I need a bit of help with Zeners

[bluefoxicy]

I'm no EE so I don't understand this !@%* please help.

http://www.geofex.com/Article_Folder...osfetfolly.htm

The B+ reducer and the standby switch (at the end) I don't understand. I've been all over Google and found some interesting stuff..

AN007 - High Power Zener Diode

That seems to be the B+ reducer.

Too bad nothing I've read has actually explained how to clamp a DC circuit with a Zener (I want to get a 36V, 24V, 18V, 12V, and 9V source out of a rectified, filtered 36VCT transformer), or how these things actually work. :\

Any takers?

[R.G.]

I'm no EE so I don't understand this !@%* please help.

http://www.geofex.com/Article_Folder...osfetfolly.htm

The B+ reducer and the standby switch (at the end) I don't understand. I've been all over Google and found some interesting stuff..

AN007 - High Power Zener Diode

That seems to be the B+ reducer.

Too bad nothing I've read has actually explained how to clamp a DC circuit with a Zener (I want to get a 36V, 24V, 18V, 12V, and 9V source out of a rectified, filtered 36VCT transformer), or how these things actually work. :\

Any takers?

I'll try to help if I can. The B+ reducer is takes advantage of the fact that a zener does not conduct at all until the voltage across it is over it's specified voltage, then conducts a lot; also that most power MOSFETs do not conduct at all until their gate is about 3-4V higher than their source, then they conduct about 1A/V.

Ignoring the MOSFET for the moment, what happens as B+ rises from 0V? Nothing happens until the B+ gets to at least the zener votlage, and then the zener conducts, letting current into the 2K resistor. The resistor then responds by V=I*R to voltage across it, and as the B+ continues to increase, any voltage greater than the zener voltage appears across the 2K resistor.

If the voltage were allowed to continue to increase, it would be fatal for the 2K, but now we think about the MOSFET. The MOSFET does nothing until the voltage across the 2K hits its threshold voltage, then it begins allowing current through. But in letting current through, it steals current from the high side of the zener that would otherwise go through the 2K, so the voltage across the 2K must go down. This is a kind of negative feedback and it settles where the MOSFET lets just enough current flow through the zener/2K resistor to keep the zener conducting a little. The 220 ohm resistor in series with the gate of the MOSFET is a gate stopper, preventing radio frequency oscillation.

The zener/2K/MOSFET loop works so the MOSFET conducts just enough to keep the zener conducting, so the voltage across the MOSFET must be the zener voltage plus the threshold voltage on the gate. As B+ increases even more, the MOSFET source floats up, but it still keeps about the zener voltage plus 3-4V across its drain-source, no matter how much current flows through it. So the MOSFET voltage is effectively subtracted from the B+.

The standby switch is even simpler. Most standby switches open up the B+ to the output tubes. But you can do the same thing by opening up the cathodes of the power tubes with a switch. Either way, current can't flow through the output tubes and no sound comes out. Using a real metal contact switch can be difficult with high voltage/high current tubes, so we substitute a power MOSFET for the switch. The MOSFET either conducts a whole lot (when it's gate is pulled high) or conducts nothing at all (when its gate is pulled low).

The zener diode and resistor to B+ is just a way to make a source of 12Vdc out of B+. The 12K resistor limits current through the switch to ground. When the switch to ground is closed, the MOSFET gate is pulled to ground through the 220 ohm gate stopper, for RF oscillation again. The 12K is also pulled to ground, and 1ma of current from the zener diode/B+ power supply is pulled through it. With 0V on its gate, the MOSFET cannot conduct, and so no current (except for the 10K, we'll get to that) can flow, and no audio happens, just like with a mechanical switch.

When the switch to ground is opened, the 12K resistor charges up the MOSFET gate, which is a very, very high impedance, to 12Vdc from the 12V zener/B+ supply, and the MOSFET conducts very heavily, "shorting" the cathodes of the power tubes to ground and letting the amp work.

Now - why is that 10K there? We have a perfectly good imitation of a metal-contact switch in the MOSFET and the 10K ruins it.

Well, almost. With 10K on the cathodes, the power tubes can't conduct much current, but they do a little. There is a problem with tubes running for long periods of time with voltage across them but no current. The cathodes and plates can get "poisoned" after long times like this. The 10K lets them conduct enough to prevent poisoning. But it's probably not necessary, since poisoning is a slow process. You'll probably wear out the tubes first.

What is still not clear?

[R.G.]

