Problem with Dynalo
Nov 19, 2005 at 5:18 AM Post #16 of 35
Quote:

Originally Posted by robzy
  1. Volume pot only works for first few degrees, then the volume is on the highest it can be - which is no louder than what the source is itself (And ive double checked that ive done the pot correctly)
  2. Output transistors get very hot. But some more than others. One row i cant hold my finger on for more than a couple of seconds
Rob.



Ok that's a start. Check the feedback resistors are the right way around (did that mistake once), and make sure no more of the output tranny's are wrong. Also in a cascode the output transistors are very dependant on the preciding ones.
 
Nov 19, 2005 at 9:06 AM Post #17 of 35
Quote:

Originally Posted by dgardner
I don't know what technique you are using to measure this but it is too high. You can expect to see about 140mA per rail. Even with the current sources running a little too high, you should see no more than 170mA per rail.

Sounds silly, but check the board to make absolutely sure you have all the other resistor *values* correct. Look at the markings carefully. It is sometimes easy to be off by a power of 10 on something.

Report back the curent draw on the negative rail if you get a chance...

EDIT: Report back the voltage you see across any one one of the 24.9ohm resistors on the output stage. Measure one on each side of the amp.



I am getting 220ma on one rail, 170ma on the other. After time though, they rise to 240ma and ~190ma.

All resistors are correct, except R19 and R49 are 500ohm instead of the 215ohm quoted in the schematic because i did not use the diodes there.

Voltage accross 24.9ohm resistors are between .6 and .7, but they seem to be steadly increasing, in ~20 seconds one of them went from low .6 to high .6

Quote:

Originally Posted by Garbz
Ok that's a start. Check the feedback resistors are the right way around (did that mistake once), and make sure no more of the output tranny's are wrong. Also in a cascode the output transistors are very dependant on the preciding ones.


Feedback resistors are the right way around - just double check. I have also just checked for the third time the output trannies.

Thanks,
Rob.
 
Nov 19, 2005 at 3:03 PM Post #18 of 35
Ok excuse the obvious question but check that you jumpered the diodes, you can't outright remove them.

Also I'd remove all the adjustment pots. And the 15k resistor running from the servo output to the 7.5k resistors. Sorry I can't be more specific i'm not reading from the same schematic as you.

This will eliminate the possible effect of any aditional biasing, and the servo acting up. From there if it's still not working I'd check the 1k (they should read 1001 on visheys, not 1000!) and also the 500s and 5k (one again the last significant digit is the power so 5k = 5001 not 5000)
 
Nov 19, 2005 at 3:49 PM Post #19 of 35
Here is brief recap of the currents and how to check them:

Section - Resistor, Current, Voltage Drop
Current Source - 500, 0.0020, 1.0000
JFET Stage - 5000, 0.0010, 5.0000
VAS Stage - 510, 0.0045, 2.2995
Output Stage - 24.9, 0.0150, 0.3735

One key thing to remember is that any currents that are too high will effect the next stage down the line. Excess currents problems become cumulative. Also, all the currents are carefully balanced in relationship to each other to achieve the proper operation.

So, the best way to correct the problem is to fix it at the source. Make sure that you have NO MORE that 1.0 v across the 500 ohm resistors. Said another way: confirm your current source magnitude by computing I=V/R, Where I is the current, R is the total resistance, and V is voltage across the resistor. Change resistors and/or pots until you compute 2mA or slightly less.

See This Post for additional information.

Assuming all the other parts are populated correct, the other measurements will almost always fall directly into place..
wink.gif
 
Nov 20, 2005 at 12:51 AM Post #20 of 35
Quote:

Originally Posted by Garbz
Ok excuse the obvious question but check that you jumpered the diodes, you can't outright remove them.

Also I'd remove all the adjustment pots. And the 15k resistor running from the servo output to the 7.5k resistors. Sorry I can't be more specific i'm not reading from the same schematic as you.



Haha, yup, i jumpered them. Else i wouldnt be getting a voltage drop reading accross the 500ohm resistor i have there.

Quote:

Originally Posted by dgardner
One key thing to remember is that any currents that are too high will effect the next stage down the line. Excess currents problems become cumulative. Also, all the currents are carefully balanced in relationship to each other to achieve the proper operation.


Okay then, so my first issue is getting 2ma accross the 100k pots and 100o resistors.

This is where my lack of basic knowledge sort of comes into play. I believe that when you measure the restance accross the 500ohm resistor - when it is isntalled - you get the combined paralel resitance of the pot and the resistor.

I also believe that changing the value of that resistance, does not chance the voltage drop over it, only the current. This is because no matter what that resistence is i am measuring 1.22v.

Following that logic... V=IR:

1.22 = (0.02)R
R = 650

So i need the combined value of the resistor and the pot to be 650 ohm. Impossible when the resistor is 500ohm and the pot is 100k ohm :p

So now, correct me if im wrong, but this could all be fixed by replacing the 500o resistor with a 1k resistor and then setting up the pot so i read 650 ohm accross it?

