Kenwood tech question

[calman46]

I have a working but very dirty Kenwood KR9600 . When recapping an amp board I noticed two modifications. First one is a 17k res was put in series with the C of Qe3 and the second mod is a small Pf cap is across the B&C of Qe8 on the green side of the board. The power modules may have been replace there was a lot of heat sink grease of the power module. I can see the small cap on the green side of the other board but not the 17k res. I haven't remove that board yet but It looks like both sides were done. Any ideas why this was done ,could be a factory modification? I have attached the amp board schematic and circuit description sheet . I have the SM, email me if you want it. thanks

[ecluser]

In my opinion, the 17k resistance changes nothing. Qe3 is a 2.5mA current source (.67V / 270 Ohms)for the differential input pair. Without resistance, the collector-emitter voltage in Qe3 is close to 66V. This transistor will dissipate about 170mW. This transistor can support much higher voltage than this. With a 17k resistance, the voltage drop in the resistance will be about 42V (for 2.48mA). This will left 24V between the collector and the emitter of Qe3, and he will dissipate about 60mW. The 2SC1775 is a 90V, 300mW transistor.

[calman46]

so the circuit is just "running" on less voltage. Will this decrease the the output ? Would you recommend changing it back to its original state. This mod may have been done to make the power modules run cooler ?

[ecluser]

This resistance has no impact on the power module. It only changes the power dissipated by Qe3.

In fact I would rather recommand a change in Re4 and Re5 by two resistances of equal values. Here is why...

The two collectors of Qe1 and Qe2 are connected to the base of Qe4 and Qe5 respectively. Since the base-emitter voltage of Qe4 and Qe5 are identical (OK, may be a few 10mV difference but it is not important), this forces the two collector voltages of Qe1 and Qe2 to be the same.

On the schematic you posted, we can see 22.4V at both collectors. Also, the diode De3 is a 24V Zener. It is easy to compute the current in Qe1 =(24 - 22.5)V / 910 = 1.65mA For Qe2, we have (24 - 22.5)V / 390 = 3.85 mA
We must understand that these voltages are approximative, may be ± 0.5V, but the collectors are forced by the second differential stage to be at the same voltage.

It is not a good idea to have such a large difference in the collector currents of the input differential stage. It generates second order harmonic distortion.

The total current in this pair passes by Re6. This gives close to 1.5V drop in this resistance. It is more than the forward voltage drop in De1 and De2. I don't see how the diodes De1 and De2 can regulate this current in this situation. From the data sheet of 1S2076, the voltage drop in each of the two diodes is only 0.3V for a forward current of 0.8mA (This 0.8mA comes from 65V / 82k).

Can you measure the exact voltage drop in Re4, 5, and 6?

[Tube_Dude]

The 17k resistor as a virtue of isolating the critical feedback summing point (the emitters of the LTP input transistors ), from the non linearity of the collector capacitance of Qe3 (the current source transistor).

[EchoWars]

Interesting...I've worked on a few 9600's (and own one), but don't think I've ever noticed the collector resistor mismatch on Re4 and Re5. I will say that there are several significant differences in the modules between the actual unit and what is shown in the manual. I should have documented the changes.

I can't help but wonder if Kenwood purposefully mismatched collector currents to force offset to zero...I dunno. If the gain of the input pair was high enough (and sourcing good NPN small signal transistors with gains of 1200 and more isn't too difficult) then it would not be a problem even if the collector currents were equal. I haven't had my 9600 apart in ages, but next time I might look into this and see if the schematic matches reality...

