Zener diode questio...

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Zener diode question

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(@voltage)
Member
Joined: 3 years ago
Posts: 193
Topic starter

For starters, I have never used a zener diode in my life so be easy on me. 🤣 So I was studying some youtube videos about diodes just to learn more in my quest to become smarter in electronics that I am now, when I had a great idea. I already knew about rectifier diodes and bridges etc and how they worked but I watched a very good video about zener diodes. This may save me with a little issue I wanted to solve on a very very simple circuit I have used in the past using rechargeable AA 1.2 Eneloop batterys in a dual battery holder for a 2.4V output. Seeing that someone could easily use regular 2- 1.5V batteries in the same holder that would interfere with my 2.4V requirement and change things maybe to the point of ruining something or just otherwise over volt my circuits original design.

My idea was to put a 2.4V 500mW zener on the input to keep the voltage at 2.4 volts no matter the battery type installed. Then I realized the zener has a forward voltage drop of about 0.7 volts (also known as knee or cutoff voltage I think). So I found I can get a zener as low as 2.0V but if I subtract the 0.7 forward voltage from my 2.4V input my circuit suffers as that brings it down to 1.7V so that won't work.

So I think using a zener for my purpose would be a fail, correct?

Thanks

Thanks,
Voltage

(@voltage)
Member
Joined: 3 years ago
Posts: 193
Topic starter

Ok, I am talking to myself already after watching the video again and think maybe I was right all along and can use a 2.4V zener and my project will work fine. I may be wrong but I am thinking out loud. Then if the 2 or 3.1V from the AA batteries is used it should clamp the voltage to 2.4V. Seems too simple so there must be an issue.

Thanks,
Voltage

(@voltage)
Member
Joined: 3 years ago
Posts: 193
Topic starter

I just realized this can be easily solved by breadboarding the circuit and test it in real time. I will order some zeners and figure it out as I don't have any zener diodes in stock and if I do they are likely some value that won't apply to this situation. Comments are still welcome. 😋

Thanks,
Voltage

(@davee)
Member
Joined: 3 years ago
Posts: 1851

Hi @voltage,

Sorry, but there is indeed an issue with your proposal. If the batteries are reasonably well-charged, then (at a guess ... I am not trying it!), your diode will probably glow for a second or two and maybe fall apart, and you might see a puff of magic smoke. Probably not want you wanted.

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

So let's go back to basics ... Ohm's Law ...

(diagram is from Wikipedia, which has more details at https://en.wikipedia.org/wiki/Ohm%27s_law)

The voltage source V can be produced in numerous ways ... for example, the 3V battery you mentioned.

The resistor shown has a Resistance R.

Then the current flow will be given by I (in Amps) = V (in volts) / R (in Ohms)

If the resistor value R = 900 Ohms, then with the 3V battery source for V:

I = 3 / 900 = 0.01 Amps = 10 milliAmps.

----------

Sorry if the above seems too easy, but please have a little patience. I am now going to suggest a simple addition circuit to your circuit, which will work, albeit maybe not achieve your final aim.

Again Wikipedia has a page with more detail, https://en.wikipedia.org/wiki/Zener_diode,

which included the following diagrams:

The first diagram illustrates the basic circuit.

• Uin on the left represents a voltage source of Uin Volts. In your case, a 3V battery.
• R is the resistor. Let's assume the resistor has a resistance of 900 Ohms again.
• D is the Zener Diode. In your case, a 2.4V Zener diode.
• Uout is the output voltage, which when measured with a multimeter will (approximately) match the Zener voltage. e.g. 2.4V

So this appears to be what you wanted.

Unfortunately, it is necessary to understand how the circuit works, because it might not do the job you have in mind.

First look at the left half of the second picture snipped from Wikipedia's Zener diode page, which shows a graph curve of a zener diode, plotting the current as a function of increasing 'negative' voltage.

This curve is drawn for a 3.4V Zener diode, but the curve for 2.4V Zener diode would be the same shape, except that the sharp increase in negative current would occur at (about) -2.4V instead of -3.4V.

So looking at the curve, start at the origin of the axes. Here, the voltage across the diode is 0V, and the current is 0 A.

