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I have to drop 14.5 ~ 12V to 5V. Using 7805 will dissipate much heat. So,using below circuit will ease 7805 to drop less voltage and will generate less heat.

Using 7.2V 1W zener and 100 ohm 1W resistance :

enter image description here

What are the negative impact of adding zener-resistor combination at input stage of the circuit?

Will it add any efficiency in circuit. I have read that the 7805 are already replacement for zener-resistor combination, but I have seen people using these at input stage of 7805.

Thoughts?

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    \$\begingroup\$ Using your extra zener + resistor will dissipate more heat. You're just moving heat dissipation from the 7805 to the resistor. The zener is just extra power consumption + heat dissipation. Use a heat sink on your 7805, or use a buck converter. \$\endgroup\$ – marcelm May 4 '17 at 17:52
  • \$\begingroup\$ @marcelm Using zener + resistor will not protect against spikes? \$\endgroup\$ – RS System May 4 '17 at 17:55
  • \$\begingroup\$ You said nothing about spikes before. Also: how many volts will your resistor drop when your load is drawing 125mA? \$\endgroup\$ – marcelm May 4 '17 at 18:11
  • \$\begingroup\$ You can place a 1W or 2W 5V1 zener in series (ie instead of the series resistor R). This will reduce the stress on the linear regulator but still heat is a problem. The power dissipated by zenet will be 5.1×0.15 = 0.8W which is quite high and enough to heat up the zener extremely. \$\endgroup\$ – Rohat Kılıç May 4 '17 at 18:24
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Why bother with a zener....

The zener itself consumes current so you are in fact increasing the power usage not reducing it. Further, the ZENER would need to be a high current device so that the series resistor does not have to be so high that it affects the regulator. That's even MORE power to dissipate and lose.

If you MUST take some of the load off the regulator, and your load is 150mA, just use a suitable rated resistor to drop the 12V down to just above the minimum voltage required by the regulator at that current. Perhaps a 1W 30R. Or use a high current zener in place of the resistor.

It won't make it more efficient, and you will still have to get rid of as much heat, but it won't be worse and heat-sinking may not be required.

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Without a zener your 7805 will dissipate about 0.94W worst-case. (0.005 * 12 + 0.125 * (12-5)). A bit of copper on PCB area will dissipate that much heat, depending on your maximum ambient temperature. You can use the TO252 version.

Or you could add a zener in series with the input. A 2.7V 1W zener will dissipate 350mW worst case, leaving less than 600mW so a TO-220 would typically not need a heatsink at all (depending on the maximum ambient temperature).

In general it's better to use a switching supply where possible, however your 125mA output current is not all that much, so a linear supply might work out a bit cheaper (ignoring lifetime energy costs). You could use a 34063 for example, though the inductor will be larger and more expensive than for a more modern chip.

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The overall power creating heat in your circuit is MORE with the Zener diode added. With just the series resistor and the 7805 (no zener) the overall power dissipation will be the same as it would be without the resistor. Usage of either or both of the components will NOT make the overall thing more efficient.

If you really want to be more efficient and reduce heat generation deploy a switching regulator or module design. With careful component selection you should be able to achieve efficiency of around 90% or better.

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