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Not sure what kind of diode to use in amplifier output circuit

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haskellbob

New Member
Hello.

I have an old Heathkit 5-watt monaural FM receiver, the AR-27. I need to replace the diode that isolates the two output transistors (schematic attached).

The original one was a small "can" type; both leads come out of the same side. It was small and mounted to the same heat sink as the output transistors.

But I can't find diodes of that type. Can I just use any old regular silicon diode? What amperage would I need it to have? etc. etc.

Any help will be appreciated!

Thanks,

Bob H.
 

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ericgibbs

Well-Known Member
Most Helpful Member
Hello.

I have an old Heathkit 5-watt monaural FM receiver, the AR-27. I need to replace the diode that isolates the two output transistors (schematic attached).

The original one was a small "can" type; both leads come out of the same side. It was small and mounted to the same heat sink as the output transistors.

But I can't find diodes of that type. Can I just use any old regular silicon diode? What amperage would I need it to have? etc. etc.

Any help will be appreciated!

Thanks,

Bob H.
hi,
The diode forms part of the output transistor bias chain, I would use a 1N914 or 1N4148 silicon signal diode.
Make sure its in physical contact with the heatsink, as the temperature change in the diode will move the operating point of the bias.

After fitting the diode, check for any signs of abnormal heating of the output transistors and any change in sound quality.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
I would disagree, the exact diode type is absolutely critical, as it alone sets the bias for the output stage. The wrong diode could easily kill the output transistors.

Personally I'd replace the diode, and it's series resistor, with a far better Vbe multiplier, which allows you to easily adjust the bias - it's a MUCH better bias stabilising scheme.
 

bountyhunter

Well-Known Member
The original one was a small "can" type; both leads come out of the same side. It was small and mounted to the same heat sink as the output transistors.
.
Whatever diode (or VBE multiplier circuit) you use, don't fail to heed your own words: that thing is mounted to the heatsink for a reason. The temp of that diode has to match the transistors to keep the bias current correct. If not, it will screw up the tracking and mess up a bunch of things, some of which might be disasterous.

EDIT TO ADD: I reviewed the circuit and you may notice how klugy it is: they have one diode and a 4.7 Ohm resistor "matching" into two VBE's of a PNP and NPN output transistor, both of which have 0.51 Ohm emitter degeneration resistors.

The translation is, I don't think the diode's VBE is super critical because this is definitely not a precision circuit and the total of resistive degeneration in both chains would make it somewhat forgiving of VBE mismatch. I'd still put the diode on the heatsink.


ADD: I estimate about 100 mA bias current through the diode, so a 1N914 can't be used (too low current). Just about any silicon diode rated 1A or more is probably good.
 
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Nigel Goodwin

Super Moderator
Most Helpful Member
The translation is, I don't think the diode's VBE is super critical because this is definitely not a precision circuit and the total of resistive degeneration in both chains would make it somewhat forgiving of VBE mismatch. I'd still put the diode on the heatsink.
Like I said above, using the exact correct diode is absolutely critical, overwise thermal runaway is almost guaranteed. I repaired a LOT of these types of poor designs back in their day, if you didn't use the correct diode you just trashed it again.

I fully agree on mounting the diode on the heatsink though.
 

bountyhunter

Well-Known Member
Like I said above, using the exact correct diode is absolutely critical, overwise thermal runaway is almost guaranteed. I repaired a LOT of these types of poor designs back in their day, if you didn't use the correct diode you just trashed it again.

I fully agree on mounting the diode on the heatsink though.
I don't think the diode is super critical because the circuit uses a single diode VBE and a series 4.7 Ohm resistor "matched" against two transistor VBEs and a couple of small 0.5 Ohm resistors in their emitters. If you read the schematic, the voltages at the nodes are there. I believe the diode in question and the series 4.7 resistor have 1.2V across them total. That means the resistor has about 0.5V across it. The output transisitors also have 0.5 Ohm emitter resistors to provide degeneration. That's why I think the diode vBE is not that critical, but you could increase the 0.5 Ohm resistors or adjust the 4.7 ohm resistor to dial the diode in:

If the "ideal" current for the transitors bias is known, just put a new diode in and adjust the 4.7 Ohm resistor as needed to dial the output idling current in if you really want it to be dead on. It's VBE should track OK if it is on the heatsink.