Now scan to the left ... until the voltage reaches -3.4V (or -2.4V if it is a 2.4V Zener diode), there is only a tiny negative current flow, which for most purposes can be regarded as zero current.

BUT, as the negative voltage exceeds the -3.4V (e.g. moves to -3.5V), then the negative current increases dramatically. Oversimplifying, the diode appears to be determined to draw enough current to stop the voltage going beyond its Zener diode voltage, even if it self-destructs by overheating with resultant power flow.

-----------

Obviously, causing it to self-destruct is not a good design plan. So instead, a resistor R is introduced into the circuit, as shown above, to limit the current flow to a safe value.

-----------

Now back to some simple arithmetic, to understand what is happening. Using your requirements, of a 3V battery, a 2.4V Zener diode, and perhaps R = 900 Ohm resistor.

When this circuit is wired up, the voltage across the Zener diode will be 2.4V, because the 3V supply is in excess of the 2.4V diode breakdown voltage, and the Zener will draw enough current to bring the voltage across it to 2.4V. However, because there is a 900 Ohm resistor between the 3V battery and the Zener diode, the current will be limited, and may be calculated by Ohm's law.

As the battery voltage is 3V, and the voltage across the diode is 2.4V, then the voltage across the resistor must be (3V - 2.4V) = 0.6V

And the value of the resistor is 900 Ohm, so by Ohm's law, the current will be

I = 0.6 (V) / 900 (Ohms) = 0.000667 Amps = 0.667 milliAmps.

At this point, it is also possible to calculate the power dissipation in the Zener diode, to make sure it will not overheat.

Power (in Watts) = Voltage (in Volts) * Current (in Amps)

Power (Watts) = 2.4 (V) * 0.000667 (Amps)

or more conveniently:

Power (milliWatts) = 2.4 (V) * 0.667 (milliAmps) = 1.6 milliWatts

Which is safely far below the maximum limit of a 500 milliWatt Zener diode capability that you suggested.

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

So is this a useful circuit for you?

The simple answer is, I don't know!!

The reason for this ambiguous answer, is that you did not state the current or power demand of the load you were going to connect to the output.

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

Checking this Resistor + Zener diode circuit with a typical multimeter will show the output is 2.4V as expected, because the current draw of the meter will only be a few microAmps.

But imagine the load rsistance is 100 Ohms. Then the circuit will consist of a potential divider consisting of a 900 Ohm resistor and a 100 Ohm resistor across the 3V battery, so that only 1/10th of the battery voltage is across the 100 Ohm resistor, which is itself in parallel with the Zener diode.

1/10 th of 3V, is 0.3V. So the Zener diode has only (-) 0.3V across it, and so it will not conduct a significant current. But more importantly, the load will only get 0.3V, and 2.7V will be dropped across the 900 Ohm Resistor R. So although, the circuit will not self-destruct, it will not work either.

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

Can this be fixed?

The first possibility is to reduce the value of the resistor R, so that it will allow more current to flow, before the voltage drop across it is too high.

And since the power dissipation in the Zener diode was only 1.6 milliWatts, compared to 500 milliWatts maximum, there is clearly considerable flexibility to try this approach. Essentially, propose a new value for R, say 100 Ohms, and repeat the above calculation with new values.

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

Another consideration, is how long must a battery last? A good, fresh AA battery might have a capacity of 1000 mAmpHr. (This is a wild guess for this discussion, not an accurate value.)

Hence, with the current draw of 0.667 milliAmp, the life might be 1000/0.667 = 1500 hours .. about 2 months

In some situations, this might be acceptable, but not if the requirement is for one replacement every 12 months!

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

In summary, using a resistor and a Zener diode is a well known and simple method of obtaining a reference voltage within a circuit, albeit the voltage tolerance might be (something like) +/- 5%, which is too imprecise for some tasks.

However, it is often too inefficient as a power supply for a load requiring a substantial amount of power, especially if the power source itself is limited, such as when it is a battery.

