Whitby, near Toronto. Sayal electronics is my go to.
Put a capacitor in backwards and it fixes the problem???!!!
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Whitby, near Toronto. Sayal electronics is my go to.
Not made anymore... found this on Audiokarma..."appears to be a..Si, ¼W, 2V zener. According to some sources, it can be replaced by BZ102/2V1, BZV86/2V0, BZX75/C2V1, ZTE 2, and others." Others have run into this problem if finding a replacement.
Or this...
Replacement diode stack. I used three 1N4148, cause I had in hand in that time. Did the trick for me.
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Ylli
Super Member
BZX75/C1V4 looks to be the best match, but as you say, that one seems to be unobtainium too.
General consensus seems to be to replace it with a series connected pair of 1N4148's. But I did a sim on it, and just dropping in the 1N4148's leaves things a bit short on bias. That is assuming the 1N4148 model in the sim program is accurate. Adding a series resistor with the 1N4148's can get the voltage up to where it needs to be but the adjustment becomes very touchy (they usually are but the sim seems to indicate this would be more touchy than usual). That touchiness can be reduced a bit by adding a resistor in parallel with the adjustment pot.
Anyway, I offer the following for your consideration....
Disconnect the KB-269 from its terminal strip at the heat sink. In its place, add a series network of qty 2 1N4148 diodes and a 100 ohm resistor. Observe polarity as shown in the schematic (or by reference to original part). The diodes should be positioned so that they make thermal contact with the heat sink in the same manner as the KB-269 did, but be careful not to allow the leads to short to the heat sink or to each other. The resistor can be positioned anywhere. Then add a 2.2K resistor from the high side of the pot to the low side of the pot. We will want to take some measurements before reinstalling the output transistors, as the real world is not always the same as a simulation. You would have the assortment below in case we needed to tweak some values.
The 1N4148's are available at your store, https://secure.sayal.com/STORE2/View_SHOP.php?SKU=247726 would be my choice.
For the resistors, they have an assortment that contains the required values. https://secure.sayal.com/STORE2/View_SHOP.php?SKU=223948
In the pix, ignore the component identifiers - the sim program assigns them and I didn't go through and edit them. The pot is VR606 and Q1 is actually TR605.
Wow, this is a lot of effort for a 50 year old $5 receiver! I feel bad that you have put in so much effort! But I appreciate it all, and I love not seeing stuff go to the dump.
I will pick up the parts tomorrow or Tuesday. I will probably semi- assemble it and take a picture to verify that I am doing it correctly. So the diodes and resistor are going along the blue and purple wires, where kb269s were? Crowded, but I'll figure it out.At the same time I'll show you a picture of how I'll solder the resistor in parallel with VR606.
Thanks again, I'll hit sayal tomorrow.
I will pick up the parts tomorrow or Tuesday. I will probably semi- assemble it and take a picture to verify that I am doing it correctly. So the diodes and resistor are going along the blue and purple wires, where kb269s were? Crowded, but I'll figure it out.At the same time I'll show you a picture of how I'll solder the resistor in parallel with VR606.
Thanks again, I'll hit sayal tomorrow.
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Ylli
Super Member
Sounds good. I'm glad you don't consider it a high value item, as I *hate* making changes to circuitry. After all, a highly paid engineer did the original design, and supposedly he/she knew how to do it. But when parts become unavailable sometimes you have to make changes.
When putting the resistor in series with the diodes, do it in such a way as it will be easy to change it. You may actually want to put just the diodes up by the heat sink and put the added series resistor in line with one of the wires at the circuit board - just lift one of the wires and add the resistor in between the wire and the board. Remember that when things are in series, it does not matter what order they are in We may have to juggle some resistor values to get the adjustment range where we want it.
Are we having fun? Yes!
When putting the resistor in series with the diodes, do it in such a way as it will be easy to change it. You may actually want to put just the diodes up by the heat sink and put the added series resistor in line with one of the wires at the circuit board - just lift one of the wires and add the resistor in between the wire and the board. Remember that when things are in series, it does not matter what order they are in We may have to juggle some resistor values to get the adjustment range where we want it.