Too bad nothing I've read has actually explained how to clamp a DC circuit with a Zener (I want to get a 36V, 24V, 18V, 12V, and 9V source out of a rectified, filtered 36VCT transformer), or how these things actually work.

Sorry - I quit too soon.

Zener diodes are a device that conducts no current until the voltage across it increases to it's rated voltage. At that point, the internal junction starts leaking current. It's kind of like a bucket filling with water. If the height of the water is like voltage and water coming out of the bucket is like current, nothing happens as the water rises in the bucket, until it hits the edge of the bucket. Then water overflows the bucket and the water level (voltage) gets no higher even though lots of current (water flow) goes out of the bucket.

As for what you want to do, you're in for some difficulties. You don't say how much current you want from the various voltages. Unless the currents are pretty small, you're going to have some problems with using zeners to make those voltages. Here's why.

The power anything produces as heat is equal to the voltage across it times the current through it. In a zener, for voltages below the rated voltage, no current flows, so power is zero (ideally; there's always some leakage, slop, etc.). When the zener starts conducting, it no longer limits current, and it allows very large currents to flow if the power source feeding it is not current limited. Zeners are rated in terms of their voltage and their power. A 10V zener which is rated for 1/2W can only conduct up to 0.5/10 = 0.05A. At that current the power is P = Volts x amps = 10V*0.05A = 0.5W. Any more and the zener burns up.

This is why you can never, ever use a zener without some form of current limiting. For simple power supplies, this is a resistor. To design this kind of voltage limiter, you figure out what voltage you have, what the zener voltage is, and subtract. This is the excess voltage that we have to keep from killing the zener. We can figure out what the max current for the zener is by dividing its power rating by its voltage. We then have to calculate a resistor that will limit the current into the zener to less than the disaster current while the resistor has the difference between the bulk DC power and the zener voltage across it.

In your case, you're rectifying a 36V transformer. That's going to give you 36V*1.414 = 50.9Vdc. Call it 50V since there's some rectifier drops I've ignored. To get DC levels of 36, 24V, 18V, 12V, and 9V out, the voltage difference is 14, 26, 32, 38, and 41V.

Zeners come in 1/2W, 1W, and 5W packages standard. Some may come in higher, lower, or in-between values, but these are common. If you want to make 9V with a resistor and zener from 50Vdc, you can only make 122ma of 9Vdc with a 5W zener. Smaller zeners get you 24ma for a 1W zener and 12ma for a 1/2W zener. It's also quite wasteful, because the resistor is dissipating more power than the zener is.

Notice that I have not included any load current. I have, in effect, assumed the load current may be as little as 0ma. If that's not true, and the load will never, ever go below some minimum current, then you can allow the resistor to let that minimum current through, and have the zener only "eat" currents between the load minimum and the load maximum. This can help. But as an EE, I think it's a very poor practice, because ensuring a load never goes to zero is chancy. Are you **sure** your bass player won't ever unplug things? And if the resistor/zener are calculated to have the zener survive with a minimum load, when the load is unplugged, the zener becomes a Darkness Emitting Diode - also known as a DED.

Zener power supplies are tricky, picky, and unreliable.

Ugh! Zeners! Ptooey!

[bluefoxicy]

Sorry - I quit too soon.

... hah. Well hello. Yeah, I'd assumed the stuff you wrote expected at least a basic understanding of more than just Ohm's Law. Unfortunately, that's exactly what I don't have.

Unless the currents are pretty small, you're going to have some problems with using zeners to make those voltages. Here's why.

Yeah, I thought there was a way to do this with a transistor that can pass lots of current, and a Zener to limit it... which is exactly what I don't understand, htf does that work?

AN007 - High Power Zener Diode

That makes no sense to me.

To get DC levels of 36, 24V, 18V, 12V, and 9V out, the voltage difference is 14, 26, 32, 38, and 41V.

So it's a drop of $E and not a limit of $E? I thought Zeners were used to i.e. deal with a floating power supply, like a battery (slowly drops voltage as it dies)....

Zener power supplies are tricky, picky, and unreliable.

Ahh... I'm trying to build a prototyping board with a variable load and might be using extra leads (like the 18V and 9V sources at once), which will mess with things. I was trying to get a stable voltage out with a zener, as here:

Zener Diode Voltage Regulator - Electric Circuit

From what you've said, and from looking at this thing and using my brain a bit more, I think I get it... the Zener blocks i.e. 12V of whatever, so if you pass 14V it shorts 2V to ground and 12V goes to the circuit; if you pass 40V it shorts 28V to ground and 12V goes to the circuit. And of course, lots of current. Right?