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

I am sorry this is rather long, and may require you to re-read it several times, accompanied by copious amounts of your favourite beverage, but I hope it will eventually become clear, as it illustrates several points which are fundamental to understanding how circuits actually work. Do not be surprised if you find it tricky for a while ... you are not alone in the respect.

If there is something that is not clear, please ask. Hopefully someone on the forum will be able to clarify the picture. (And given the length of the sermon, it is probable I have left in a couple of confusing typos.)

Good luck and best wishes, Dave

Voltage reacted
(@voltage)
Member
Joined: 3 years ago
Posts: 193
Topic starter

Posted by: @davee

Hi @voltage,

Sorry, but there is indeed an issue with your proposal. If the batteries are reasonably well-charged, then (at a guess ... I am not trying it!), your diode will probably glow for a second or two and maybe fall apart, and you might see a puff of magic smoke. Probably not want you wanted.

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

So let's go back to basics ... Ohm's Law ...

(diagram is from Wikipedia, which has more details at https://en.wikipedia.org/wiki/Ohm%27s_law)

-- attachment is not available --

The voltage source V can be produced in numerous ways ... for example, the 3V battery you mentioned.

The resistor shown has a Resistance R.

Then the current flow will be given by I (in Amps) = V (in volts) / R (in Ohms)

If the resistor value R = 900 Ohms, then with the 3V battery source for V:

I = 3 / 900 = 0.01 Amps = 10 milliAmps.

----------

Sorry if the above seems too easy, but please have a little patience. I am now going to suggest a simple addition circuit to your circuit, which will work, albeit maybe not achieve your final aim.

Again Wikipedia has a page with more detail, https://en.wikipedia.org/wiki/Zener_diode,

which included the following diagrams:

-- attachment is not available -- -- attachment is not available --

The first diagram illustrates the basic circuit.

• Uin on the left represents a voltage source of Uin Volts. In your case, a 3V battery.
• R is the resistor. Let's assume the resistor has a resistance of 900 Ohms again.
• D is the Zener Diode. In your case, a 2.4V Zener diode.
• Uout is the output voltage, which when measured with a multimeter will (approximately) match the Zener voltage. e.g. 2.4V

So this appears to be what you wanted.

Unfortunately, it is necessary to understand how the circuit works, because it might not do the job you have in mind.

First look at the left half of the second picture snipped from Wikipedia's Zener diode page, which shows a graph curve of a zener diode, plotting the current as a function of increasing 'negative' voltage.

This curve is drawn for a 3.4V Zener diode, but the curve for 2.4V Zener diode would be the same shape, except that the sharp increase in negative current would occur at (about) -2.4V instead of -3.4V.

So looking at the curve, start at the origin of the axes. Here, the voltage across the diode is 0V, and the current is 0 A.

Now scan to the left ... until the voltage reaches -3.4V (or -2.4V if it is a 2.4V Zener diode), there is only a tiny negative current flow, which for most purposes can be regarded as zero current.

BUT, as the negative voltage exceeds the -3.4V (e.g. moves to -3.5V), then the negative current increases dramatically. Oversimplifying, the diode appears to be determined to draw enough current to stop the voltage going beyond its Zener diode voltage, even if it self-destructs by overheating with resultant power flow.

-----------

Obviously, causing it to self-destruct is not a good design plan. So instead, a resistor R is introduced into the circuit, as shown above, to limit the current flow to a safe value.

-----------

Now back to some simple arithmetic, to understand what is happening. Using your requirements, of a 3V battery, a 2.4V Zener diode, and perhaps R = 900 Ohm resistor.

When this circuit is wired up, the voltage across the Zener diode will be 2.4V, because the 3V supply is in excess of the 2.4V diode breakdown voltage, and the Zener will draw enough current to bring the voltage across it to 2.4V. However, because there is a 900 Ohm resistor between the 3V battery and the Zener diode, the current will be limited, and may be calculated by Ohm's law.

As the battery voltage is 3V, and the voltage across the diode is 2.4V, then the voltage across the resistor must be (3V - 2.4V) = 0.6V

And the value of the resistor is 900 Ohm, so by Ohm's law, the current will be

I = 0.6 (V) / 900 (Ohms) = 0.000667 Amps = 0.667 milliAmps.