As for thermal tracking, this design is going to be poor at best because a single diode's TC will never track the TC of the two output transistor's VBE junctions. That's why there is so much resistive degenertaion in both legs of the bias chains.

You could get much better tracking using two diodes made up of the same type of output transistors connected as diodes in the left hand bias chain.
 
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tunedwolf

Well-Known Member
I too repaired loads of amplifiers using this type of biasing scheme back when it was common, Nigel is absolutely correct. The exact diode is absolutely critical, if not replaced, the amp will work great for a few days then pop goes the output transistors and emitter resistors and drivers...and anything else that takes it's fancy for that matter! There was no real way to tell what caused the sudden instability, loud passage of music, blue moon in the sky etc. It just happened, it was quick and deadly to the amp. The old Teleton & Tandberg amps were a real sod for just that very reason.

rgds
 

bountyhunter

Well-Known Member
OK, but I can't see why he couldn't install a similar diode and then adjust the external resistors to set the output idling current to the correct value. I have taken a lot of temp data on TC of various sizes of diodes, and they don't vary a whole lot: maybe from 2.2mV/C down to 1.8mV/C. I do believe the initial VBE value of the diode can vary quite a bit based on current rating, doping, etc and that could affect the bias set point. I still think you can compensate for a different initial VBE by tweaking the 4.7 Ohm resistor and set the correct output idling current. That's what the circuit theory says. You can measure the idling current across the 0.51 Ohm resistors and see what it is and adjust the 4.7 resistor based on that. I believe if you get that current right, the design should work as original. Considering the fact he has said he has no idea what the original device was and no way to find out, that seems like the best course available.

I suspect I know the failure mode you describe: if the diode was such that the VBE was a lot bigger than nominal, it would increase the output idling current (heating the transistors) and could cause thermal runaway, especially if pushed hard. I believe if the external resistors are adjusted to force the output current to the correct value, that won't happen. Just my opinion.
 
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Nigel Goodwin

Super Moderator
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I suspect I know the failure mode you describe: if the diode was such that the VBE was a lot bigger than nominal, it would increase the output idling current (heating the transistors) and could cause thermal runaway, especially if pushed hard. I believe if the external resistors are adjusted to force the output current to the correct value, that won't happen. Just my opinion.
If you're redesigning the amplifier to that extent, why not do it properly and just stick a Vbe multiplier in?.

Like tunedwolf says, those old amplifiers were a real pain, you had to stock the exact correct diode for every single amplifier you repaired.
 

bountyhunter

Well-Known Member
If you're redesigning the amplifier to that extent, why not do it properly and just stick a Vbe multiplier in?.

Like tunedwolf says, those old amplifiers were a real pain, you had to stock the exact correct diode for every single amplifier you repaired.
I am not sure I understand. A VBE multiplier would have a little different characteristics than a discrete diode. If the point is to match the original, that would seem to be more risky to me. It's been a while, but I seem to recall VBE multipliers work best when they are set up to create at least two or three VBEs worth of voltage across the VCE of the transistor, so it will be in it's linear region. In this case, you only want one or a little over one VBE, I think that puts the transistor pretty close to being in saturation which changes the current draw of the circuit. Anyway, that was an interview question once so it's what I remember. If you push the VBE multipliers down towards sat, they don't act like VBEs down there.


My point was that tweaking a resistor is easy to do. If you are tweaking "down" you can just solder a resistor across the terminals of the 4.7 resistor already in place and dial it until you get the current you want. To tweak up, you desolder one end and stand it up and solder a small resistor in series down to the board.

Installing a VBE multiplier would entail making actual changes to the board or somehow wiring it in which is harder to do. I just figured changing a resistor value was easier and probably more likely to end up close to where the original design wanted to be.
 
