It tested fine originally. I may have damaged it in my attempts to isolate the faults when I was making smoke! I hope Q801 is not bad too.
What you need to be very aware of is that a fault that raises the 15 volt regulated rail could cause other problems as it effectively runs everything on that line in an overvolts condition. That hopefully is where the 18 ohm might come into play and help minimise any damage.
I managed to damaged Q802. The search for a replacement is starting. I know the gig by now...
There is something I quite don't understand:
The output of bridge rectifier gives me 31.5 V. This voltage then goes through R807, the 2.2 K. The voltage is now 23.3 V. Using the formula v=ri, I should rather have 31.5 - (2200 ohm*0.00096mA=2.1V)=29.4. Current seems to be 4 time higher in this part of the circuit because the drop is 31.5 - 8.2 =23.3V.
In reality, I need a drop of 15.9V. I'm wondering how this amp ever got this value when it came out of the factory!
Here is an improved schematic view to help the discussion:
There is something I quite don't understand:
The output of bridge rectifier gives me 31.5 V. This voltage then goes through R807, the 2.2 K. The voltage is now 23.3 V. Using the formula v=ri, I should rather have 31.5 - (2200 ohm*0.00096mA=2.1V)=29.4. Current seems to be 4 time higher in this part of the circuit because the drop is 31.5 - 8.2 =23.3V.
In reality, I need a drop of 15.9V. I'm wondering how this amp ever got this value when it came out of the factory!
Here is an improved schematic view to help the discussion:
These are text book generic circuits. TIP41C (I know its a different package) or similar for Q801 and any suitably rated small signal for Q801 such as a BC546, 2N5551 etc. Watch the pinouts though.
It doesn't matter what the bridge voltage is 25 volts, 35 volts, 50 volts. The regulated output will not change.
That's not really correct 🙂 the voltage is applied to the resistor. The load current (everything powered off the 15 volt rail) draws current and that causes a volt drop across the 18 ohm. The current transistor can supply (from Collector to Emitter) is determined to some extent by the available base current. Although the 2.2k turns the transistor on there isn;t enough current gain in the transistor and not enough base current (via the 2.2k) to deliver unlimited current.
So what happens is is the load draws current. That causes the voltage on the collector to fall because of R18. The transistor may or may not appear as fully conducting at that point. That depends on the gain of the transistor. The base voltage will always be around 0.6 volts higher than the emitter.
So we could have +32 volts on the bridge. The emitter could be at 12 volts because the load is drawing 1.1 amps. That 1.1 amps is dropping 1.1*18 = 19.8 volts across the 18 ohm. The transistor is fully on (lets assume) because the 2k2 is supplying 12.6/2200 = 5.7 milliamps into the base. The 12.6 volts is the emitter volt plus 0.6v.
We therefore have 32 volts bridge, 12 volts on the emitter and 12.6 volts on the base. The 2k2 drops 19.4 volts and the 18 ohm drops 20 volts. The transistor has virtually no voltage dropped across C and E because it is fully on.
In practice the base current may not turn the transistor fully on and so some voltage would be lost across C and E. That depends on the transistor and its gain.
Q802 in the full circuit conducts just enough to maintain the set 15.9 volts no matter what the bridge voltage and no matter what the load.
This is awesome! You are putting so much efforts in explaining this to me that I ought to have it working. My biggest problem right now is my local store won't help me much in finding a replacement for Q2. I may have to go online and pay the $20 CDN shipping for a $0.43 part.
Out of curiosity, does your software allow to enter real values? This design was originally made for a zener diode of 8.4V which is not standard nowadays. The zener I have has a very low voltage of 7.9V. I was recalculating R3 and found that I needed a little above the 6.8 K resistor.
My current values:
R1 = 17.5
R2 = 2.13 K
R3 = 6.76 K
R4 = 9.0 K
D1 = 7.9 V
I see that LTSpice is free. I will try it by myself if I have time.
Out of curiosity, does your software allow to enter real values? This design was originally made for a zener diode of 8.4V which is not standard nowadays. The zener I have has a very low voltage of 7.9V. I was recalculating R3 and found that I needed a little above the 6.8 K resistor.
My current values:
R1 = 17.5
R2 = 2.13 K
R3 = 6.76 K
R4 = 9.0 K
D1 = 7.9 V
I see that LTSpice is free. I will try it by myself if I have time.
Firstly, if you click my signature line it will take you to the LTspice thread. The software is a doddle to install and then you can just 'click and run' a simulation file. I'll attach the one for this power supply in case you do want a play. Just click the file and it will open in the program. Click run on the top line and away it goes.
The transistors should be easily replaceable with generic parts, the only issue being a different package outline.
If the TO66 packaged Q801 is mounted without an insulating kit (meaning the case which is the collector is in direct contact with the heatsink) then you can just screw a flatpak type like a TIP41 onto the heatsink, two wires to base and emitter and its good to go. The metal tab of the flatpak is the collector. If it is insulated then you would need an insulating kit (mica washer and insulating bush) for the flatpak.
Anything you want 🙂 We can swap the Zener in the simulation for a voltage source and dial in any value we wish. Instead of increasing the 6k8 you could alternatively just add a second resistor in parallel with the 9k1.
The transistors should be easily replaceable with generic parts, the only issue being a different package outline.
