Wurlitzer 206

[woodyc]

Any thoughts about the connection of the output stage bases to the speaker? This puts about 100mA thru the load. And seems kind of pointless?

[woodyc]

And by the way, somebody had a post about a cathode follower as a first stage -- this is an amp that does that. The reed bar puts out a good signal but it's capacitive and high impedance.

[loudthud]

In that circuit the output capacitor does double duty as the driver stage bootstrap. 100mA shouldn't affect the speaker. You only see this on low cost designs.

[woodyc]

In that circuit the output capacitor does double duty as the driver stage bootstrap. 100mA shouldn't affect the speaker. You only see this on low cost designs.

Yeah, it raises the input impedance of the power stage -- but the amp works the same without it. My question is... what were they hoping to gain by doing that?

[loudthud]

The bootstrapped current source has been used since the late 60's and is still used today. On what planet does the amp work "the same without it"?

[woodyc]

The bootstrapped current source has been used since the late 60's and is still used today. On what planet does the amp work "the same without it"?

Earth? Referring to the 206 schematic, consider lifting the bottom end of R7 off the speaker terminal and grounding it -- let's say through an 8 ohm resistor so the dc doesn't change. What practical effect do you think that will have on the operation of the amp?

[Steve Conner]

Well, it is a bootstrap. If you disable it, the bottom half of the output stage will clip sooner, so the output power will be reduced.

[woodyc]

Well, it is a bootstrap. If you disable it, the bottom half of the output stage will clip sooner, so the output power will be reduced.

I don't follow why that connection should affect power stage clipping, or why it would only affect the bottom transistor when you have a complementary pair with the same signal on the bases?

Also this amp doesn't operate anywhere near power stage clipping -- the weak link in terms of signal size is TR7 and the way its connected to the driver.

[J M Fahey]

I don't follow why that connection should affect power stage clipping, or why it would only affect the bottom transistor when you have a complementary pair with the same signal on the bases?

Yes, you will get the "same signal on the bases" ; only problem is that the positive peak will clip at about 20V ; while the negative side will clip at around, say, 10V .......... *if* connected straight to ground instead of bootstrapped by connecting it to the *hot* speaker terminal.
Talk about assymmetry.
Or to see it from another point of view: the top half of the wave is fed from a constant current supply: Tr8, while the lower half is not: R7 (200 ohms) .... *but* bootstrapping turns it also into a good imitation of a constant current source, that's what the designer intended.
He saved a somewhat large electrolytic and a power resistor by making the coupling cap (which, by the way, is pitifully small by modern standards) do double duty and allowing for those 100mA through the speaker.
Please don't be harsh with him, 40 years ago this was acceptable.

Also this amp doesn't operate anywhere near power stage clipping

Don't get what you mean. The amp will operate to the volume set by the Musician, according to his needs.
If in a bedroom, maybe it won't matter (if he uses 1 watt ) but in a stage, club, or church, etc. it is *not* the same to have 25 clean watts available, as having only 6 clean watts.
Specially in a piano-family keyboard, where distortion is ugly.

-- the weak link in terms of signal size is TR7 and the way its connected to the driver.

Sorry, don't get it.
There is not *voltage* gain involved in Tr7's operation, it works as an audio modulated current source for Tr8, which is *the* main voltage gain stage in that amp architecture.
By the way, do you have one of those Wurlitzers?
Love them.

[woodyc]

...Don't get what you mean. The amp will operate to the volume set by the Musician, according to his needs.
If in a bedroom, maybe it won't matter (if he uses 1 watt ) but in a stage, club, or church, etc. it is *not* the same to have 25 clean watts available, as having only 6 clean watts.
Specially in a piano-family keyboard, where distortion is ugly...

Juan thanks for your thoughts.

I'll have to think more about your theory of operation of the power amp.

But about the output...