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Nigel Goodwin

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Most Helpful Member
I am not sure I understand. A VBE multiplier would have a little different characteristics than a discrete diode. If the point is to match the original, that would seem to be more risky to me. It's been a while, but I seem to recall VBE multipliers work best when they are set up to create at least two or three VBEs worth of voltage across the VCE of the transistor, so it will be in it's linear region. In this case, you only want one or a little over one VBE, I think that puts the transistor pretty close to being in saturation which changes the current draw of the circuit. Anyway, that was an interview question once so it's what I remember. If you push the VBE multipliers down towards sat, they don't act like VBEs down there.
You actually need a little over 2 x Vbe, in order to turn both transistors slightly on. It's a perfectly normal and common method, and a transistor tracks the transistors far better than a diode does.

You can certainly vary the resistor to set the quiescent current, but it still doesn't track correctly, the diodes used are VERY specific and specialised.
 

bountyhunter

Well-Known Member
You actually need a little over 2 x Vbe, in order to turn both transistors slightly on. It's a perfectly normal and common method, and a transistor tracks the transistors far better than a diode does.

You can certainly vary the resistor to set the quiescent current, but it still doesn't track correctly, the diodes used are VERY specific and specialised.
As I said, i can't follow your reasoning where you said I should not redesign it by changing the 4.7 resistor, since putting in a 2VBE circuit will require that as well. The initail design uses a single diode and a 4.7 Ohm resistor to force the voltage across the two output trans VBE's. I certainly agree a BETTER design would be to have two matching diodes (or diode connected transistors) in the left chain to set the bias, especially since that is the design I use when I build a discrete output stage.

If you are putting in a 2VBE "multiplier", it could work but you would be forced to change the 4.7 Ohm resistor to make it work as designed since you have replaced one diode with two. Adding a single diode and tweaking the 4.7 ohm resistor was my original suggestion for the simple fix (see above). If we are going to a two VBE redesign, I would use a diode connected NPN and PNP transistor (similiar to the output transistors) in a TO-220 package so the bias diodes so they can be screwed to the heatsink. That would also necessitate changing the 4.7 resistor. If the transistors are similar, you could probably use a value of one Ohm in the left chain since that resistor should equal the total of the resistance in the output transistor emitters for the currents to balance. For safety, you could go up to maybe 2 Ohms left chain and two 1 Ohm resistors in the output transistor emitters and get more degeneration, but that isn't necessary if the transistors are the same..

That design would track better than a VBE multiplier, since the base current in a multiplier is an error term in the total current following the VBE curve over temp. The second error is that the gain of transistor changes with temp, again changing the base drive which is an error term in the total bias current. I'm not saying the VBE multiplier won't work, just not sure about how well over temp. If you want it to track, the simple way is to use a second set of transistors (same as output) diode connected in the left side bias chain as they will track the output transistors perfectly if they are screwed to the heatsink. I think using two diode connected transistors is a better design if we are going to change it, the cost is one extra transistor used which isn't all that pricey.
 
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Nigel Goodwin

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Most Helpful Member
Vbe multipliers work excellently, which is why most quality amplifiers use them - but most importantly it's so simple, and dead easy to adjust.

The reason for not altering the 4.7 ohm is because you're not using a special diode designed for this purpose, and it WILL blow the output stage.

Those of us from that era have been through all this decades ago - you either fit the correct diode (I used to stock six or so different types back then, used in the various anmplifiers we repaired), or the repair bounces back.
 

bountyhunter

Well-Known Member
If you say so, but if I was worried about it, I would just use the solurtion I suggested above: get two extra output transistors and diode connect each one (tie the collector and base together) and use them as diodes in the bias chain. Match the series resistor in the bias chain with the total resistance in the output emitters and I GUARANTEE they will track because both sets of transistors have the same electrical characteristics. That way I know I would never have to fix it again. If the manufacturer wasn't trying to save twenty cents, they would have used the same thing. It is a standard design procedure.

As far as VBE multipliers go, I am not saying it would not work, but we used them a lot in IC design and we documented that they don't behave exactly like series diodes because they do have a base drive component. We used them to save die area compared to a series of diodes, but they didn't track the VBE curve as precisely as diodes do. Might work here, but if you want it bulletproof, use diode connected transistors.
 
