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I see a lot of information regarding Zener diode voltage regulators but how about a regular diode voltage regulator?

What would be the potential drawbacks of using something like this on the picture below:

schematic

simulate this circuit – Schematic created using CircuitLab

Where Rload should get whatever voltage is across the 1N4001 or whatever diode we use?

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  • \$\begingroup\$ That's not a voltage regulator, it's just a voltage divider until the voltage drop across Rl starts getting close to 0.5 volts. \$\endgroup\$ – EM Fields May 5 '16 at 19:44
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    \$\begingroup\$ in the circuit as drawn, Rload only sees 5 mV, so no regulation is taking place. \$\endgroup\$ – markrages May 5 '16 at 19:45
  • \$\begingroup\$ That's my point. Suppose we had a diode that has a forward voltage of 3.3V then we can connect a load that requires 3.3V up in parallel with it? \$\endgroup\$ – Alex H May 5 '16 at 22:30
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    \$\begingroup\$ You're kidding me, stop calling me names I'm trying to ask a legitimate question. The 3.3V comes from my comment of "suppose we had a diode that has a forward voltage of 3.3V" the 1N4001 might have something like 0.7V but the logic still applies. \$\endgroup\$ – Alex H May 5 '16 at 23:20
  • \$\begingroup\$ Yes, in theory this will work about the same as a Zener diode. But as someone else said, with 1-ohm and 1000-ohm resistors the diode won't conduct unless your input is over 700-ish volts. \$\endgroup\$ – user253751 May 6 '16 at 1:57
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  1. The choice of voltages is very limited. For a silicon diode it will be around 700 mV.
  2. The voltage is not well regulated. The forward voltage of a diode is reasonably invariant with current over some range, but it's not as good as most zeners.
  3. The temperature dependency will be higher. This is particularly compared to a zener roughly in the 6 to 6.5 V range. There are two opposite temperature effects that cross over at about that voltage, reducing the overall temperature dependence.

One place where a diode as a shunt regulator can be useful is when you want to generate a voltage that offsets another diode or the B-E drop of a bipolar transistor. In that case the temperature dependency can actually be beneficial.

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In the circuit you show, the voltage across the load will be about 5mV, with or without the resistor. It won't start to regulate until the load resistor is more than about 135 ohms.

In this circuit with a 1K series resistor to +5, the diode acts a bit like a voltage source with about a 12 ohm resistor in series.. for load resistance that is relatively high.

That dynamic resistance increases with the load current- with a 270 ohm load it will be more like 24 ohms (calculated from 52mV/Idiode- the diode ideality factor of about 2 multiplied by the thermal voltage at room temperature). That means that with a 270 ohm load the ripple rejection is only about -20dB. Load regulation will be similarly poor, but zeners for less than a few volts are horrible in performance or nonexistent so you can't really compare it with a zener. You won't find a 0.6V or a 1V zener.

The diode has a temperature coefficient of about -2mV/°C.

So, it's quite usable in some undemanding applications- for example to determine the current through an LED or to set an overcurrent limit on a power supply, but it won't be mistaken for a precision reference.

This type of circuit (where the zener might be something else) is called a 'shunt regulator' and is mostly used for relatively low currents and certain other special applications. It suffers from a number of disadvantages such as inefficiency if the input voltage is close to the output voltage and the load is variable (because that series resistor has to be very low and the wasted power with a light load is very high).

A popular example of a chip that does this well is the TL431 (or TLV431). Using a couple of resistors you can set an accurate voltage from about 1.25V to some tens of volts depending on the chip. The chip needs 1mA or less to operate.

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That's not a voltage regulator, it's just a voltage divider until the voltage drop across Rl starts getting close to 0.5 volts.

As you've drawn it, the load is tied high side,in parallel with D1, and your circuit won't even start to regulate the voltage across the load until the supply gets to 700 volts or so, like this:

enter image description here

More tomorrow...

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  • \$\begingroup\$ "... as we discovered on the train, tomorrow never happens. It's all the same f***** day, man" --Janis Joplin Did you do that with LTSpice? +1 \$\endgroup\$ – Spehro Pefhany May 9 '16 at 19:34
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    \$\begingroup\$ @SpehroPefhany: Quite a touching reference. Sweet, beautiful, raunchy Janice. I miss her... Yes. I learned how to invoke multiple plot panes from someone here, and it was like a new day, glomming an n channel autoranging scope. \$\endgroup\$ – EM Fields May 9 '16 at 21:18
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As others pointed out, that simply won't work as your load current is far too high. To get 0.7A to your 1R load you'd need to use a 6R series resistor or less.

This kind of regulator would be appropriate if need a few milliamps.

I use a zener occasionally when I need 9V or something I need very little current from.

I also occasionally use a LED when I need a proscribed voltage drop with small currents as zeners do not work well with very low currents.

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