BAJA Silent Selector

[bajaman]

you are using momentary (non latching) switches I assume. If not, you should be.
bajaman

[bajaman]

time to get a new drill bit
bajaman

[Barcode]

one thing I can see straight away - the extreme right diode is the wrong way around.
bajaman

DUH!!!!!! Okay, I'll fix that and report back with results. Thanks!

EDIT: Okay, flipped it right way round. No dice. any other ideas?

[bajaman]

okay - do the momentary switches cause the indicator lights to change as expected or not??
bajaman

[Barcode]

okay - do the momentary switches cause the indicator lights to change as expected or not??
bajaman

no, the lights don't come on at all. I'm going to hook up the input jack and test it so i can tell whether the path is at least changing. then it would just be an LED problem. I'll let you know.

[bajaman]

Are you sure the indicator LEDs are connected the correct way around
If not sure, then reverse their polarity and try again.
bajaman

[Barcode]

Are you sure the indicator LEDs are connected the correct way around
If not sure, then reverse their polarity and try again.
bajaman

Eureeka! Actually, The border around the image on the pcb i'm guessing the pnp melted a little too much, the border was shorting some connections. Now i have working LEDs, my only issue now is that the combine works, and output a works, but (again, this is without testing for actual signal, just based on LEDs) i can't get "B" only. Again, i'll wire it up to an input hack so i can see if the LED behavior actually matches what is going to be heard.

[bajaman]

output b will only work on it's own if the combine switch latches the cmos chip.
let me know if you are still having problems.
cheers
bajaman

[Barcode]

i'm SO close all the LEDs are working properly now, except that when single output A is selected, the "B" LED is still slightly on. I'm thinking I just need to raise the value of the LED resistor on that side to fix it, but any input from you on that one?

[Barcode]

i'm SO close all the LEDs are working properly now, except that when single output A is selected, the "B" LED is still slightly on. I'm thinking I just need to raise the value of the LED resistor on that side to fix it, but any input from you on that one?

Okay, definitely something amiss. The path doesn't switch from one to the other. "A" works fine, but when i switch to "B" I just get a louder version of "A." ????

[bajaman]

wtf - the bottom left most diode to the left of the 2N5088 and next to the electrolytic should be a 150 ohm 1/2 watt resistor - replace that and the circuit will work as designed.
bajaman

[Barcode]

wtf - the bottom left most diode to the left of the 2N5088 and next to the electrolytic should be a 150 ohm 1/2 watt resistor - replace that and the circuit will work as designed.
bajaman

wow. Just wow. Okay, I'll fix that and get back with you

[bajaman]

If you do not have a 150 ohm half watt, you can always use two 330 ohm 1/4 watt resistors in parallel with each other ( or a 330 ohm and 270 ohm in parallel ).
The higher wattage is required, because when the 2N5088 transistor turns on the complete 9 volt power supply voltage is dropped through this resistor. ( 9/150 = 60mA of current flowing - wattage is voltage times current or 9 times 60mA = .54 watts !!! - that is why it is a half watt resistance )
cheers
bajaman

[Barcode]

If you do not have a 150 ohm half watt, you can always use two 330 ohm 1/4 watt resistors in parallel with each other ( or a 330 ohm and 270 ohm in parallel ).
The higher wattage is required, because when the 2N5088 transistor turns on the complete 9 volt power supply voltage is dropped through this resistor. ( 9/150 = 60mA of current flowing - wattage is voltage times current or 9 times 60mA = .54 watts !!! - that is why it is a half watt resistance )
cheers
bajaman

Hmm, this is interesting. When you get time, is it possible you could explain exactly how all the stages of this circuit function? It is quite the unique design (at least to me) and I find this one fascinating enough that I'm not satisfied by "painting by number" on it.