At this point, it is also possible to calculate the power dissipation in the Zener diode, to make sure it will not overheat.

Power (in Watts) = Voltage (in Volts) * Current (in Amps)

Power (Watts) = 2.4 (V) * 0.000667 (Amps)

or more conveniently:

Power (milliWatts) = 2.4 (V) * 0.667 (milliAmps) = 1.6 milliWatts

Which is safely far below the maximum limit of a 500 milliWatt Zener diode capability that you suggested.

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

So is this a useful circuit for you?

The simple answer is, I don't know!!

The reason for this ambiguous answer, is that you did not state the current or power demand of the load you were going to connect to the output.

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

Checking this Resistor + Zener diode circuit with a typical multimeter will show the output is 2.4V as expected, because the current draw of the meter will only be a few microAmps.

But imagine the load rsistance is 100 Ohms. Then the circuit will consist of a potential divider consisting of a 900 Ohm resistor and a 100 Ohm resistor across the 3V battery, so that only 1/10th of the battery voltage is across the 100 Ohm resistor, which is itself in parallel with the Zener diode.

1/10 th of 3V, is 0.3V. So the Zener diode has only (-) 0.3V across it, and so it will not conduct a significant current. But more importantly, the load will only get 0.3V, and 2.7V will be dropped across the 900 Ohm Resistor R. So although, the circuit will not self-destruct, it will not work either.

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

Can this be fixed?

The first possibility is to reduce the value of the resistor R, so that it will allow more current to flow, before the voltage drop across it is too high.

And since the power dissipation in the Zener diode was only 1.6 milliWatts, compared to 500 milliWatts maximum, there is clearly considerable flexibility to try this approach. Essentially, propose a new value for R, say 100 Ohms, and repeat the above calculation with new values.

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

Another consideration, is how long must a battery last? A good, fresh AA battery might have a capacity of 1000 mAmpHr. (This is a wild guess for this discussion, not an accurate value.)

Hence, with the current draw of 0.667 milliAmp, the life might be 1000/0.667 = 1500 hours .. about 2 months

In some situations, this might be acceptable, but not if the requirement is for one replacement every 12 months!

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

In summary, using a resistor and a Zener diode is a well known and simple method of obtaining a reference voltage within a circuit, albeit the voltage tolerance might be (something like) +/- 5%, which is too imprecise for some tasks.

However, it is often too inefficient as a power supply for a load requiring a substantial amount of power, especially if the power source itself is limited, such as when it is a battery.

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

I am sorry this is rather long, and may require you to re-read it several times, accompanied by copious amounts of your favourite beverage, but I hope it will eventually become clear, as it illustrates several points which are fundamental to understanding how circuits actually work. Do not be surprised if you find it tricky for a while ... you are not alone in the respect.

If there is something that is not clear, please ask. Hopefully someone on the forum will be able to clarify the picture. (And given the length of the sermon, it is probable I have left in a couple of confusing typos.)

Good luck and best wishes, Dave

Hi Dave,

Wow! Thanks for the ever so thorough answer to what I thought was a simple question. Like I said, I never dealt with zener diodes before as far as using them in a circuit. The load is very simple on my circuit as it is 3 LED's wired in parallel with an added one in the circuit to see if it is powered on or off. The 3 LED's have a 1.5V forward voltage and draw a max of 100mA but I run them at 60-75% for longevity as not to stress them. So figure 75mA for this discussion. The batteries are typically going to be 2-1.2 rechargeable AA batteries and they will be recharged as needed based on the length of time they are on. If someone were to mistakenly use two regular AA batteries I wanted to cover for that in case that would cause a problem and burn out the LED's. I did figure the difference in the resistors I would use if I designed it for a 3V battery vs the 2.4V setup running them both at 75% and the 3V version would require 3 - 20r resistors, while the 2.4 uses 3 - 12r resistors. I am pretty sure with the right fooling around I could get it adjusted where they would be bright enough with one size resistor for both scenarios without worrying about damaging them with the use of 3V, so I should reconsider that too. I will read you amazing reply over and over until it makes perfect sense but here is a link to the video where I watched and got the idea.