Rob.
 
Nov 20, 2005 at 1:08 AM Post #21 of 35
It's true the formula for resistors in parallel is 1/Rt = 1/R1 + 1/R2 +...+1/Rn
You can't get 650ohm from a 500ohm and a 100k. The highest you can get from this combination is 497

The pots are designed for offset correction, which is why I sugested removing them. Start by trying to build an amp which works properly but gives you offset. Tackle one problem at a time. The simpler things are the easier it is to trouble-shoot.

The other option would be the option I took. This is not populating the resistors in parallel and putting a 1k pot in instead to adjust the bias from 0-1k.
 
Nov 20, 2005 at 1:19 AM Post #22 of 35
Quote:

Originally Posted by Garbz
The pots are designed for offset correction, which is why I sugested removing them. Start by trying to build an amp which works properly but gives you offset. Tackle one problem at a time. The simpler things are the easier it is to trouble-shoot.

The other option would be the option I took. This is not populating the resistors in parallel and putting a 1k pot in instead to adjust the bias from 0-1k.



Would it not just be easier to replace the 500ohm resistors with 1k ohm resistors? That would easily be able to give me 650ohm and everything will be happy.

I personally dont believe one extra pot or resistor really makes the amp that much more complicated, so im happy to leave it in. (Especially since they work in tandom).

So, its correct that i need 650ohm there in order to create the 2ma of bias current - right?

Rob.
 
Nov 20, 2005 at 1:21 AM Post #23 of 35
Quote:

Originally Posted by robzy
This is because no matter what that resistence is i am measuring 1.22v.


Yes, the voltage will always be around V(LED)- V(BE). You have 1.8 volt LEDs installed. Quote:

Originally Posted by robzy
So i need the combined value of the resistor and the pot to be 650 ohm. Impossible when the resistor is 500ohm and the pot is 100k ohm .


Precisely. You need around 650 ohms. Easiest way to get this is to raise the fixed resistor from 500ohms to 681ohms (consult the 1% preferred values table). Next, since you have a 100K pot, you need to dial it down to around 15Kohms. 15K in parallel with 681 results in 651 ohms. ( formula for R1 parallel to R2 is (R1*R2)/(R1+R2) ) If you have a 20K pot it might have more adjust ability, but you can stick with the 100K. I believe you can preset the total R value to 650 with the board powered off. Just hook your meter across the 681 resistor and dial the pot till the meter reads 650 ohms. Do this for all four locations. Then hopefuly you are done and can get on with some fun listening.
280smile.gif
 
Nov 20, 2005 at 1:36 AM Post #24 of 35
Quote:

Originally Posted by dgardner
Do this for all four locations. Then hopefuly you are done and can get on with some fun listening.
280smile.gif



The appropriate word(s) to be used here are "Woohooo" :p

Thanks a heap dgardner and Garbz!

I have a feeling i will be back though. I get the feeling that the left channel is not going to work straight off (because of the fact i cant get any dc offset reading on it at all).

Anyway, ill report back later,
Rob.
 
Nov 15, 2009 at 8:14 AM Post #25 of 35
Old thread, but I recently found enough parts around to build a second dynalo board (built with the original schematic). Unfortunately, for the new board, only one channel works. I get the following voltages for the various stages on the faulty board:

CCS: 1.072V; 1.09V
JFET: 5.33V; 5.41V
VAS: 2.2V
Output: All around 0.48V

DC Offset is 0.01V cold and settles below my multimeter's range (was 0.36V before I paralleled pots to each of the 500 ohm resistors).

These measurements are quite similar for the the operational channel. Only difference between the channels is that the non-working side uses salvaged k389 and j109 from another project, while the working one uses nos i had around. Output transistors are all Y grade, and same batches. Any ideas as to my problem?
 
Nov 15, 2009 at 2:49 PM Post #27 of 35
If the voltage readings are the same and the DC offset is normal, are you sure you hooked up the pot and jacks right? Is the dead channel dead silent?

Can't you measure the voltage across the resistors for all stages LTP, VAS and output so we know if the currents are OK.
 
Nov 15, 2009 at 7:59 PM Post #29 of 35
I will try to get some photos later, but I currently do not have a camera. The channel is indeed silent, and measurements from those test points checked out. I am running a Welborne PS1 at +/-16.4V. I have checked and rechecked that the pot and jacks are wired correctly (even removed the pot to make sure that wasn't the problem). I am really quite puzzled. I'm assuming the normal operating currents rule out dead transistors?
 
Nov 15, 2009 at 8:14 PM Post #30 of 35
I would assume that normal operating voltages would rule out either dead transistors or certainly any transistors in the wrong place. You can probably assume that the currents are correct as well. If you have a scope, you could trace the music signal thru the amp. Possibly even with an AC voltmeter set on mV. I suppose you soldered the dual FETs in instead of socketing them? Swapping those between channels would be a start.
 

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