Diode drop to match transistor VBE drop question

**See below post for actually accurate information regarding this question**

So my inquiry is specifically in regards to Electrodyne line amps. They have a push pull output, and use two diodes between the power supply rail and the collector to the "bottom" transistor (the "top" transistor's collector is connected directly to the +V input).

Many of the diodes in this console are way out of spec, and some even measure almost like wire (in both directions).

Is it important that the voltage drop across the two transistors match the diode drops of the two diodes exactly?

Am I making sense? I don't have the schematic right in front of me to maybe be more clear about this.

Without seeing the schematic I can't answer intelligently.. but that won't stop me.  ;D

The most common failure mode for diodes is as a dead short, so if you measure low ohms across a diode in both direction, it is not going to function as either a steering diode, or for make a useful voltage drop.

JR

Thanks, I'll have to peek at the schematic again.

How much do voltage drops across different types of diodes differ? Or drops across transistor junctions?

(silicon) transistors and diodes, will generally be in the .3 to .6V range.. It varies with current and other factors.  There are lower forward drop diodes made from materials other than silicon, but they are not the most common.

All (good) diodes will only conduct in one direction.

JR

At first I thought you meant the base bias diodes. Look at the voltage across the emitter resistors. If less than 30mV, you have crossover distortion, and if too high the transistors cook.

It is astonishing how often two unselected rectifiers bias two medium transistors well-enough.

BUT:

> two diodes between the power supply rail and the collector to the "bottom" transistor (the "top" transistor's collector is connected directly to the +V input).

That's quite strange. A wire-short diode in that position would seem to lock-up the output, if not smoke the power supply? Or it may be an anti-kickback diode, normally reverse bias and not affecting operation (if not shorted). Or it may be something else?

And I dunno what "almost like wire" really is. In-circuit you may be measuring other stuff.

I'll be the coward and ask for a schematic before I attempt intelligence.

> All (good) diodes will only conduct in one direction.

Well, a Zener will often show 0.6V one way and 7V the other way. (Any diode will conduct backward, if you have enough voltage!) And since DMM ohms tests are usually <200mV, diodes often don't conduct enough to be sure about.

OK you got me...    but who says zeners are "good" diodes..  ;D ;D

JR

Ok, so I definitely wasn't remembering correctly.


By measure like wire, I mean like John said, short in either direction. This measurement is in circuit, but one end open (in the console channels anyways).

I'm still learning my transistor theory, I have a fuzzy idea of what these diodes are doing in the circuit. These diodes are obviously matching the VBE drop across the two transistors, but what exactly is the ultimate function they're serving? And how important is it to match the VBE drops with the diode drops (I'm talking in regards to installing new transistors, not whether the old shorted ones will work fine or not)?

Thanks!

based on the schematic, look at Class AB push-pull outputs..

Well I guess it could be class B as well depending on what biasing is coming from that opamp.

Anyway, the diodes and the resistors create a small voltage bias on the bases of the BJTs to keep them in the class.  I bet those diodes are attached to the BJTs somehow too since they should be thermally coupled to the BJTs.

bobschwenkler said:

Ok, so I definitely wasn't remembering correctly.

These diodes are obviously matching the VBE drop across the two transistors, but what exactly is the ultimate function they're serving?

And how important is it to match the VBE drops with the diode drops (I'm talking in regards to installing new transistors, not whether the old shorted ones will work fine or not)?

Thanks!

Click to expand...

There is a list of things I don't like about that circuit.. mainly things left poorly controlled for wrt production variances.

To answer your question of what are the diodes doing.. it kind of depends on what the -V supply is?  The transistor stage appears to be class AB but the class A bias voltage is not necessarily defined by the diodes if the resistor divider formed by R3/R4 is less than two diode drops across R3 at nominal AC 0v output.  Since the -V is not visible on that schematic I will use that as an opportunity to not answer, my guess is that the diodes are not conducting at equilibrium.

One thing that is likely is for large positive signal swings, long enough for the cap C2 to charge up, the voltage across r3 could increase and the diodes would then conduct and clamp at 2 diodes, or roughly 2x Vbe. 