The supply voltage and output transistor ratings are misleading. I have a couple of Wurlitzer amps here, but no complete pianos, so I can't get an exact number on the actual power. But I'd guess that if you banged on a chord continuously with the volume cranked, 2 Watts might be an upper limit? The reason for the low output is that most of these pianos have internal speakers and its very easy to get the reeds feeding back. The 206 uses the same amp board with an 8 Ohm/5W resistor in place of the speaker, but 5W is way more than they needed. Another clip-related thing thats different here is that the reed/hammer mechanism has a soft attack -- there isn't a giant transient at the start of every note like a guitar. The reason I mention all this power stuff is that it may be the reason that you can make changes to the amp (like grounding R7) that make no audible difference.

[Steve Conner]

Here's how I understand it. This amp runs off a single supply. Let's say 24V. So at idle, the output stage emitters sit at 12V. The speaker terminal sits at 0V, so the output capacitor is charged to 12V.

Now let's try reproducing a big negative transient. The emitters go down to maybe 1V. The speaker terminal is forced down to -11V. 11 volts are available for forcing current through R7, whereas if it were connected to ground, there would be less than 1V across it. The result is that the lower output transistor has 11 times more base drive available.

I don't own a Wurlitzer 200, but I have played a few, they sound great!

[woodyc]

Here's how I understand it...

Steve, you're right. I had to do a simulation to see it. And I had to do about six more simulations to see exactly how it works. The bootstrap does raise the clip level of the lower transistor by a small amount. It does this in a distorted way, but most of the distortion is removed with the negative feedback.

I don't know if anybody noticed this, but the way this amp works is really bizarre...

At idle TR9 carries about 100mA and TR10 is off so for small signals TR9 is a single-ended class A amp. (If you check the bias string and max current in TR8, its clear only one output transistor can ever be on.) When TR9 reaches cutoff, TR10 turns on to take over for the cut off peak. This creates an asymmetry which affects the bias of TR7 since the feedback is DC. The inclusion of positive feedback just adds to the intrigue.

P.S. The thing I said earlier about TR7 being the weak link was complete BS -- the only reason you see clipping there is because it's in the feedback loop. Sorry, internet.

[woodyc]

The bootstrap definitely gives you more output, But I think the "explanations" above are incorrect. I'm not sure I can explain it any better, but its a really interesting technique so I'll try...

Firstly, all they are doing with the bootstrap is increasing the AC impedance of R7.

This does not mean that the base current in TR10 is impeded. It means that more of the AC current in the bias string will flow through the bases, making the driver much more effective (about 3x in this case.)

While technically the current coming out of TR10 base does go through R7, its just replacing current that is being removed by the driver. So R7 becomes irrelevant AC-wise.

The reason this gives you more power is that without the bootstrap the max ac current in the bias string was just not capable of pushing enough current through the bases -- most of it was going thru R7.

While there is definitely some asymmetry here, it seems like the principal effect of the bootstrap is similar at both bases. So the thing Steve said about only the bottom half clipping, or what Juan said about only the top half being fed by a current source seem incorrect to me.

But maybe I'm just not getting it?

Also if what I'm saying is correct, replacing R7 with a choke should have about the same effect as the bootstrap?

[J M Fahey]

A transistor collector is a current source, meaning current does not depend on voltage applied (in this case it depends on base current only).
Just look at transistor collector current vs collector voltage graphs.
You'll see they are practically horizontal, meaning they don't change with applied voltage; hence, constant current.
This falls out when voltage is so low as to cause transistor saturation, usually 1 or 2 volts, so worst case you "lose" a couple volts.
On the contrary, current through a resistor *does* depend on applied voltage, the so called Ohm's Law.
I guess you can understand and differentiate between such different behaviors I just stated.
In a nutshell: in a transistor, current does NOT depend on applied voltage, hence constant current; on a resistor, current DOES depend on applied voltage.
If you do not get this, please re read it until you do.
No explanation will make sense to you otherwise.
We must agree on certain basic premises or it's like speaking different languages; words will be heard, but convey no meaning.
Thanks.

[woodyc]

Juan, the point of disagreement has nothing to do with whether or not a collector is a current source. Its this...

...the top half of the wave is fed from a constant current supply: Tr8, while the lower half is not: R7 (200 ohms)...

In your language, I would say that both bases are fed by the same current source. The problem is that without the bootstrap, a lot of the AC driver current passes through R7 instead of the bases. Giving R7 a higher AC impedance just makes the driver more effective -- at both bases.