FWIW, the 1S2076 diode is a plane-Jane silicone job, with Vf of about the same as any other small-signal diode (.6V at 1mA). Perhaps you were looking at the elevated temperature curve (Ta=125°C).

http://documentation.renesas.com/eng/products/diode/rej03g0559_1s2076.pdf

[ecluser]

FWIW, the 1S2076 diode is a plane-Jane silicone job, with Vf of about the same as any other small-signal diode (.6V at 1mA). Perhaps you were looking at the elevated temperature curve (Ta=125°C).

http://documentation.renesas.com/eng/products/diode/rej03g0559_1s2076.pdf

Thank you EchoWars for this precision. I was not looking to the bad curve, I was directed to a bad datasheet for the 1S2076 from this link

http://www.datasheetarchive.com/preview/133054.html

I saw the part numbers didn't match, but I assumed this was a replacement.

I should have double check this. It is my error, I assume!

It is reassuring to see a nice 1.2V drop across the two diodes.

I thing Kenwood mismatched the collector currents to force the output offset voltage to zero because the feedback resistor Re15 (18k) to Qe2 is much lower than the base resistance Re3 (56k) for Qe1. Since the first differential pair is directly connected to a second differential pair with no emitter degenerative resistors, the collector current variations in the first diff. pair is very, very small.

This circuit is a classic example of the ultra high open loog gain with a lot of feedback.

[calman46]

I'm putting in new big filter caps one had leaked ,then I will take some voltage reading. Is this a practical mod ? Why lower the power of Qe3 ?
tube dude ,huh ? right over my head. I am listing to a Moscode 600 love them tubes.

[ecluser]

I think the only purpose of this 17k resistor is to make the transistor operate in a more comfortable area.

Check the attached file. You will see that without resistor, the transistor is operating close to the 300mW limit.

With a 17k resistor, the operating point is much more comfortable for this transistor. It changes nothing to the current source because an ideal current source has an infinite resistance. The exact value of this resistance is not important. You don't need a 1% metal film here... a 10% carbon resistor 1/4 Watt is perfect for this application.

Also, the current in this transistor is constant DC current. It is his function, to deliver constant current source to the differential pair. You may forget the parasitic capacitance when dealing with DC current.

Making Re4 = Re5 should, in principle, reduce the second order distortion. In this particular design, it will be at the expense of a small (a few mV)DC offset at the output of the amplifier.

If it was my amplifier, I would favor a reduction in distortion. But it is your amplifier...

[calman46]

Thanks for the explanation . I'm going to leave it as is for now .I probably wouldn't hear the difference anyway. thanks again ........ I'll be back

[calman46]

Ok ,the 4 new filter caps are in the power supply has been recapped and so has both amp boards. It powers up fine and I tuned in a local FM station ,sounds good with cheap headphones. But nothing is prefect ,the DC offset is to high R=80mV. L=199mV .I read the posts on this and the offset is control by the differential amp. I've been checking voltages Qe1 &Qe2 are ok (a little high) all the rest are close except I did find that the voltage from Qe6 collector is low I have 3.2v on both channels the schematic has 9.2V could Qe6 cause this high Dc offset ?

[EchoWars]

No...it is the job of Q1 and Q2 to zero out offset.

I use Zetex ZTX694B trannies in the 9600. I match pairs with over 1200 gain, and offset is under 3mV.

[calman46]

I'll get some thanks

[EchoWars]

If you go with the Zetex transistors, pay attention to the lead arrangement. The stock transistors are BCE, and the Zetex are CBE. This means some creative lead arrangment when installing the new transistors. Check that you get this right before you power up, then re-check, and then check once more.

You can opt for the pseudo-Japanese transistors from Fairchild, such as the KSC1845, which have a BCE layout and would ease installation a little bit, but the best gains you'll get with these devices will be in the 750 to 800 range (as if that is bad :) ) rather than the gains well over 1000 with the Zetex devices.

The name of the game with diff-input pairs is high gain, which keeps base currents low, which means low offset.

[calman46]

So you really can't tell if the transistors are operating correctly by the test point voltages , though mine are a little high . The Base on Qe1 -.08 . I'll watch the pinouts on the Zetex thanks. I might remove that non factory mod resistor on the C of Qe3 and recheck the voltages .