So in order to use that topology at all, the voltage has to be above but CLOSE to the voltage I want, and I just shunt the excess to ground (i.e. short it out). :s

Thanks, you've clarified some things. Now if I can only figure out how to build this power supply...

[R.G.]

So it's a drop of $E and not a limit of $E? I thought Zeners were used to i.e. deal with a floating power supply, like a battery (slowly drops voltage as it dies)....

Oh. You need a new picture, and a few new concepts.

1. Ideal voltage source. This is an imaginary, perfect device that can put out its rated voltage **NO MATTER WHAT LOAD** is attached to it. If you have an ideal voltage source of 12Vdc, and you connect a 12K resistor across it, then 12v/12K = 0.001A (1mA) flows. If you connect a 1 ohm resistor to it, then a current of 12V/1 ohm = 12Amperes flows. If you connect across it a copper buss bar with a resistance of 0.001 ohm, then 12 thousand amperes flow.
A car battery is as close to an ideal voltage source as we get in the common world.
2. The sum of voltages around an electrical loop is always zero. This is a fancy and formal way of saying that if you have a source, then the voltage from that source is equal to the sum of the voltages dropped in things connected to the source. Let's go back to the 12V ideal voltage source. We will connect two resistors across it, one a 1K and one a 2K resistor. Starting at the bottom/minus of the battery, we go up to the top and gain 12V. Then we go through the 1K resistor, then the 2K resistor, and arrive back at the bottom/minus of the battery.This rule tells us that the 12V is equal to the voltages dropped across the resistors. We can figure that out from ohm's law, because the total resistance is 3K, and so the total current from the battery must be 12V/3K = 4mA.

That same current must flow through both the 1K and 2K resistors - it has to, they're in series - so the voltage dropped in the 1K is 1K * 4ma = 4V, and it's most positive on the side touching the +12V terminal. The voltage across the 2K must be 2K*4ma = 8V, and it's most negative on the side touching the battery negative. The sum of the two voltages, 4V and 8V is equal to the 12V battery, as the rule says it must be. And the voltages are polarized so that from 12V at the top end, the voltage drops 4V to +8 at the top of the 2K, and then to 0V at the bottom of the 2K and the bottom/minus of the battery.

3. Real world voltage sources may always be modelled as an ideal voltage source in series with an impedance (think "resistance" until you understand impedance better.) You can think of every battery and power supply in the world as being an ideal voltage source with some resistor in series with it. The difference between voltage sources is the value of that internal resistor you can't get at.

But we can always calculate this resistance by measuring the open circuit voltage of the source and the short circuit current. This is because the open circuit voltage is measured with zero current flowing, so the voltage drop across the internal resistor is zero. The short circuit current is the current which is caused by the ideal voltage source driving just the internal resistor. So we can calculate the internal resistor as the open circuit voltage (that is, the same voltage as the ideal voltage source inside) divided by the short circuit current, which is limited only by the internal resistance.

For instance, if you take a fresh 9V battery and hook a current meter directly across it, you can often measure currents of up to 1A. That tells us that the internal resistance (plus any resistance of the current meter, which we're right now going to ignore!) is 9V/1A = 9 ohms.

So remember: Every voltage source in the real world can be thought of as an ideal voltage source in series with some internal resistor (which can be calculated) and the voltages across parts connected to a voltage source must equal the voltage source.

So let's hook up some stuff. Let's take a 12V car battery and hook a 9V zener across it. What happens?

There is a small explosion as the parts of the zener fly across the room, possibly some smoke and flames. The 12V car battery, which can supply maybe several hundred amperes for an instant, tries to force the 9V zener to have 12V across it. What that *really* says is that the current is only limited by the internal resistor of the car battery, and that is so small that hundreds of amperes flow and the zener dissipates 9V times 100's of amperes, or about a thousand watts per 100A.

That current flows because the only resistance limiting current was the internal resistance of the battery. Actually, the zener and the wires leading to it have some resistance, too. But not much.

Remember a zener diode is a diode that lets no current flow until its breakover voltage is reached, then it lets almost any current flow for voltages even slightly higher. The zener does not limit current. It only withstands voltage.

To make a zener supply, you need (1) a voltage source higher than the zener voltage, (2) a zener, and (3) some way to limit that current. In the case of our 12V car battery and a 9V zener, we need ... a resistor! This is in addition to the battery's internal resistance, because the battery internal resistance is too small, and does not limit the current enough.

