# Electronic – a modern 2sd669 transistor replacement

amplifieranalogsoundtransistorsvintage

I have an old Nad 302 integrated amp where one of the 2sd669 transistors is blown. I Cannot find any suggestion on modern replacements. To be honest I am a software engineer and analogue circuits is a bit out of my comfort zone.

To be precise it is the Q411 according to the service manual.

What would be a currently (2021) available equivalent, or better replacement?

The transistor isn't critical.

A $$\25\:\text{W}\$$ output into $$\8\:\Omega\$$ means a peak $$\Q_{415}\$$ emitter current of $$\\frac{20\:\text{V}}{8\:\Omega}=2.5\:\text{A}\$$. (Resistive load assumption, which isn't entirely accurate.) $$\Q_{411}\$$ only has to supply its peak base current. $$\Q_{415}\$$ minimum $$\h_\text{FE}=50\$$. So this means $$\50\:\text{mA}\$$, worst case. Although I do not like this design, the maximum current pulled through the $$\V_\text{BE}\$$-multiplier by $$\R_{447}\$$ and $$\R_{449}\$$ is $$\19.5\:\text{mA}\$$. (Or should be so long as the $$\25\:\text{W}\$$ is respected.) That means a maximum drop across $$\R_{437}=68\:\Omega\$$ of $$\\approx 1.33\:\text{V}\$$ due to the $$\V_\text{BE}\$$-multiplier. But as the rail is $$\+38\:\text{V}\$$ (with ripple? haven't checked that) and a peak base drive voltage to $$\Q_{407}\$$ of $$\\le 23\:\text{V}\$$, there's $$\15\:\text{V}-1.33\:\text{V}\$$ or $$\\ge 13\:\text{V}\$$ of available headroom for $$\R_{437}\$$ needed voltage drop to supply $$\Q_{411}\$$'s base. Enough that you could easily saturate $$\Q_{411}\$$ and not have to worry about it. So there's really no need for high values of $$\h_\text{FE}\$$ for $$\Q_{411}\$$. I'm sure anything better than a minimum of 50 would be fine. It must also squeak by standing off at least $$\33\:\text{V}-\left(-20\:\text{V}+2\:\text{V}\right)\$$ or $$\V_\text{CBO}\ge 51\:\text{V}\$$. That said, the 2SD699 is rated for $$\V_\text{CBO}=180\:\text{V}\$$. So, you may be able to get by with anything rated $$\V_\text{CBO}\ge 80\:\text{V}\$$. Maybe a little less if you are stuck looking around, though of course you'd probably prefer something rated as high as what they already used. But at least you have an idea about what you might be able to get by with, if you need to.

In short, you want the same package (TO-126 or TO-225AA or equiv) and you want it to be an NPN. Since the peak collector current shouldn't be more than about $$\50\:\text{mA}\$$, I'd double that for a cushion at $$\100\:\text{mA}\$$ which is... pretty much any BJT at all. So that's not going to be an issue and you can just not worry about it, for now. I'd probably make sure that $$\80\:\text{V} \le V_\text{CBO}\le 200\:\text{V}\$$. With larger values being preferred while realizing that lower values may be acceptable and work well. A lot of BJTs may fall under your need here, so this is one parameter to keep in mind as you look.

I'd assume that the worst-case dissipation requirements are going to be met so long as you select a similar package. But the worst case instantaneous dissipation will be about half-way through its active part of the half-cycle. That dissipation value is so small that I'm not at all worried. The package is over-kill.

## Notes

If you are interested, I'll sketch out some detail about my thinking process. This exposes any errors I may have made (which may hurt how others see me should I care about that) but it also may teach (should I be lucky enough to have avoided making serious errors.)

This site is more about electronic design as I see it and less about giving simplistic answers. (More about learning to fish than about being given a fish.)