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mneary

New Member
It really depends on the OP's objective. If it's authenticity, then he needs a diode identical to the original. Maybe there are markings that will help us find one.

If he wants almost authentic, but is willing to tolerate poor reliability, then a heatsink-mounted diode with a proper adjustment to R128 might meet his objective. Set R128 so that the output stage current is 5-10ma (<10mV from the top of R129 to the bottom of R130) with no audio. Crossover distortion will be just as annoying as it was when the radio was new, and it will only die occasionally.

If he wants reliable, then two diode-connected transistors is great. But if it doesn't have to appear authentic, he could replace the whole amplifier section.
 

bountyhunter

Well-Known Member
It really depends on the OP's objective. If it's authenticity, then he needs a diode identical to the original. Maybe there are markings that will help us find one.

If he wants almost authentic, but is willing to tolerate poor reliability, then a heatsink-mounted diode with a proper adjustment to R128 might meet his objective. Set R128 so that the output stage current is 5-10ma (<10mV from the top of R129 to the bottom of R130) with no audio.
That's a surprising value. I thought the output idling current would be higher? Based on schematic, I think the current flowing down the diode chain is supposed to be about 100mA. I would have guessed the output current would be in the same ballpark. Is there a reason it's so low?
 

mneary

New Member
I proposed an idle current of 5 to 10 mA since I haven't seen the heat sink. If it was a 5W amplifier, even in the pre-FTC days, the heat sink was doubtlessly designed to safely dissipate 3W or more. In retrospect, I agree an idle of 10mA (150mW) is much too conservative.

Your suggestion of 100mA (1.5W) idle is much better. My value would probably result in distortion so annoying that the radio would never be used again. :eek: Thanks for correcting me.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
That's a surprising value. I thought the output idling current would be higher? Based on schematic, I think the current flowing down the diode chain is supposed to be about 100mA. I would have guessed the output current would be in the same ballpark. Is there a reason it's so low?
100mA is FAR too high for such a small amplifier - easiest technique is to fit a Vbe multiplier (like I've said all along). Set the bias to zero with the preset, then input a VERY low level sinewave and monitor the output with a scope. Adjust the preset until the crossover distortion just disappears, then check the actual current (it will be a LOT less than 100mA). Leave the amp to warm up, and keep checking it.
 

mneary

New Member
easiest technique is to fit a Vbe multiplier
OP hasn't stated whether he wants to improve the amplifier or preserve it.

In either case, he seems to have disappeared.
 

bountyhunter

Well-Known Member
100mA is FAR too high for such a small amplifier - easiest technique is to fit a Vbe multiplier (like I've said all along). Set the bias to zero with the preset, then input a VERY low level sinewave and monitor the output with a scope. Adjust the preset until the crossover distortion just disappears, then check the actual current (it will be a LOT less than 100mA). Leave the amp to warm up, and keep checking it.
If you read any or all of my posts, I never said I knew it should be 100mA, in fact I said the opposite. What I said was that based on circuit values, the current in the diode side of the bias chain will be in that ballpark and it's really obvious why: you have a diode in series with a 4.7 Ohm resistor tied across the two output transistor VBEs and their 0.5 Ohm resistors. The two transistor VBEs plus a small resistor drop equals about 1.4V. The diode will eat about 0.7V or maybe 0.8V, so that leaves about 0.6 - 0.7V across the 4.7 Ohm resistor which is a bit over 100 mA.

I don't know the exact value of output current, but I do know in MOST cases, the two currents are in the same ballpark to get decent thermal tracking. Maybe not here, I realize this is a screwed up design.

Your idea of tweaking the output current up to just enough to eliminate visible crossover is interesting, but I was taught that you don't see visible distortion until THD is about 3% and I doubt anybody wants to listen to that. I would have thought you would want to dial it up to make the sine wave look good then give it some more to make sure you don't go back down into calss B at low signal levels.

If it was mine, I would use diode connected transistors in the bias chain and not a VBE multiplier which does not behave as close to a diode as a DCT does. I'm not sure exacvtly what output current is best, I would assume somewhere between about 30mA and 100mA maximum. I have not seen the heatsink so I am not sure how much heating 100mA would create.
 
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