For a circuit like this where the drivers are not doing heavy lifting and/or biased up with a lot of class A current, I would be very tempted to just add a small resistor in series with one transistor emitter (or perhaps one in each). Even something as small as a 5 ohm resistor will dramatically reduce the sensitivity of how accurately the base voltage needs to track the actual transistors.  Vbe voltage increases will increase the class A current exponentially, while having a resistor in series with the emitter makes it more linear and less likely to run away thermally.   

If that node is driving the outside world, you can add a couple more diodes in connection with the emitter degeneration resistors to current limit the output and prevent failures due to accidental short circuits, etc.

JR

Svart said:

based on the schematic, look at Class AB push-pull outputs..

Well I guess it could be class B as well depending on what biasing is coming from that opamp.

Anyway, the diodes and the resistors create a small voltage bias on the bases of the BJTs to keep them in the class.  I bet those diodes are attached to the BJTs somehow too since they should be thermally coupled to the BJTs.

Click to expand...

The diodes are not physically coupled to the BJTs.

it kind of depends on what the -V supply is? 

Click to expand...

This whole console runs off of +24, so this circuit is sitting between that supply rail and ground.

The transistors are not physically coupled to the BJTs.

Click to expand...

BJT= Bipolar Junction Transistor

I think you meant that the diodes are not physically coupled to the BJTs?

hmm.  usually in class AB you couple them so that the BJT currents don't runaway and cause a meltdown.

Oops, typo. I did mean that.

Do diodes tend to have a positive tempco then? Would it be increasing resistance, or increasing voltage drop?

if it's a zener, then below 5.6v rating, it's negative, above, it's positive.

Silicons are generally negative.

It's changes in currents.  It's not resistance or voltage drop(other than those noticed via Ohm's law).

bobschwenkler said:

it kind of depends on what the -V supply is? 

Click to expand...

This whole console runs off of +24, so this circuit is sitting between that supply rail and ground.

Click to expand...

OK doing the math in my head,, if opamp is sitting at 1/2 of 24v supply, that resistor divider would result in approx 2V.. so the two diodes will clamp that to 1.2V and will definitely be conducting with approximately 1 mA.. If the transistor junction was exactly the same as the diode junction, we would expect 1 mA of class A current in the transistors. Real life isn't quite that neat (unless perhaps you made your diodes out of some left over transistors). But actual current should be in the low mA ballpark +/- a bunch.

The voltage drop and temp co's of the diodes and the Vbe junctions are nominally the same.. i.e. as temperature increases the forward voltage decreases for the same current, or current increases for same voltage. The concept is to have the same number of diode junctions and base-emitter junctions, so any thermal variation with temperature will track and cancel out.

Learning that those diodes are actively biasing the transistors makes me a little more nervous.. especially if you substitute newer and/or different parts. Please add 5-10 ohms in series with at lease one transistor emitter. The mVs of drop across the R will mitigate thermal effects or poor matching.

======

OK for the TMI portion of this post, thermal runaway is caused when say the diodes are not at the same temperature as the transistors. If the transistors heat up, their forward voltage vs, current drops (or current increases for same voltage). .... but, as that current increases, the device power dissipation increases so it heats up even more, then draws even more current, then heats up even more, until the magic smoke inside comes out.

If the diodes setting the base drive voltage are thermally coupled to the transistors and also heat up, the drive voltage will drop and device current remains roughly constant.

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For a low power buffer it isn't worth the trouble to try for thermal perfection, just degenerate the emitters with several ohms and move on..

Thermal runaway in power amps can lead to spectacular failures..  but it is managed with thermal compensation of base drive voltage, emitter degeneration, and adequate heat sinking.  Typical resistor values in power amp device emitters are in the < .5 ohm ball park.

JR

If R4 is over about 6V, the diodes are doing most of the biasing. Now we know R4 is nearly 24V/2, so the diodes should dominate. So why is R3 there?

> These diodes are obviously matching the VBE drop across the two transistors

I would say they are NOT matching the Vbe. That was the idea, but it just is too primitive to be maintainable.

> I'm talking in regards to installing new transistors

Why? Are they blown? Is anybody surprised?