Start watching at the 9 min 45 seconds spot and you will see what I seen.

Thanks for the response,

Dave

Thanks,
Voltage

(@davee)
Member
Joined: 3 years ago
Posts: 1851

Hi @voltage,

I have only looked at the first part of your timed reference, and saw the bit with this diagram:

Which, I assume, is where you (understandably) got confused.

This guy appears to be talking sense, but (also understandably), is leaving out some of the details.

When we are discussing any issue, which might be politics, dressmaking, or electronics, if we know the person we are talking to has some knowledge in that subject, then we will concentrate on the main point and assume the other person will use their knowledge to fill in the missing detail.

With electronics, (and maybe the other topics as well), part of this implied detail includes realising that every component has implicit characteristics as well. For example, every resistor obviously has a 'resistance' characteristic, but it also has some inductance and some capacitance. These 'extra' characteristics are not 'optional'. Indeed, the manufacturer MAY have tried to design the part with minimal inductance and capacitance, but if you look hard enough, they will be there.

Now in many cases, the 'extra' characteristics are small enough to be ignored, but sometimes they become large enough to be visible, and occasionally, even useful, to the point of being essential.

-------

In this context, the discussion starts with a voltage source, which in your circuit is a battery, and in the circuit in the sketch, where the presenter is referring to the right hand half of the circuit sketch, then the voltage source is the secondary winding of the transformer.

Start by considering a battery made of 2 AA alkaline cells in series. If they are new and unused, a voltmeter might show a voltage of about 3.2V, when they are not connected to a load. If a similar battery, which had been in use for a while, but not yet completely exhausted, then the voltage might be 3V.

Now connect a 3V (tungsten filament) bulb that draws about 0.3A when the voltage is 3V, from each of the batteries, and measure the voltage again.

The new battery will probably cause the bulb to burn brightly, and the voltage will fall slightly to (say) 3.1V.

However, the used battery will struggle to produce more than an orange glow, and the voltage might fall to about 1.5V.

The explanation for these observations is that the battery consists of a 'perfect' voltage source, in series with an 'invisible' resistor, typically called the 'internal resistance'.

For reasons associated with the chemistry of the battery, the 'perfect voltage source' of a new battery is showing a voltage of about 3.2V, whilst the used battery is down to 3.0V. The voltmeter only draws a minuscule current, so that the voltage drop across the  'internal resistance' is negligible, and the voltmeter shows the voltage of the source, unaffected by the internal resistance.

However, when the bulb is connected, the voltage drop from 3.2V to 3.1V for the new battery, and from 3.0V to 1.5V for the used battery is due to the internal resistance of the battery, which is dropping 0.1V and 1.5V in the respective cases.

Now, you should understand that the 'perfect voltage source' and the 'internal resistance' in series are not real components ... if you dissected the battery cells, you would never find them! Instead, they are mathematical models that can (roughly, at least) enable a circuit designer to calculate how a circuit will behave. The calculations might be simple 'hand calculations' or built into a simulator programme, like LTSpice.

Of course, for something like a battery, these 'extra characteristics' are not constant values, because the values change as the battery ages. The 'extra characteristics' of most electronic components tend to remain (approximately) unchanged for the life of the component, but there are some components, such as electrolytic capacitors, that also 'age' with time and usage.

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

Applying the above principle to the transformer secondary in the video explanation, the windings of the transformer will probably be made of a very thin wire, which will have an appreciable resistance. This and other characteristics of the circuit will mean that the transformer output will have an appreciable 'internal resistance', so that if the Zener diode starts to conduct, then the output voltage from the transformer will fall. Hence, the Zener diode will only be required a limited amount of power to prevent the voltage exceeding the safe limit.

(I didn't carefully listen to the discussion on the video, but I think I heard that a common fault was finding the Zener diode had failed, suggesting the designer might have been overconfident about the design characteristics of the circuit. Designing reliable circuits is harder than just designing one that works during a bench test!)