That said, here are my thoughts in the order in which I had them:

1. The continuous power rating is $$\25\:\text{W}\$$. This means that the peak voltage into the load is $$\\sqrt{2\cdot 8\:\Omega\cdot 25\:\text{W}}=20\:\text{V}\$$. At least $$\3\:\text{V}\$$ of overhead is required (without even looking at the schematic.) I observe a rail of $$\+33\:\text{V}\$$ for $$\8\:\Omega\$$ and $$\+28\:\text{V}\$$ for $$\4\:\Omega\$$. Obviously, they've added extra headroom for some reason I do not yet entirely follow. (This is my first look at the schematic.)
2. I notice on the schematic some voltage values, which I take to be quiescent values. So I decide to do a once-over on those to make sure I don't find anything shocking there that needs more attention.
3. I note that the power stage is Darlington-arranged. They are specifying $$\V_\text{BE}=600\:\text{mV}\$$, quiescent. I think that's kind of low. But I'll bite, for now.
4. I find the $$\V_\text{BE}\$$-multiplier, $$\Q_{409}\$$, and try to work out its quiescent current. (That's how they work. And by the way, they are also usually thermally coupled to the power BJTs in the stage, as well. I expect it to be on some aluminum strip with the others. Also note the dashed line around it on the schematic. Certain, now.) In working upwards towards the $$\+38\:\text{V}\$$ rail, I see $$\Q_{405}\$$, $$\Q_{407}\$$, and $$\R_{437}\$$ there. Nothing immediately tells me what the current should be from that mess.
5. I look at the other end of the $$\V_\text{BE}\$$-multiplier and see two resistors in crazy-orientation (stuffed to make the schematic on one page but not to make it very readable!): $$\R_{447}\$$ and $$\R_{449}\$$. I see that's how it's established and then compute $$\\frac{-28\:\text{V}-\left(-1.2\:\text{V}\right)}{2\,\cdot\, 1.2\:\text{k}\Omega}\approx -11\:\text{mA}\$$. So that's the $$\V_\text{BE}\$$-multiplier current.
6. To double-check, since I also know this same $$\V_\text{BE}\$$-multiplier current must also pass through $$\R_{437}\$$, I compute $$\11\:\text{mA}\cdot 68\:\Omega\approx 750\:\text{mV}\$$ for its voltage drop. Okay. That's no problem. There is tons of voltage headroom around that resistor. So far, okay.
7. I go look at the 2SC3519 datasheet and see that the minimum guaranteed $$\h_\text{FE}=50\$$ at one set of conditions. I go look at Figure 6 and see that maybe $$\h_\text{FE}=25\$$ is indicated over its entire temperature range (very cold, likely.)
8. I go look at the 2SC2837 datasheet and see that the minimum guaranteed $$\h_\text{FE}=90\$$ at one set of conditions. I go look at Figure 6 and see that maybe $$\h_\text{FE}=50\$$ is indicated over its entire temperature range (very cold, likely.)
9. So I'm not worried. It's hot at the power stage. So I'll stick with $$\h_\text{FE}=50\$$ as the going value for what I need to find to replace it.
10. I compute the maximum needed base current for $$\Q_{415}\$$ as $$\\frac{2.5\:\text{A}}{h_\text{FE}=50}=50\:\text{mA}\$$. So this is all that $$\Q_{411}\$$ has to supply.
11. I now go look at the 2SD669 datasheet and look at Figure 6 to see that the minimum guaranteed $$\h_\text{FE}=150\$$ at $$\I_\text{C}=50\:\text{mA}\$$. The table says the minimum $$\h_\text{FE}=60\$$ at anywhere near my needs. I doubt the designers required the Figure 6 value and, given the voltage overhead I see for $$\R_{437}\$$ I know that anything with $$\h_\text{FE}\ge 60\$$ is totally safe.
12. I've gotten a basic feel for the circuit. I'm curious about (and don't like, as initial impressions go) why they just used $$\R_{447}\$$ and $$\R_{449}\$$ as pull-downs and I'm worried about what that will mean for the $$\V_\text{BE}\$$-multiplier when the output is driving hard towards the negative output end as they must also supply base current for $$\Q_{413}\$$. So, assuming about $$\2\:\text{V}\$$ headroom required for $$\Q_{413}\$$ and $$\Q_{417}\$$, I estimate about $$\-20\:\text{V}-2\:\text{V}-\left(-28\:\text{V}\right)=6\:\text{V}\$$ across those resistors, or $$\\frac{6\:\text{V}}{2\,\cdot\, 1.2\:\text{k}\Omega}=2.5\:\text{mA}\$$. Okay. This scares me. It's a bit worrisome as the $$\V_\text{BE}\$$-multiplier also needs something to work with. Further, when quiescent I already know it gets $$\11\:\text{mA}\$$. So we are getting close to a 10:1 variation and the multiplier doesn't have a collector resistor to help it maintain itself properly over that kind of range.
13. So I think about this for a moment. I have already worked out the fact that $$\Q_{411}\$$ needs to supply a peak of $$\50\:\text{mA}\$$, likewise for $$\Q_{413}\$$, and I've found that $$\Q_{411}\$$'s $$\h_\text{FE}\ge 60\$$. I've not looked at the PNPs, but I don't have time to worry about it for now. So I just figure it will be slightly less and call $$\Q_{413}\$$'s $$\h_\text{FE}\ge 50\$$ and work out that $$\Q_{413}\$$ will need about $$\1\:\text{mA}\$$ at its base. Okay. I can breathe again. That's too close for comfort in my world to the available $$\2.5\:\text{mA}\$$. But at least it checks out. So I'm good enough for now. I just don't like the design so much, anymore.
14. This does mean that there is only about $$\1.5\:\text{mA}\$$ left over for the $$\V_\text{BE}\$$-multiplier at this side of the swing, and given the opposite end of the swing that does mean well more than a 10:1 variation in its current. Without the collector resistor it needs to compensate, it's going to vary how it separates the two output driver quadrants throughout a single cycle. But that's not death. It's just annoying to me. It will work.
15. $$\V_\text{CBO}\$$ is now of interest to me. This is pretty easy. At worst, they have only $$\+38\:\text{V}-\left(-28\:\text{V}\right)=66\:\text{V}\$$ of maximum tension running around in the box. So I know the BJTs they selected are more than good enough. Frankly, this just means to me that I can pick $$\80\:\text{V} \le V_\text{CBO}\le 200\:\text{V}\$$ and will probably be okay.
16. I now completely understand why all that extra voltage headroom was added. Yes, it wastes power like a banchee. But they needed it to squeak by with $$\R_{447}\$$ and $$\R_{449}\$$ supplying the negative half-cycle operation. The resulting design is wasteful. But they saved a few pennies and it probably will work just fine (in some part because they added $$\C_{427}\$$.)
17. I also now believe that this was not their first design. I just said they saved a few pennies by providing extra voltage headroom so they could just use those resistors and make it work over the rated spec range of power. But this meant that there is a larger power supply, more dissipation in the box, etc. Which all costs even more money. Also, they included not one, not two, but three positive rails. Note that there are two negative rails. The two rails for each polarity may make sense because they wanted to support two different speaker loads. But a third positive rail? Why the expense? My suspicion is that they didn't get it right the first time and that they had to work out something to patch up the earlier design. To know better, I'd have to go through the entire design and see if I can find sufficient bread crumbs of a redesign to make a good case. Lacking that, I'm at least suspicious, now.