[bajaman]

After you get it working - I will give you a breakdown on how it works - a very ingenous use of a CMOS 4069 hex inverting buffer.
There are 6 completely independent inverting buffers inside a 4069 - an inverting buffer has the ability to change the state of the DC votage applied to it's input. For example: +9 volts at the input will give zero volts at the output, and zero volts at the input will give+9 volts at the output.
In this particular circuit, two sets of two buffers are arranged back to back in parallel out of phase with each other. This particular building block is known affectionately as a "flip flop". They are situated under the SWITCH and COMBINE momentary switch headers n the schematic. Okay - notice the 100k resistors and 1uf electrolytic capacitors hanging off each of the momentary switch and inverter outputs. These form a CR charging circuit - 0.1 seconds to charge the electrolytic up to +9 volts.
Still with me I hope.
Referring to the bottom SWITCH when it is open (at zero volts) we have +9 volts at the flip flop output (pins 10 and 13).
Okay - now this + 9 volts flows through a 100k , 1N4148 and 1k2 resistor and turns the "A" led on. Likewise optocoupler LED 4 is turned on and because the + 9 volts encouters an inverter - optocoupler LED 3 is turned off (zero volts).
This gives a low resistance to the optocoupler in series with the signal voltage and an extremey high resistance to the optocoupler in parallel with the "A channel" output.
The selector is set to output signal unimpeded througfh output "A".
What happens when the momentary SWITCH is closed? Well - the 1uf capacitor which is charged up to +9 volts is now applied to the inverter input and the output now becomes zero volts, effectively reversing the previous state conditions.
Now inverter (9 input and 8 output) outputs +9 volts and in doing so turns LED "B" and optocoupler LED 2 on. It also turns the 2N5088 transistor on, which because the collector is now at zero volts, turns off optocoupler LED 1. Now signal voltage is sent unimpeded through the "B channel" output. The electrlytic capacitor next to the SWITCH contact is now of course drained (0.1 second) and at zero volts. So what happens when we press the switch again - that's right, the output becomes +9 volts again because we have connected a zero volt pulse to the flip flop inverter input (pin 11), and the "A channel" is energised once again.
See if you can figure the rest - including the combine out.
cheers
bajaman

[Barcode]

After you get it working - I will give you a breakdown on how it works - a very ingenous use of a CMOS 4069 hex inverting buffer.
There are 6 completely independent inverting buffers inside a 4069 - an inverting buffer has the ability to change the state of the DC votage applied to it's input. For example: +9 volts at the input will give zero volts at the output, and zero volts at the input will give+9 volts at the output.
In this particular circuit, two sets of two buffers are arranged back to back in parallel out of phase with each other. This particular building block is known affectionately as a "flip flop". They are situated under the SWITCH and COMBINE momentary switch headers n the schematic. Okay - notice the 100k resistors and 1uf electrolytic capacitors hanging off each of the momentary switch and inverter outputs. These form a CR charging circuit - 0.1 seconds to charge the electrolytic up to +9 volts.
Still with me I hope.
Referring to the bottom SWITCH when it is open (at zero volts) we have +9 volts at the flip flop output (pins 10 and 13).
Okay - now this + 9 volts flows through a 100k , 1N4148 and 1k2 resistor and turns the "A" led on. Likewise optocoupler LED 4 is turned on and because the + 9 volts encouters an inverter - optocoupler LED 3 is turned off (zero volts).
This gives a low resistance to the optocoupler in series with the signal voltage and an extremey high resistance to the optocoupler in parallel with the "A channel" output.
The selector is set to output signal unimpeded througfh output "A".
What happens when the momentary SWITCH is closed? Well - the 1uf capacitor which is charged up to +9 volts is now applied to the inverter input and the output now becomes zero volts, effectively reversing the previous state conditions.
Now inverter (9 input and 8 output) outputs +9 volts and in doing so turns LED "B" and optocoupler LED 2 on. It also turns the 2N5088 transistor on, which because the collector is now at zero volts, turns off optocoupler LED 1. Now signal voltage is sent unimpeded through the "B channel" output. The electrlytic capacitor next to the SWITCH contact is now of course drained (0.1 second) and at zero volts. So what happens when we press the switch again - that's right, the output becomes +9 volts again because we have connected a zero volt pulse to the flip flop inverter input (pin 11), and the "A channel" is energised once again.
See if you can figure the rest - including the combine out.
cheers
bajaman