BTW, you might see discussions using the term 'internal impedance' instead of 'internal resistance'. Simplistically, they are discussing the same thing ... 'resistance' refers to DC only situation, whilst 'impedance' refers to an AC situation.  The maths of impedance calculations is a bit more tricky, but the general effect is the same.

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

Summarising so far, the circuit sketch on the right-hand side, in the video consists of a transformer secondary, a Zener diode and a 'hidden' internal resistance in series, the last of which reduces the amount of current the Zener diode will need to pass to limit the output voltage.

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

Curiously, the circuit on the left hand half of the screenshot shows a second Zener diode in series with a resistor. This message is already too long, but these two components, plus the attached transistor, shows a classic circuit block for a voltage regulator, which is more efficient and less demanding upon the Zener diode.

Oversimplifying, the resistor and Zener diode pass a small current, with the voltage at the Zener's anode corresponding to the component's Zener voltage ... 12V in this case.

This 'reference' 12V source is then connected to the base of the transistor and determines the output voltage to be about 0.6V less than the base voltage. Hence, the resistor and Zener diode are providing a low current 12V reference, and the transistor is doing the 'heavy work' of supplying current to the load, at the required voltage.

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

Hope this helps ... sorry, you may need to read this more than once as well.

Best wishes, Dave

Voltage reacted
(@voltage)
Member
Joined: 3 years ago
Posts: 193
Topic starter

Posted by: @davee

Hi @voltage,

I have only looked at the first part of your timed reference, and saw the bit with this diagram:

-- attachment is not available --

Which, I assume, is where you (understandably) got confused.

This guy appears to be talking sense, but (also understandably), is leaving out some of the details.

When we are discussing any issue, which might be politics, dressmaking, or electronics, if we know the person we are talking to has some knowledge in that subject, then we will concentrate on the main point and assume the other person will use their knowledge to fill in the missing detail.

With electronics, (and maybe the other topics as well), part of this implied detail includes realising that every component has implicit characteristics as well. For example, every resistor obviously has a 'resistance' characteristic, but it also has some inductance and some capacitance. These 'extra' characteristics are not 'optional'. Indeed, the manufacturer MAY have tried to design the part with minimal inductance and capacitance, but if you look hard enough, they will be there.

Now in many cases, the 'extra' characteristics are small enough to be ignored, but sometimes they become large enough to be visible, and occasionally, even useful, to the point of being essential.

-------

In this context, the discussion starts with a voltage source, which in your circuit is a battery, and in the circuit in the sketch, where the presenter is referring to the right hand half of the circuit sketch, then the voltage source is the secondary winding of the transformer.

Start by considering a battery made of 2 AA alkaline cells in series. If they are new and unused, a voltmeter might show a voltage of about 3.2V, when they are not connected to a load. If a similar battery, which had been in use for a while, but not yet completely exhausted, then the voltage might be 3V.

Now connect a 3V (tungsten filament) bulb that draws about 0.3A when the voltage is 3V, from each of the batteries, and measure the voltage again.

The new battery will probably cause the bulb to burn brightly, and the voltage will fall slightly to (say) 3.1V.

However, the used battery will struggle to produce more than an orange glow, and the voltage might fall to about 1.5V.

The explanation for these observations is that the battery consists of a 'perfect' voltage source, in series with an 'invisible' resistor, typically called the 'internal resistance'.

For reasons associated with the chemistry of the battery, the 'perfect voltage source' of a new battery is showing a voltage of about 3.2V, whilst the used battery is down to 3.0V. The voltmeter only draws a minuscule current, so that the voltage drop across the  'internal resistance' is negligible, and the voltmeter shows the voltage of the source, unaffected by the internal resistance.

However, when the bulb is connected, the voltage drop from 3.2V to 3.1V for the new battery, and from 3.0V to 1.5V for the used battery is due to the internal resistance of the battery, which is dropping 0.1V and 1.5V in the respective cases.

Now, you should understand that the 'perfect voltage source' and the 'internal resistance' in series are not real components ... if you dissected the battery cells, you would never find them! Instead, they are mathematical models that can (roughly, at least) enable a circuit designer to calculate how a circuit will behave. The calculations might be simple 'hand calculations' or built into a simulator programme, like LTSpice.