You "can" set a BJT current by setting the base-emitter voltage. Much as we bias tube amps with -27V at the grid. BUT we do not know what Vbe is at our desired current, except it is probably between 0.5V and 1.0V. The "safe" way is to throw in a whomping big resistor, dropping much more than 1V. If we hold the base up at 6V, and have a 1K emitter resistor, we know the current can not be as high as 6mA, and won't even be as low as 5mA: this is "exact" for our purpose.

But this is a Power Output stage (not shown in your photo, but you know it is and I can guess). Big emitter resistors are wasteful.

What they surely wanted was 1mA-10mA idle and 50mA peaks.

They could get the idle if the diodes and transistors had similar Junction Area. As John says, if you make the diodes out of transistors, the same transistors as the outputs, current WILL match. The 11V/5.1K= 2mA flowing through diodes to R4 would be a happy bias.

But they didn't (it cost more). It looks like they used too-small diodes. On one hand, the current ratio should be as the area ratio. OTOH, little diodes are sometimes pushed hard for their area, and you don't really know what area you will get.

And a mere 20mV difference of voltage drop means 2:1 difference of current; 60mV means 10:1 difference in current.

For more fun, the diode or transistor drop changes 2mV per degree C. When it gets hot, you must reduce Vbe bias or the current will soar. Thermal Runaway.

While the diodes are supposed to compensate that, hard-worked transistors distant from their bias diodes won't track and will idle very hot.

I said 60mV gives 10:1 change of current. If you use an emitter resistor which drops 60mV, bias and temperature sensitivity is much less. It's still on a razor edge, but a blunter edge.

But what REALLY bothers me: no over-current protection ! ! ! What happens when (not if) the output is shorted? The pull-up current can be M1's output current (surely >10mA) times Q1 beta (surely ~~100), or 1 AMPere. And at short, the voltage is essentially 24V/2 or 12V. 1A*12V is 12 Watts. These are half-watt parts. Expected longevity is about a blink.

> short in either direction

After Q1 Q2 melt, there's various ways to have large current in the diodes. In power systems the usual course is the silicon melts to a short, the short flows huge current, the short burns-off leaving an open junction. But in lower-power systems, it may stay short.

So you are replacing transistors which should be perfectly good, you have apparently shorted diodes, in a plan with crappy bias and no short protection.

Do you want lifetime employment? Or do you wish to use hind-sight to reveal things the original designer could have done differently, make this module as blowup-proof as most modern gear is?

Since it is on blueprint, I fear it is Vintage. Since it hasn't been chucked, I fear it has Cherished Sound. So I hate to go changing it "better" and lose "the sound". OTOH, a blown module gives no sound at all. And I can't see any euphonic sound in this plan; I think it was a quick hack.

Woh-ho!? These transistors are TO-66 metal diamonds? I shudda checked that first.

That's still no excuse for sloppy bias and lack of short-protection. They have a 25 Watt rating but that's on the infinite heatsink which is always out of stock. If they are naked or on card-sized sinks, I'm sure I can melt them.

Anyway: the plan below is good design, widely used, notably in Jensen 990. The exact values could be different. I have no clue if 1N4007 will really bias-up an unknown pair to a happy bias (but it will tend toward cool rather than hot). The current-limit scheme is not obvious, but it is in there.

The ? resistors are for cowards. Assuming the DC level stays near 12V (you overload with AC signal, not DC failures), then 33 ohm resistors here will limit transistor abuse to less than 0.55 Watts and 0.25A, which should be tolerable all morning. However it reduces headroom.

+1 to PRRs suggested circuit.

I have used variants of this many times with good results.

I don't know that the diodes need to be high current, I have used small signal diodes without problem.

JR

Shoot. Another typo (I must have been really spaced out yesterday). I meant to say replacing the diodes, not the transistors.

Most of the diodes in these modules are unmarked white ones similar in size to 1N400x, but there are 1N4148s in one module. As mentioned, many of the 4148s in the channels in this (1968, and barely maintained) console are failing.

These transistors are not heat sinked. One in one module has actually seen some hot times; the PCB beneath and just around has started to carbonize.

I'll look into the mods mentioned. Thanks so much for passing on the knowledge!

Also, what are factors that contribute to diodes going bad? Are transistors susceptible to the same manner of failure (going short)?