Are we having fun? Yes!
Ok, got the components. Resisters are non polar, correct? So that's easy. Diodes have polarity, right? So, to be clear, how should the diodes be attached in relation to the blue and purple wires?(I'm not replacing both channels right?)
And in the VR606, which pins do I attach the resistor to?(again, I'd hate to make a mistake now)
And in the VR606, which pins do I attach the resistor to?(again, I'd hate to make a mistake now)
Ylli
Super Member
Resistors are non-polar. You are working only one channel. A diode has an anode and a cathode.
You want to connect two diodes in series with the cathode of the first soldered to the anode of the second. Keep the distance between the first and second to a reasonable length, but you do not want to over stress the leads and physically break the diodes. Carefully desolder and remove D601 and put this diode pair in it's place. Then on the other end of the purple wire, lift it from the board, and install the 100 ohm resistor between the board and the wire.
Let's leave the shunt resistor across VR606 off for now, and take some measurements. Turn VR606 fully CCW.
Measure the voltage from point A to chassis ground
Measure the voltage from point B to chassis ground
Measure the voltage between point A and the emitter of TR605.
Then a picture or two of your work may be helpful.
You want to connect two diodes in series with the cathode of the first soldered to the anode of the second. Keep the distance between the first and second to a reasonable length, but you do not want to over stress the leads and physically break the diodes. Carefully desolder and remove D601 and put this diode pair in it's place. Then on the other end of the purple wire, lift it from the board, and install the 100 ohm resistor between the board and the wire.
Let's leave the shunt resistor across VR606 off for now, and take some measurements. Turn VR606 fully CCW.
Measure the voltage from point A to chassis ground
Measure the voltage from point B to chassis ground
Measure the voltage between point A and the emitter of TR605.
Then a picture or two of your work may be helpful.
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At 3/4 rotation (which was roughly its original setting), at 2v, it jumps around a lot, around 1400
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Ylli
Super Member
Changes look OK for now (after reversing those backwards diodes 
.
If your meter reads "1", that is an over range and you need to change the meter setting. If it is on the 2V range and you get a "1", then you need to change the meter to the 20 volt range. You want the meter on the most sensitive range you can while not getting an over range reading ("1"). Here, you are going to want to use the 20 volt range.
. If your meter reads "1", that is an over range and you need to change the meter setting. If it is on the 2V range and you get a "1", then you need to change the meter to the 20 volt range. You want the meter on the most sensitive range you can while not getting an over range reading ("1"). Here, you are going to want to use the 20 volt range.
Ylli
Super Member
Good believable numbers, but too low. I though that might be the case - please change the 100 ohm resistor to 470 ohms (yellow-violet-brown) and repeat the measurements. I'm looking for the point 'A' to TR604 emitter voltage to be less than 1.9 volts with the pot fully CCW, and around 2 with the pot about 75%.
First thing Thursday morning. Shouldn't the higher ohms lower the measurement? I seem to be missing some fundamental understanding.
Ylli
Super Member
There is a current flowing from the anode of the first diode, through the second diode and then thru the resistor to the base of the transistor. Some of that base current is bypassed by the adjustable resistor VR606.
So in this series string we have two diodes (for a drop of 0.6 each or 1.2 volts), the drop across the added resistor, and the base emitter drop of the transistor (0.6 volts). Without the added resistor that adds up to about 1.8 volts. We need a bit more than that to turn on the following stages. We get that extra voltage via the drop across that added resistor - the higher that resistor value, the more voltage will drop across it; hence the higher the bias voltage (collector (or 'point A') to emitter of TR605).
Clear as mud?
So in this series string we have two diodes (for a drop of 0.6 each or 1.2 volts), the drop across the added resistor, and the base emitter drop of the transistor (0.6 volts). Without the added resistor that adds up to about 1.8 volts. We need a bit more than that to turn on the following stages. We get that extra voltage via the drop across that added resistor - the higher that resistor value, the more voltage will drop across it; hence the higher the bias voltage (collector (or 'point A') to emitter of TR605).
Clear as mud?