Of course, for something like a battery, these 'extra characteristics' are not constant values, because the values change as the battery ages. The 'extra characteristics' of most electronic components tend to remain (approximately) unchanged for the life of the component, but there are some components, such as electrolytic capacitors, that also 'age' with time and usage.

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

Applying the above principle to the transformer secondary in the video explanation, the windings of the transformer will probably be made of a very thin wire, which will have an appreciable resistance. This and other characteristics of the circuit will mean that the transformer output will have an appreciable 'internal resistance', so that if the Zener diode starts to conduct, then the output voltage from the transformer will fall. Hence, the Zener diode will only be required a limited amount of power to prevent the voltage exceeding the safe limit.

(I didn't carefully listen to the discussion on the video, but I think I heard that a common fault was finding the Zener diode had failed, suggesting the designer might have been overconfident about the design characteristics of the circuit. Designing reliable circuits is harder than just designing one that works during a bench test!)

BTW, you might see discussions using the term 'internal impedance' instead of 'internal resistance'. Simplistically, they are discussing the same thing ... 'resistance' refers to DC only situation, whilst 'impedance' refers to an AC situation.  The maths of impedance calculations is a bit more tricky, but the general effect is the same.

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

Summarising so far, the circuit sketch on the right-hand side, in the video consists of a transformer secondary, a Zener diode and a 'hidden' internal resistance in series, the last of which reduces the amount of current the Zener diode will need to pass to limit the output voltage.

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

Curiously, the circuit on the left hand half of the screenshot shows a second Zener diode in series with a resistor. This message is already too long, but these two components, plus the attached transistor, shows a classic circuit block for a voltage regulator, which is more efficient and less demanding upon the Zener diode.

Oversimplifying, the resistor and Zener diode pass a small current, with the voltage at the Zener's anode corresponding to the component's Zener voltage ... 12V in this case.

This 'reference' 12V source is then connected to the base of the transistor and determines the output voltage to be about 0.6V less than the base voltage. Hence, the resistor and Zener diode are providing a low current 12V reference, and the transistor is doing the 'heavy work' of supplying current to the load, at the required voltage.

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

Hope this helps ... sorry, you may need to read this more than once as well.

Best wishes, Dave

Hi Dave,

I have read it word for word and I understand now that a zener is not the answer to my project without complicating it more than necessary and likely impossible in the end due to lack of voltage/current delivered. I did realize the example I showed you was using a transformer for the source voltage and also that it was AC, so I was expecting that to be a likely issue when it came to assuming it would work with DC voltage. I have also watched some more good videos that dumb it down to the basics and ohms law so I can clearly see why my initial thoughts were wrong. I knew it may have been too easy and too good to be true. I guess that is how I learn sometimes, by having an idea, doing some quick research and falling for the easiest answer and hoping for the best. 🤔 But the good thing is now I will keep studying and learning all about zener diodes until I feel if and when to use them, and also how to check them when trying to repair something. So it's a win win situation. I also understand the reference to the transistor use in that video being used for the heavy lifting. Same as a low voltage primary circuit switching a relay that handles the high voltage and current in a control panel for example. I learned a lot between your help and some added studying because of your help to make me even more interested. I appreciate the time you took to write the long detailed replies. My little project will be fine without the zener but in the future I will be a little smarter. 😀

Thank you,

Dave

Thanks,
Voltage

DaveE reacted
(@davee)
Member
Joined: 3 years ago
Posts: 1851

Hi @voltage,

Thanks for your reply. I am sorry it is a bit of a negative tone, but sometimes reality can be tough.

For a bit of entertainment, I decided to a simple simulation of a circuit, using LTSpice (which can be downloaded for free, from the analog.com website). Simulations can be a bit tricky to start with, but can also be informative, and the circuit never dies with a puff of magic smoke, even with a million watts dissipation in a 0.5W transistor.

The circuit is essentially the same one as shown on the left-hand side of the video, adapted to supply 100mA at 2.2V output.