If the TO66 packaged Q801 is mounted without an insulating kit (meaning the case which is the collector is in direct contact with the heatsink) then you can just screw a flatpak type like a TIP41 onto the heatsink, two wires to base and emitter and its good to go. The metal tab of the flatpak is the collector. If it is insulated then you would need an insulating kit (mica washer and insulating bush) for the flatpak.
Anything you want 🙂 We can swap the Zener in the simulation for a voltage source and dial in any value we wish. Instead of increasing the 6k8 you could alternatively just add a second resistor in parallel with the 9k1.
Thank you again for this. Very interesting. The simulation is producing the opposite of my calculations. If I reduce R3 to 6K2 ohms, I get closer to 15 volts (14.97V). Using 6k8 ohms, I get 15.56V and 7k5 ohms will give me 16.25V.
The current resistor of 6760 ohms will regulate to 15.5V which is probably quite acceptable.
The current resistor of 6760 ohms will regulate to 15.5V which is probably quite acceptable.
Try this one as it might be easier to quickly see what is happening. Its the same simulation but set up to show DC voltages. The raw input is now fixed.
The small window that will appear when you run it shows all the DC conditions at each node. Just close that window. If you click any point on the diagram having run it you will attach a voltage marker at that point. If you change a part or value just rerun the sim to recalculate the new values.
Reducing R3 means the output voltage has to be a fraction lower to reach that all important point to turn on Q2. Alternatively raising R4 will have the same effect.
Also remember that when talking about minor changes in voltage you will find real world component tolerances also play a part. In particular the exact Zener voltage in use and the base/emitter volt drop of the transistors.
The small window that will appear when you run it shows all the DC conditions at each node. Just close that window. If you click any point on the diagram having run it you will attach a voltage marker at that point. If you change a part or value just rerun the sim to recalculate the new values.
Reducing R3 means the output voltage has to be a fraction lower to reach that all important point to turn on Q2. Alternatively raising R4 will have the same effect.
Also remember that when talking about minor changes in voltage you will find real world component tolerances also play a part. In particular the exact Zener voltage in use and the base/emitter volt drop of the transistors.
Thank you Mooly for this. This is a very helpful tool indeed.
I am about to wrap up this phase of the project. I now have regulated voltage!
I found a small transistor for Q2 (BC549B) which is under spec but given the current and voltage the simulator gives me, I think it is fine.
With a zener of 7.9 V, I get a regulated voltage of 14.6 V, with another zener I had of 8.3 V, I get a regulated voltage of 15.4V. I think I will leave it like this unless my mentor tells me it needs to be closer to 15V!
I am about to wrap up this phase of the project. I now have regulated voltage!
I found a small transistor for Q2 (BC549B) which is under spec but given the current and voltage the simulator gives me, I think it is fine.
With a zener of 7.9 V, I get a regulated voltage of 14.6 V, with another zener I had of 8.3 V, I get a regulated voltage of 15.4V. I think I will leave it like this unless my mentor tells me it needs to be closer to 15V!
The BC549 should be fine in practice. The absolute value of the regulated supply is probably not that significant in itself although like you, I would go with the 15.4v 'version' as that probably comes closer to the original design. That has to be the simple answer 🙂 and the one that probably covers everything but for the sake of completeness......
....the only area I can think of where it could in theory make a difference (and you would have to study the circuit and all the rails to see) would be on the tuner stages and the alignment of them. It all depends how sensitive the tuner alignment is to different rail voltage and even whether that particular rail voltage is even a factor or not. I'm assuming here that the tuner alignment is done in situ with the unit functional as it comes off the production line and that the tuner stages are not preassembled and already aligned modules... if that makes sense.
Whatever you do though, do not touch anything in the tuner and IF stages, don't even physically move any parts in that area as even that can alter the alignment due to mutual coupling between parts being altered.
Bottom line is I'm sure its all fine 🙂
....the only area I can think of where it could in theory make a difference (and you would have to study the circuit and all the rails to see) would be on the tuner stages and the alignment of them. It all depends how sensitive the tuner alignment is to different rail voltage and even whether that particular rail voltage is even a factor or not. I'm assuming here that the tuner alignment is done in situ with the unit functional as it comes off the production line and that the tuner stages are not preassembled and already aligned modules... if that makes sense.
Whatever you do though, do not touch anything in the tuner and IF stages, don't even physically move any parts in that area as even that can alter the alignment due to mutual coupling between parts being altered.
Bottom line is I'm sure its all fine 🙂
I'm concluding this discussion by saying that I believe this thread is worth a gold mine for any one starting in electronic repair. Throughout my experiences and my mistakes, I learned:
Most importantly, I learned that there are great people out there ready to give their time and to patiently share their knowledge. These people don't get much in return other than eternal gratitude. @Mooly, you are exceptional. Thank you!
- Soldering/de-soldering
- How to avoid melting wire insulation
- Fault isolation
- Transistor replacements and important figures
- Calculating voltage and current
- The role of a regulator
- Basic understanding of circuits (almost, I still not get everything 😉)
- How to use LTSpice to simulate circuits
Most importantly, I learned that there are great people out there ready to give their time and to patiently share their knowledge. These people don't get much in return other than eternal gratitude. @Mooly, you are exceptional. Thank you!