So our power supply has a voltage source, a resistor from the voltage source, and a zener back to the battery. Since the voltage across the resistor and zener together has to be the battery voltage they are hooked across, and the zener is a fixed voltage, the difference between the battery voltage and the zener voltage appears across the resistor.

So all we have to do is figure out how much current the zener can take. To do that, we take the zener's power rating and divide by the zener voltage. `1/2W zeners are common; if we have a 9V, 1/2w zener, then the current it can withstand before burning up is 0.5W/9V = 0.055A, or 55ma.

The difference between 12V and 9V is 3V across that resistor, so the resistor value is 3V/0.055A = 54 ohms.

What happens when we load down this new 9V power supply?

The zener keeps the voltage across it constant at 9V. The resistor sees a constant voltage across it, so it passes a constant current of 55ma. The load eats some of the 55ma away from the zener. If the load is 10ma, then the load gets 10ma and the zener is left with only 45ma - which is fine with the zener, it just runs a little cooler.

When the load current increases to 55ma, then the zener current reduces to 0. But the load still sees 9V, limited by the 54 ohm resistor. It's only when the load goes over 55ma that the voltage across the 54 ohm limiting resistor gets so big that the zener doesn't have 9V across it any more, so the output voltage drops. The zener can't "regulate" it because it doesn't have enough voltage to work with.

And that's how zener regulated power supplies work - the zener sets the voltage, some limiting resistor (or, in fancy ones, transistors, etc.) limits the maximum current going to the load plus zener diode. The zener eats all the current that the load doesn't want.

[Steve Conner]

LM317 Variable Power Supply

Bluefoxicy: Suggest you go build some circuits (maybe starting with the LM317 circuit above?) watch them go on fire, and sit scratching your head until you figure out why they went on fire. Repeat until the circuits work without catching fire. Get your bad self a copy of Horowitz and Hill's "The Art Of Electronics". It'll fill in the big picture for you better than any number of websites and forums ever could.

[R.G.]

LM317 Variable Power Supply

Bluefoxicy: Suggest you go build some circuits (maybe starting with the LM317 circuit above?) watch them go on fire, and sit scratching your head until you figure out why they went on fire. Repeat until the circuits work without catching fire. Get your bad self a copy of Horowitz and Hill's "The Art Of Electronics". It'll fill in the big picture for you better than any number of websites and forums ever could.

Butbutbutbut then how would I ever keep up with my typing practice??

[Steve Conner]

Sorry RG, it never occurred to me that you might actually enjoy doing this.

The Art of Electronics is still a damn good book, though.

[gbono]

Remember that a zener's voltage is specified at a given current (Izt). If you force more or less current then the Izt your Vz will vary from specified.

[Steve Conner]

Yes, for instance in RG's car battery example, Vz was probably about 11.5V until it exploded.

And likewise when sitting in a drawer, Vz is 0V. Or is it? Does the light really go out when you shut the fridge door?

[gbono]

Mmmm 11.5V??? Try more like 35V. The problem is even more obvious when nowitall designers think they can clamp the input of an op amp with a zener voltage characterized in the 10's or 100's of mA and there application is in the uA range for current.

You can always shut yourself up in the fridge to see if the light goes out.

[Enzo]

11.5. Hard to get 35v from a car battery.

If I read it right, he meant the car battery dropped from 12v down to 11.5 with the copper bus bar across it.... until it exploded.

[gbono]

Okay, duh, and my point is - zener diode voltage is specified at a given current - one should read the data sheet and avoid disappointment.

[Steve Conner]

Yes. We actually had this very problem with the undervoltage lockout circuit in a portable instrument we were designing. The UVLO reference was a 2.7v zener, but the actual voltage you got from it varied a lot with zener current, and that made the UVLO pretty hit-or-miss. Then the manufacturer got onto a new batch of zeners and it stopped working altogether!

This was a reference design from the single board computer vendor, that I just copied without testing it too carefully. I ended up replacing the offending diode with a LM385-2.5 reference, which is what should have been used in the first place: its "zener" voltage doesn't change by more than a few millivolts as the current varies from uA to mA. Moral of the story, the low voltage zeners are really awful. Anyone ever try them as clipping diodes in a Tube Screamer?

Yes Enzo, that's what I meant.

[gbono]

NXP (e.g Philips) used to make a line of "low current" zeners just for similar applications - these devices had razor sharp voltage transitions (knee voltage). Zeners also exhibit some interesting TC characteristics above and below the 6 V range. Very interesting things happen to the diode's junction around the zener voltage transistion. Do not use these in a circuit that requires very good phase noise performance.