[ecluser]

Calman,
What voltage do you have at the base of Qe3 and Qe8, and across Re6 and Re14 ?

[EchoWars]

So you really can't tell if the transistors are operating correctly by the test point voltagesNot when we're talking about an error of 100mV or so.

[calman46]

B of Qe3 is -67.9v, B of Qe8 is -67.9v. across Re14 -0.52v ,across Re6 -0.51v.
I need to open up the right power meter(not working) there some kind crap in there. If I disconnect the wires to it can I slide the back off ,its not glue on . also what is the best way to replace the function lamps ?I can get to them I just can't tell what kind if base the lamps have or if there solder in. SM is no help

[ecluser]

Ok ..... But nothing is prefect ,the DC offset is to high R=80mV. L=199mV .I read the posts on this and the offset is control by the differential amp. I've been checking voltages Qe1 &Qe2 are ok (a little high) all the rest are close except I did find that the voltage from Qe6 collector is low I have 3.2v on both channels the schematic has 9.2V could Qe6 cause this high Dc offset ?

Very amusing!

From the voltage you measured across Re6 and Re14, I computed 3V at the collector of Qe6, very close to what you get. Also, the other voltages are 1V higher than the schematic.

[whsh93a]

If you PM me I can send you a list of the lamps/types that your Kenwood 9600 takes. It's a bunch. Here's a photo of a recently relamped 9600 I did using all LEDs.

DH

B of Qe3 is -67.9v, B of Qe8 is -67.9v. across Re14 -0.52v ,across Re6 -0.51v.
I need to open up the right power meter(not working) there some kind crap in there. If I disconnect the wires to it can I slide the back off ,its not glue on . also what is the best way to replace the function lamps ?I can get to them I just can't tell what kind if base the lamps have or if there solder in. SM is no help

[calman46]

Ecluser how come I have 3.2v at the C of Qe6 and Re 11 . The schematic as 9.2v

[ecluser]

Ecluser how come I have 3.2v at the C of Qe6 and Re 11 . The schematic as 9.2v

The answer to this is in the attached file I posted. This is why I asked you measure the voltage drop across Re6 and Re14.

It is very easy to follow, simple algebra... Print the file, and follow me...

The analysis starts at Re6. You measured 510mV across the 270 Ohms resistor. This gives the Tail current in the first differential pair:

Tail current 1 = 1.9mA = 510mV / 270 Ohms

The total collector currents in Qe1 and Qe2 MUST equal this tail current.

I1 + I2 = 1.9 mA

From the second differential stage, the base voltages of Qe4 and Qe5 must be equal. The reason for this condition is that in a differential stage with no degenerative resistors on the emitter, the stage will be in strong saturation with as little as 5 to 10 mV voltage difference between the bases. This is a well known fact to every competent electronician.

This condition (Vb4 = Vb5) implies that Vc1 = Vc2 because they are connected together. Since Re4 and Re5 are connected together at the 24V zener diode, the voltage drop across Re4 and Re5 must be equal.

Re4 x I1 = Re5 x I2

If you replace I2 in the last equation by 1.9mA - I1 (from the previous equation), you have

Re4 x I1 = Re5 x (1.9mA - I1)

(Re4 + Re5) x I1 = Re5 x 1.9mA

I1 = (Re5 x 1.9mA) / (Re4 + Re5)

It is easy to solve for any values of Re4 and Re5.

In the original design, with Re4 = 910 Ohms, and Re5 = 390 Ohms, you get

I1 = 0.57 mA, and I2 = 1.33 mA

This stage is very unbalanced, with one collector current more than twice the collector current of the other transistor. Not a good idea on my opinion.