The schematic drawing is really pretty bad, as shown. Those two pull-down resistors, $$\R_{447}\$$ and $$\R_{449}\$$, look almost as though there is some mysterious signal coming in through $$\C_{427}\$$ via them into the output section. Signal is usually arranged to flow from left to right. So that's what I tend to expect when I see things laid out like that. But it's really just a pull-down with $$\C_{427}\$$ helping to sustain a more constant current in $$\R_{447}\$$ over a cycle. In fact, that capacitor detail (kind of a bootstrap concept) was almost certainly included because of the above comments I made about the negative-half-cycle. If they'd have missed including it, then I'd have stayed more leery of this schematic. But it was there, so I felt better.
(If interested, just walk your mind around this loop: start at the (+) end of $$\C_{427}\$$ (the output), then through the BE junction of $$\Q_{417}\$$, then through the BE junction of $$\Q_{413}\$$, which takes you to one side of $$\R_{447}\$$. The other side of $$\R_{447}\$$ is the (-) end of $$\C_{427}\$$. So if $$\C_{427}\$$ maintains its voltage for a moment then the current in $$\R_{447}\$$ is also maintained. Which is a constant current, therefore. So $$\C_{427}\$$ helps sustain a constant current in $$\R_{447}\$$ which compensates for all the movement going on with the output. This is why $$\C_{427}\$$ is an important addition to the circuit. It has nothing at all to do with signal.)