# Using a zener diode as a voltage regulator

Hello folks,

I’m trying to lower the circuits voltage from 5 Volts to 1.8 Volts. Is the schematic below the right way to do it?

@peekay123 you were right, 1.5V Zeners are super weird (digikey didn’t have any) so I’m settling for 1.8V. Close enought

Thanks!

VCC: 5 Volts - 2 Amps
Vo: 1.8 Volts - <10 mA
Zener: DIODE ZENER 1.8V 500MW SOD123
Resistor: 330 Ohms +/- 1%; Power: 125mW

Is there any reason you’re opposed to using an adjustable voltage regulator? I’m not entirely sure, but wouldn’t this work: https://www.sparkfun.com/products/527 ?

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Use a voltage regulator… that’s what they are for.
Both of these come in 1.8V fixed or variable versions.

\$1.25 for through-hole @1.5 A

40 cents for thru-hole @ 250mA

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@Moors7 and @Awake, thanks for the answers guys.

I don’t have a good reason for wanting to use Zeners vs Voltage Regulators. I’m just trying to find a simple way to replicate a AA battery from the 5V DC power source from my circuit. And as you probably guess, I’m very new to electronics in general

Below is the schematic from the LM317 Voltage regulator. Just to confirm, the value I will need for the resistors are:

• Vin: 5 volts
• Vout: 1.5 Volts
• R1: 240 Ohms (specified by the circuit)
• R2: 49 Ohms, right?
(49 = (1.5 / 1.25 - 1) * 240)

Thanks a lot!

I don’t think the 240 Ohms is specified. R2 is a potentiometer, allowing you to go through a big range of voltages. A quick search for “LM317 calculator” brought me here, that might help.

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Great. Thank you very much @Moors7!

@sazp96, the zener is a simple approach but requires a minimum of 10ma forward current which you barely achieve with the 330ohm resistor. I REALLY like @Awake’s MCP1700 solution and the MCP1702 has a 1.5v output with a 2.0uA quiescent current. It provides a reduced parts count from the LM317 approach and up to 250ma, though I suspect this is more than you will need.

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@peekay123, you are right, 250mA should have me more than cover.

@Awake and @peekay123, in the datasheet of the MCP1700 it says that the output voltage range is from: 1.2V to 5V, but there are not resistors in the example diagram to adjust the output?

I’m probably missing something obvious, but:

• Any idea how can I get 1.5V from the MCP1700 instead of 1.8V?
• Why does the Vin say “2.3V to 3.2V”? The spec says that the
input voltage is 2.3V to 6V.

@sazp96, you purchase the part based on the specific desired output voltage. It is not programmable.

I’m not sure why the diagram shows 3.2v since the datasheet clearly states a range of 2.3 to 6v. The schematic represents a typical only. The MCP1702 has a larger voltage range and has a part with a 1.5v output. Make sure to use low ESR ceramic caps as recommended.

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I would go with a fixed output part if you do not need adjustability. That way you can connect the center pin straight to ground.
The specific part number for a 1.5V version can be seen here:
http://www.mouser.com/ProductDetail/Microchip-Technology/MCP1700-1502E-TO/?qs=sGAEpiMZZMsGz1a6aV8DcBHlmKLodU9v0k9TvHCmYKc%3D

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So here is the new circuit:

I’m using this Voltage Regulator.

Thanks for all the help @Awake and @peekay123!

@sazp96, you are missing the 1uF capacitors on VIN and VOUT of the regulator as recommended by the manufacturer. On VOUT, you can replace the 100nF cap with a 1uF cap.

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Thanks @peekay123.

To close the loop on this thread. Here is the final circuit:

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The reduced input voltage range is to ensure the maximum device power dissipation is not exceeded with 150mA output current

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@sazp96
Why is the input voltage shown in the sample diagram smaller than the allowed voltage?
Think of it this way.
You have 5V in, and you are going to get 1.8V out. What happens to the other 3.2V? That 3.2V is ‘lost’ in the regulator.
So assume that your regulator is rated for 1 Amp output… that means that 1A is going through the regulator. 1A times 3.2V equals 3.2 Watts, which is dissipated as heat.
In the example they use a smaller input voltage because the voltage drop across the regulator is correspondingly smaller. Suppose you have an input voltage of 3V, and an output of 1.8V, That means that the regulator only ‘uses’ 1.2V, which when multiplied by 1A is equal to 1.2 Watts… much less heat than with the greater input voltage.

So the higher the difference between your input and output voltage, the hotter the regulator will get.

At low current, the heat may be irrelevant, but at rated current and high input voltage, the heat will be enough to require a substantial heat sink to dissipate all that heat generated inside the regulator.

At high currents, a ‘switching regulator’ is advised, since it produces very little heat. But the design, while not too complicated requires more parts.

There are charts on the data sheet that show how hot the regulator will get according to input voltage and current, and there is a maximum rated temperature at which you can operate the regulator. If your design will make the regulator too hot, then you need additional cooling capability with a heat sink. There are other factors that will also influence the temperature of the regulator, such as air circulation and other hot components nearby, so you should give yourself a decent margin of safety.

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Thank you so much for the great EE lesson @Awake!

I hadn’t though about the heat dissipation. Quick question, which current should I use for the head disipation math? The regulator’s max output current, or the amount of current that the device connected to the regulator will use?

Voltage regulator rating: 200mA output
Device connected to the regulator current usage: 20mA

So would the heat dissipation be: 3.2V * 0.2A or 3.3V * 0.02A?

Thanks!