A resistor might work, but only if the load is constant. (That is, the load does not change over time). If all of the sudden the load starts to draw more current, there will be a greater voltage drop across the resistor of your choice. So this means that a resistor would be the simplest solution, if an increase the "output impedance" is tolerable.

Also, if the load in non-linear over the full AC cycle, than the voltage will reflect this nonlinearity. This means that the "5Vac" output will only look like nice and sinusoidal if powering a resistive load.

===================================

The 2 antiparallel diodes is not a bad idea, as the voltage drop across a diode is almost the same regardless of current. However, there will be a bit of "crossover" distortion as the AC waveform crosses over between 0.7V, and -0.7V. During that time, neither diode is conducting, so the the output voltage will be very close to 0V. If this is not a problem, then that would totally work.

The voltage drop of a resistor is linear (i.e. totally depending on the flowing current) while the voltage drop of a diode is non-linear (i.e. inside a range of 10mA to 1A it only varies between 0.6 and 0.7V). Numbers given are only a rule of thumb. Check the data sheets and do the math.

As long as you use DC, only one diode is needed, anyway.

Why? What for? Reasons? Why is everyone posting posting incomplete questions with no details at all? We might be good but too lazy to use crystal balls or use our clairvoyant skills....

## Comments

5 years ago

A LDO regulator will do the buisness.

Answer 5 years ago

Wait, nevermind, LDO regulators are for DC.

Answer 5 years ago

A resistor might work, but only if the load is constant. (That is, the load does not change over time). If all of the sudden the load starts to draw more current, there will be a greater voltage drop across the resistor of your choice. So this means that a resistor would be the simplest solution, if an increase the "output impedance" is tolerable.

-------------------------------------------------------------

Also, if the load in non-linear over the full AC cycle, than the voltage will reflect this nonlinearity. This means that the "5Vac" output will only look like nice and sinusoidal if powering a resistive load.

===================================

The 2 antiparallel diodes is not a bad idea, as the voltage drop across a diode is almost the same regardless of current. However, there will be a bit of

"crossover" distortionas the AC waveform crosses over between 0.7V, and -0.7V. During that time, neither diode is conducting, so the the output voltage will be very close to 0V. If this is not a problem, then that would totally work.Answer 5 years ago

It is always wonderful to witness a developing prefrontal cortex.

Hoping your holidays were pleasant and fruitful.

Thank you for acknowledging the concept (with the obvious limitations).

A good test of concept would be resolved with an RMS voltmeter, especially the cross over points which may be a non sequitur fallacy.

Answer 5 years ago

IDK how I possibly missed the AC part, read the question and LDO regs are the first thing that popped up in my head lol

5 years ago

Two 1N4004 diodes inverse parallel will reduce the 7v AC by almost 1 volt AC..

Answer 5 years ago

+1

Sounds plausible, any benefit over a resistor other then Max current?

Answer 5 years ago

More current makes more resistor voltage.. Variable current will make variable resistor voltage..

While variable AC current will be about the same as long as you size the diodes for the peak current.

.

Adding resistors in parallel

lowers the voltage !!While adding diodes in parallel

does nothing to the voltage :-)BTW, adding diodes in parallel only increases cumulative current capacity.

Please, Click pic to see full image

Answer 5 years ago

I never seem to know if the image will show partial or whole ???##@@!!

Answer 5 years ago

The voltage drop of a resistor is linear (i.e. totally depending on the flowing current) while the voltage drop of a diode is non-linear (i.e. inside a range of 10mA to 1A it only varies between 0.6 and 0.7V). Numbers given are only a rule of thumb. Check the data sheets and do the math.

As long as you use DC, only one diode is needed, anyway.

Answer 5 years ago

Sweet thanks !

Answer 5 years ago

Yes, the voltage drop is kind of constant with current, over a small range.

5 years ago

Why?

What for?

Reasons?

Why is everyone posting posting incomplete questions with no details at all?

We might be good but too lazy to use crystal balls or use our clairvoyant skills....

Answer 5 years ago

Welcome to the Instructables question section...