Instead of LEDs, there is a 22 Ohm resistor, and instead of the battery, there is a voltage source that steadily increases from 2V to 3.5V, so that it is possible to see how the circuit would react to any battery voltage in that range.

(Click on image to enlarge.)

The transistor Q1, is a small, widely cloned silicon npn device (2N2222), and the Zener diode was 'tuned' to produce 2.2V output Of course, 'tuning' a Zener diode voltage is only realistically possible in a simulator.

The plot shows the voltage supplied to 'LoadResistor' in Dark Blue, and the input (battery) voltage (V1) in light green.

When V1 is in the range 2V to 3.3V, the voltage to LoadResistor is about 0.8V less than V1, due to the voltage drop across the transistor. (This voltage drop will be a bit different for each transistor, but the differences will be small.)

This means the voltage passed to LoadResistor will only reach 2.4V when the incoming voltage, V1 is at least 3.2V.

However, further increasing V1, beyond 3.2V, does not result in an increase in the voltage sent to LoadResistor.

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

I hope this helps you to understand the general picture, although sadly it does not help with your specific requirement, which could require a rather more complex circuit. It is possible you could find a switch mode power supply chip, that would include most of much more complex circuit, within the device.

Best wishes and good luck, Dave

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(@voltage)
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Posted by: @davee

Hi @voltage,

Thanks for your reply. I am sorry it is a bit of a negative tone, but sometimes reality can be tough.

For a bit of entertainment, I decided to a simple simulation of a circuit, using LTSpice (which can be downloaded for free, from the analog.com website). Simulations can be a bit tricky to start with, but can also be informative, and the circuit never dies with a puff of magic smoke, even with a million watts dissipation in a 0.5W transistor.

The circuit is essentially the same one as shown on the left-hand side of the video, adapted to supply 100mA at 2.2V output.

Instead of LEDs, there is a 22 Ohm resistor, and instead of the battery, there is a voltage source that steadily increases from 2V to 3.5V, so that it is possible to see how the circuit would react to any battery voltage in that range.

-- attachment is not available --

(Click on image to enlarge.)

The transistor Q1, is a small, widely cloned silicon npn device (2N2222), and the Zener diode was 'tuned' to produce 2.2V output Of course, 'tuning' a Zener diode voltage is only realistically possible in a simulator.

The plot shows the voltage supplied to 'LoadResistor' in Dark Blue, and the input (battery) voltage (V1) in light green.

When V1 is in the range 2V to 3.3V, the voltage to LoadResistor is about 0.8V less than V1, due to the voltage drop across the transistor. (This voltage drop will be a bit different for each transistor, but the differences will be small.)

This means the voltage passed to LoadResistor will only reach 2.4V when the incoming voltage, V1 is at least 3.2V.

However, further increasing V1, beyond 3.2V, does not result in an increase in the voltage sent to LoadResistor.

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

I hope this helps you to understand the general picture, although sadly it does not help with your specific requirement, which could require a rather more complex circuit. It is possible you could find a switch mode power supply chip, that would include most of much more complex circuit, within the device.

Best wishes and good luck, Dave

Hi @DaveE,

No problem on the way my little query ended without success, but you stopped me just in time from ordering a bunch of zener diodes of various values. 😋 I have LTSpice on my PC but have mainly used it where I have downloaded the file that someone else created for whatever the subject was at that time. But looking at you image I see what point you are making. Also, the circuit would be overkill for what I was trying to accomplish anyhow. My little circuit as it stands is as simple as they come and I thought maybe I could enhance it with a simple mod. No luck. The good news is that I learned a lot about zener diodes that I did not know before. And I also learned I have a lot of learning to do in the electronics field. I am good with electricity as far as 120/240 stuff and have built things using arduino's, and I even make pcb's, but electronics is very interesting and I am learning new stuff every day as fast as I can. And with help from people like you that makes it even better! I really appreciate the time you took to explain all of this and not make me feel like a complete dummy. 😎 I can repair electronics easier than designing them. Thanks again for all the help.

Here is an image of the control panel I designed and built from scratch for my powder coat oven that I also built from scratch. Just trying to feel better after my zener fail. JK. 😆

Thanks,
Voltage

DaveE reacted