From the computed currents in Qe1 and Qe2, it follows that the collector voltages of both transistors in the first differential stage are 23.5V if the zener is exactly 24V (I don't know what you measured at this point because you didn't reported it, but it must be very close to 23.5V, which is 1V more than the 22.5V on the schematic. Don't worry about this small difference, there is no problem with this)

I want to make two important remarks at this point:

1) The way the Tail current is shared between the two transistors (Qe1 and Qe2) is NOT dependent of the Hfe of the individual transistors. It may be a surprise to some but it is true. It is only the two resistors Re4 and Re5 that set the particular sharing of the Tail current in this design, if we neglect infinitesimal effects. The matching of the Hfe has an effect on the base currents and consequentely on the output DC Offset though.

2) The added 17 kOhms (at Qe3's collector) resistor doesn't figured in the above equations. You remember I told you this resistor changes nothing in the operation of the circuit? Nevertheless I recommand that you ad a 17k resistor (1/4 Watt is OK) in the other channel to help Qe3 to operate in a more comfortable area.

Since the bases of Qe4 and Qe5 are at 23.5V, it follows that their emmiter are at 22.9V (22V on the schematic) if we assume a 0.6V voltage drop in the BE junction.

The Tail current of the second differential stage will be

Tail current 2 = 22.9V / 5600 Ohms = 4.1 mA

Since the two collectors of the second differential stage are connected at the bases of the two transistors in the third differential stage, with no degenerative resistor on their emmiter, the same principle applies as before.

The two collectors voltages of the second differential stage must be equal.

This time, the two collector resistors are the same (680 Ohms), and they share the same current which is half the Tail current 2. The voltage drop in the 680 Ohms resistor is 1.4V

680 Ohms x (4.1 mA / 2 ) = 1.4V

This will set the Tail current 3 in the third differential stage because the voltage drop in the 91 Ohms resistor (Re13) will be 0.6V less than the voltage drop in Re8 or Re9. Remark that the exact voltage at the positive supply rail is not important. This circuit provides a great ripple rejection.

Tail current 3 = (1.4V - 0.6V) / 91 Ohms = 8.7 mA

Now, the great question is... How this Tail current will be shared between Qe6 and Qe7? We need this information to compute the current in Re11, to find the voltage at Qe6's collector.

This is where the voltage drop in Re14 comes to help. You measured 520mV across the 91 Ohms resistor at Qe8's emmiter.

This gives the current in Qe8 = 520mV / 91 Ohms = 5.7mA

This set the bias current in the power transistors module ICe1. If you refer to the schematic for the power IC in your service manual, you will see the diode resistor connection inside the IC between pins 3 and 8 of this IC. This sets the bias voltage for the drivers and power transistors in this IC.

This is also the current across Qe7.

The current in Qe6 is given by the difference between the Tail current 3 and the current in Qe7

I6 = 8.7mA - 5.7mA = 3mA

The voltage at Qe6's collector is thus given by the voltage drop in Re11 (1kOhms) by this 3mA current

Vc6 = 3mA x 1000 Ohms = 3 Volts

In this third stage also, one collector current is twice the other.

The exact value of the voltage at Qe6's collector is not what causes the DC Offset at the output of this amplifier, and by itself is not very important.

If it was my amplifier, I would change Re4, Re5, Re13, to reduce the second order distortion generated by the great unbalance in the first and third differential stages. [Re4* = Re5* = 510 Ohms, and Re13* = 68 Ohms]

I would also change Re15, Re7 and Ce3 to minimise the DC Offset with the previous change in the collector resistors in the first stage. [Re15* = Re3 assuming the two base current in the first stage are equal gives Re15*=56 KOhms. Since the other side of Re3 is connected to ground, this will force the output voltage of the amplifier at ground potential. Re7* = 1.5 KOhms to keep the same close loop gain, Ce3* = 680 uF (to keep the same low cut-off frequency in the feedback branch) or higher]

I would match Qe1 and Qe2 for equal Hfe (and preferably high) transistors.

The two back to back Tantal caps Ce2 and Ce13 must be replaced by a non polar cap (1uF is OK, this gives a 3Hz high pass filter).

I hope it may help every KR-9600 owners...

