega at 5v?
It will operate at 3.3V (1.8V - 5.5V) and you can get rid of the zener.

The resistor gets hot because that's its job.
To alleviate thermal stress from the zener.

Now, I notice the voltage regulator with input 7.4V heats up faster.

You should use an LDO regulator.
This LT3070 has a 85 mV drop out @ 5 A, and under 10 mV @ 300 mA.

Same thing applies to the voltage regulator.
If it gets too hot, you add a power resistor between the battery and the regulator.
Estimate the max current. Do not underestimate.
You want to reduce the voltage into the LM7805 by about 2V at max current.
The resistor value is 2V ÷ max current.

I use a ceramic wirewound resistor for this purpose.

That said...

You should not use linear voltage regulators in a battery powered project.

For good efficiency at low current I use a TI Simple Switcher LM46002
It's a little pricey ($2-$4) but efficiency costs.
You can use TI's WebBench to tweak the efficiency, calculate component values, and create a BOM.
WebBench will also suggest other parts.

• Thanks, I will likely use the switcher as efficiency is important for this project - removing the zener and minimizing resistors is huge improvement. I have another question though, this switcher output voltage is in range 1V-28V, and I am using 7.4V input voltage, how do I control this to maintain a 3.3V output for my esp and atmega? – Brian Jacob Sanchez Oct 18 '18 at 3:06
• This one is very simple available with a fixed 3.3V output and it has a low battery detection. ti.com/lit/ds/symlink/tps62056.pdf – Misunderstood Oct 18 '18 at 3:47
• Thanks, you have been very helpful. I think now the problem is gonna be the component availability in my area. – Brian Jacob Sanchez Oct 18 '18 at 4:22
• That would be absurd, They are probably made in Indonesia. Keep in mind you need to not allow the battery to discharge below 6V. – Misunderstood Oct 18 '18 at 5:16
• I'm starting to look for alternatives. If I use the same battery in series I would have 3.7V and double the capacity and would just use LDO 3.3V voltage regulator I'd still get a high efficiency. I'd say that would be a better setup for longer usage. Wouldn't it? – Brian Jacob Sanchez Oct 18 '18 at 5:36

Are you powering the circuit through the LM7805 when it's hooked up to the USB supply? LM7805 requires about a ~2V headroom (ie +7.0VDC in in order to regulate +5.0 out at 1A). So you may not actually be running that zener regulator at 5V input- check your 5V rail to verify. With the battery you now have 5V and that resistor is going to get warm. Note that wattage ratings on resistors often assume a pretty high heat rise, I don't know what package (SMT/through hole) you're using but it can get uncomfortably warm with a 1/2W dissipation rating.

• I did test the zener using USB power supply and you are right, it is not at 5V. My problem now as I change to a larger resistor approx 6.88 ohms @ 2W is the heat dissipation at the voltage regulator. The change in load resistance has significant change in its temperature. – Brian Jacob Sanchez Oct 17 '18 at 15:34
• One additional suggestion (I have no idea what the use of this circuit is) is to replace both the linear reguulators (7805 and zener) with some simple switching devices (or at least the zener with an LDO regulator). The problem is that no matter what the ESP8266 is consuming, the zener regulator is chewing through your battery life continuously. Even a simple linear regulator would improve your battery life (at roughly 300mA current consumption through the zener as it stands, your battery life is about 10 hours, assuming full charge and 3000mA-Hours capacity. – isdi Oct 17 '18 at 15:45