I think i kind of get what's going on. Basically, at the highest level, a cmos inverter is driving LEDs arranged in a photocoupling manner with LDRs. To cut the signal of either path, the cmos inverter is turning on an LED attached to the opposite indicated path which introduces enough resistance to kill the signal. Is that correct? I'll tackle the rest a little later on when my head is clearer

[Barcode]

After you get it working - I will give you a breakdown on how it works - a very ingenous use of a CMOS 4069 hex inverting buffer.
There are 6 completely independent inverting buffers inside a 4069 - an inverting buffer has the ability to change the state of the DC votage applied to it's input. For example: +9 volts at the input will give zero volts at the output, and zero volts at the input will give+9 volts at the output.
In this particular circuit, two sets of two buffers are arranged back to back in parallel out of phase with each other. This particular building block is known affectionately as a "flip flop". They are situated under the SWITCH and COMBINE momentary switch headers n the schematic. Okay - notice the 100k resistors and 1uf electrolytic capacitors hanging off each of the momentary switch and inverter outputs. These form a CR charging circuit - 0.1 seconds to charge the electrolytic up to +9 volts.
Still with me I hope.
Referring to the bottom SWITCH when it is open (at zero volts) we have +9 volts at the flip flop output (pins 10 and 13).
Okay - now this + 9 volts flows through a 100k , 1N4148 and 1k2 resistor and turns the "A" led on. Likewise optocoupler LED 4 is turned on and because the + 9 volts encouters an inverter - optocoupler LED 3 is turned off (zero volts).
This gives a low resistance to the optocoupler in series with the signal voltage and an extremey high resistance to the optocoupler in parallel with the "A channel" output.
The selector is set to output signal unimpeded througfh output "A".
What happens when the momentary SWITCH is closed? Well - the 1uf capacitor which is charged up to +9 volts is now applied to the inverter input and the output now becomes zero volts, effectively reversing the previous state conditions.
Now inverter (9 input and 8 output) outputs +9 volts and in doing so turns LED "B" and optocoupler LED 2 on. It also turns the 2N5088 transistor on, which because the collector is now at zero volts, turns off optocoupler LED 1. Now signal voltage is sent unimpeded through the "B channel" output. The electrlytic capacitor next to the SWITCH contact is now of course drained (0.1 second) and at zero volts. So what happens when we press the switch again - that's right, the output becomes +9 volts again because we have connected a zero volt pulse to the flip flop inverter input (pin 11), and the "A channel" is energised once again.
See if you can figure the rest - including the combine out.
cheers
bajaman

I think i kind of get what's going on. Basically, at the highest level, a cmos inverter is driving LEDs arranged in a photocoupling manner with LDRs. To cut the signal of either path, the cmos inverter is turning on an LED attached to the opposite indicated path which introduces enough resistance to kill the signal. Is that correct? I'll tackle the rest a little later on when my head is clearer

AARRGH!!!!!! I'm still having the issue even after replacing thatdiode with the appropriate resistor! In peering in the homegrown photocouplers, it appears that the top right one in the picture is never fully turning off. Still have no path switching either

[bajaman]

okay - 1 - are the indicator channel LEDs turning on and off correctlyand completely when you press the changeover switch ?
2 - does one of these indicator lights stay on all the time when you press the changeover switch after pressing the combine switch?
bajaman

[Barcode]

okay - 1 - are the indicator channel LEDs turning on and off correctlyand completely when you press the changeover switch ?
2 - does one of these indicator lights stay on all the time when you press the changeover switch after pressing the combine switch?
bajaman

"A" turns on and off as expected. combine turns both on. "B" stays on constantly, whether i've hit the combine switch or not.

[bajaman]

Okay - are you sure those two diodes near pin 7 of the cmos chip have their black bands closest to pin 7??
If so, then try changing the 4069 - these are very static sensitive, so handle with care with the power supply disconnected of course.
Once again - you are using momentary (push to make contact) switches???
Post another photo please - check for solder bridges on solder side where there should not be any!!
good luck - perseverance will get it working properly.
bajaman