PS: For a better understanding of the importance of the unbalance in the collector currents in differential stage, see figure 7 of this EXCELLENT paper about amplifier distortion

http://www.dself.dsl.pipex.com/ampins/dipa/dipa.htm#5

[calman46]

Ok I did walk through with the math (thanks for taking time) I'll re-read it again but I get most of it. I read the link also . This unbalance is not how the board should be operating .So instead of changing the resistors &cap to compensate for the change would replacing the differential pair transistors get the balance back ? something is throwing it off. Although Qe6 collector voltage is not important by itself its still off by 6 volts ,thats seems to be a lot. I'm going to take some readings on the other board and see if they match . I'll change out Ce2 &Ce13 I have some 1uf 50v BP caps.Both boards have the same mods by the way .

[ecluser]

Ok I did walk through with the math (thanks for taking time) I'll re-read it again but I get most of it. I read the link also . This unbalance is not how the board should be operating .So instead of changing the resistors &cap to compensate for the change would replacing the differential pair transistors get the balance back ? something is throwing it off. Although Qe6 collector voltage is not important by itself its still off by 6 volts ,thats seems to be a lot. I'm going to take some readings on the other board and see if they match . I'll change out Ce2 &Ce13 I have some 1uf 50v BP caps.Both boards have the same mods by the way .

Changing only the differential pair transistors will never correct the unbalance between the two collector currents. My analysis shows this in the equation

Re4 x I1 = Re5 x I2

If you replace the transistors for higher Hfe, the DC Offset will be lower because the base currents will be lower and the Offset comes from the base current flowing across the base resistors. But higher Hfe will never balances the collector currents.

In this design (called Differential-input, Differential-output mode), connected directly to a second differential stage, the only way to balance this stage with resistors is to have equal value resistors at the collectors (you may also make a current miror but this mod is more difficult).

This circuit is unbalance by (bad) design because they chosed to adjust the collector resistors in the same ratio (910 / 390) as the ratio of the base resistors (56k / 18k), to minimise the DC Offset at the output of the amp. They didn't care about the distortion. The value of the feedback resistors Re15 and Re7 were dictated by the value of the capacitor Ce3. Maybe there was no room for a bigger capacitor than the 220uF / 10V cap on the board, I don't know. But today's caps are smaller and you should find enough room for a larger (in uF) cap.

I repeat, the voltage at Qe6 collector is not important. If you change the value of Re6 or Re14 (I don't recommand that you change it), this will change the voltage at Qe6's collector.

The large DC Offset you have may comes from different Hfe in the first stage, or leaky caps at the bases of the first differential stage. Don't neglect Ce3.

[calman46]

I see you're correcting the unbalance (by bad design). I have some ZTX694B on the way for Qe1 & Qe2 I should have the resistors . Is 1000uf ok for Ce3 ? this value I have .

[EchoWars]

Is 1000uf ok for Ce3 ?Part of the power-up process is charging all these caps. Stick a 1000µf cap in for Ce3, and the relay will engage before this cap is fully charged, and you'll have rather annoying DC thumps at the speakers as the relay engages. There's nothing wrong with the 220µf stock value, but if you feel a need to go larger, 330µf, or maybe even 470µf ought to be fine. I don't think I'd go larger than that, simply because of the time needed to get the sucker charged.

[ecluser]

Part of the power-up process is charging all these caps. Stick a 1000µf cap in for Ce3, and the relay will engage before this cap is fully charged, and you'll have rather annoying DC thumps at the speakers as the relay engages. There's nothing wrong with the 220µf stock value, but if you feel a need to go larger, 330µf, or maybe even 470µf ought to be fine. I don't think I'd go larger than that, simply because of the time needed to get the sucker charged.

Sorry, it's my mistake...

If you change Re7 and Re15 for the values I suggested, you should REDUCE the cap value by a factor 3 and not rise it by the same factor, if you want to keep the same cut-off frequency (time constant if you like).

Try it with the 220 uF cap (the time constant with this cap and 56k in place of the 18k resistor (Re15) is about 12 seconds), or reduce to 68 uF.

Sorry for my error...

[calman46]

I changed Ce3 to 470uF and Ce 2 & Ce13 to 1uf @50v BP . The offset was at 17mV but didn't hold ,its at 199mV + now. Should I go ahead with the resistors change or wait from the replacements for Qe1 &Qe2 ? I'll put Ce3 back to 220uf

[calman46]

I pulled Qe1 & Qe2 and tested the gain and Qe1 is 375 and Qe2 is 68 . I have a meter with hfe option don't know how accurate it is . I also check the resistors Re4 &Re5 and the values on schematic match the board .

[ecluser]

Great mismatch !

Start by replacing them. You may try any transistors in your stock with close and reasonable hFE, and they will be much better than this "pair". Vce is only 30V, Ic less than 2 mA, not difficult to find a good replacement...

The meter will not tell you the exact hFE in the actual circuit because the collector current in the meter is different than the one in the circuit. But if you test two transistors with your meter, and the hFE readings are very close, they will be close also in the actual circuit. You can trust your meter for the purpose of matching small signal transistors.

For the offset which is growing up, I suspect a leaky cap. Make sure Ce3 is in the good direction, that is his + side toward ground and his - side toward the base of Qe2. The leakage of electrolytic caps is much higher if they are reverse polarised.

There is good chance the 470 uF cap is more leaky than the 1uF cap. If polarities are good, and you have growing offset, remove completely the input cap to isolate the input from any external DC input signal. It will rule out the small possibility of a bad 1 uF cap. If it doesn't cure the problem, put the 1uF NP cap in place and put back the original 220uF cap.

[calman46]

I have some ZTX696B Vcbo =180v, Vceo =180, Vebo =5v, Ic 0.5A.I have some 2N5210 also but the Vcbo is 5volts less than the 2sc1345 (Vcbo=55v). I'm going with the ZTX696B there gain measures high 900's not a perfect match though. Either one work here ,right? Did you used the schematic voltage of 22.5v at Qe1 collector to get the 30v Vce ? The gain of some of the 2n5210 is 350-400.

[ecluser]

Either transistor will work there. The gain of the 2N5210 is high enough.

I said Vce less than 30V because the collector will never be higher than + 24V (the 24V zener (De3) sets this upper limit), and the emitter voltage will never be lower than minus 1V.

You know why some prefer to use very high hFE transistors in the differential stage?

The higher the hFE, the lower is the base current for a particular collector current. With a lower base current, the voltage drop in the base resistances will be lower and consequently the output offset voltage. In your amplifier, for instance, if the voltage drop is the same in Re3 than Re15, the output voltage will be at the same potential as the other side of Re3 (the ground), plus the difference in the Vbe drop in Qe1 and Qe2 (Vbe1 - Vbe2), which is very small and is always neglegted in analysis, we assume Vbe1 = Vbe2.

Higher hFE gives good conscience to mask the unbalance in the collector currents in this stage. In general, those who switch to very high hFE transistors in the differential stage don't care about the unbalance in the collector current. As long as they measure low offset voltage at the output of the amplifier, they think everything is OK.

My attitude about this topic is different. I am more interrested to improve the balance in the differential stage than to reduce the offset voltage at the expense of the balance, as it is in the original design of this amplifier.

[calman46]

The ZTX696B did the trick the offset at 1.2 mV ,it fluctuates some but I only had the amp for a few minutes . So Kenwood went with the low DC offset instead of balance C current .

[calman46]

I let the amp run and the offset is fluctuating around 5mV. Would putting Ce2 &Ce13 back to original value of 2.2 uF lower the offset or stabilize it and are BP replacement caps a better choice here ? I took a part the non-working power meter and cleaned out the crap ,it now tests good (needle moves ) .

[ecluser]

If you read this 28 years old , but still excellent, paper:

http://web.archive.org/web/20030807122631/http://capacitors.com/picking_capacitors/pickcap.htm

You will declare war to tantal caps in your audio gears...

To be sure your 1uF NP is good, remove it and operate the amp without input capacitor (open, not replace by a jumper !), as I explained at the end of my #30 post here. See if the offset stay stable.

What is the actual Ce3 ?

[EchoWars]

Perhaps he is expecting a 'set' value for offset, which is rarely the case. Thermal drift will cause offset to 'float' by a few mV...slowly rising and then falling.

[calman46]

He's not expecting a set value for the offset just to slow down the fluctuating some. Actual Ce3 is a new 220uf 50V cap (high temp).It seems to be better now it went has high as 8mV and then down to 4.5mV . I'm just trying to tweak it but I can live with 5mV offset. What do think the offset was when it came out of the factory?

[ecluser]

I hope you can live with 5 mV offset. It represents no more than 4.2 microwatt of power in a 6 Ohms load. You won't cook your woofer with this !

Glenn is right, it is normal to have some thermal drift.
Vbe and hFE vary with temperature and collector current.

In case you are still concerned about the 9.2V figure on the schematic, across Re11, I want to assure you that it is impossible to have this. I computed backward, from the third diff. stage to the first, assuming 9V at Qe6's collector, and the bias current in Qe8 from the voltage drop you measured in Re14. I found 40V at Qe1 and Qe2's collectors. But the zener has only 24V.

[calman46]

Thanks , One more tech question please, why did you want Non-polarize caps for Ce2 &Ce13 ? Would using 2.2uf(original value) polarized caps make a difference here ? The tanal caps were polarize (which are gone).

[markthefixer]

:lurk: :thmbsp: :D

[EchoWars]

Thanks , One more tech question please, why did you want Non-polarize caps for Ce2 &Ce13 ? Would using 2.2uf(original value) polarized caps make a difference here ? The tanal caps were polarize (which are gone).If you have some decent non-polar caps, you don't need both Ce2 and Ce13...replace one of them with a clipped lead from a resistor...

[ecluser]

If you read the paper I suggested on my post 35, you will see that the two back to back tantal caps approach is a very imperfect (there should be at least a negative bias between the two caps, which is not the case in this amp) attempt to make a non polar cap from two polar caps.

For Ce13, you need a non polar cap because the input voltage is an AC signal that may vary from -1.5V to +1.5V approximately, according to the sensitivity spec (1VRMS) for this amp. A non polar cap here is a must.

For the value, when you put two caps (say C1 and C2) in series like this, the net capacitance C of this combo is given by

C = C1 x C2 / (C1 + C2)

In this particular case, the net capacitance is only 1.1uF. This is for this reason I suggested to replace this bad combo by a 1uF non polar cap.

For Ce3, a non polar cap would be fine because the base voltage for Qe2 is the same amplitude as Qe1. But polar caps are smaller and much cheaper than non polar flavor, for the same capacitance. Also, you need a much larger capacitance on this side because Re7 is much lower than Re3. This is a place where manufacturers cut construction costs.

One of the benefits about the mods I suggested on this thread, is that if you replace the resistors as I suggested, you may also replace the 220uF polar cap by a 68uF non polar one (10V should be OK). You should find room for this lower capacitance non polar cap on the board.

I often bypass this large cap by a good film cap, that I glue (a drop of silicone adhesive or a drop of glue from a hot glue gun) on the copper side of the printed circuit board. A rule of thumb is to use a film cap of any value between 1/100 and 1/1000 of the capacitance of the feedback capacitor. Usually, there is room for this.

[calman46]

thank you very much. great explanation

[calman46]

I finally put it back together .I used a satin hammered finish spray paint on the case came out nice .This unit is big and heavy .The light is not good for pic here but I posted some anyway. Thanks for all the help . I started on a DBX-BX3 ,this could be tough the service manual is not for beginners ,not much info.