[klnglim]

Thanks guys! More of this in the morning.

Hey kinglim, that lighting idea is good, I was wondering if it might help all of us if you could take some good closeups of the board top and bottom to reveal the traces. This was a problem are for us at the start of this thread and with all the components in place it isn't easy to see them all.

Maybe Maxx would let you put them on the first page.

That's the way how I search and fix those damage parts of the board. What you need is LED flash light with flat surface like glass

 

[klnglim]

Let's say the signal swing at the junction of RL and R16 is 50VPP. This would mean the differential (meaning, actual voltage seen by the load) voltage across the load would be an astounding 100VPP, completely crushing your ears and burning to death your headphone speaker coils. So in other words, enough voltage.

This voltage is between the mentioned junction and circuit ground. The path to ground from that junction that goes thru the feedback resistor R16 is either 390k + 10k or 80k + 10k depending on the setting. So 400k or 90k.

Let's use Ohm's Law: 50V divided by 400k means current of 0.125 mA. Combined dissipation thru that path is then 0.125 mA times 50V which means 0.00625 W. Divide that by 400k and multiply by 390k and you get the dissipation for R16 at 390k as 0.006 Watts.

For the 90k scenario: 50 divided by 90k means current of 0.55 mA. Combined dissipation thru that path is then 0.55 mA times 50V which means 0.0275 W. Divide that by 90k and multiply by 80k and you get the dissipation for R16 at 80k as 0.0244 Watts.

In both cases you'll be more than fine with 0.25W resistors. However I recommend simply using 1W no-name carbon film resistors for all spots that dissipate under 0.75 W. That covers usually 95% of resistors inside any circuit. Why mess with different types.

As I am lack of electronic experience especially calculation,further more there are too many choices of resistor to replace and confuse about which is the best. Anyway, thanks for the answer.

 

[MrCurwen]

As I am lack of electronic experience especially calculation,further more there are too many choices of resistor to replace and confuse about which is the best. Anyway, thanks for the answer.

Indeed.

Equations used in my post were Ohm's law, dissipation, and simple proportional calculation.

Ohm's law in conjunction with Kirchof's law are the most important to understand, they will come up everywhere all the time. Luckily they are very simple and easy to understand.

Ohm's law is simply 1A of current flowing thru a 1 ohm resistor will cause 1 V of voltage to appear across the resistor.

Or, if you apply a 1 V voltage across a 1 ohm resistor, you will cause 1 A of current to flow thru the resistor.

Or, if you have 1 A flowing thru a resistor and you have 1 V of voltage across the resistor, the resistor must be 1 ohms.

These three variations will help you determine a great number of things in electronics. Memorize them.


Dissipation is simply voltage (in volts) times current (in amperes, so 1mA is 0.001 A), and the result is watts. If you have AC, you must first calculate RMS values. DC is most useful though, you rarely need to calculate AC dissipations.


Kirchof's law simply means that the same amount of current that leaves the power supply must return to the power supply.

A common application of this is that if you have a current path within a circuit, let's say you have two resistors in series between B+ and ground, they both must have the same amount of current thru them.

No extra current cannot appear in the middle of that resistor series, and no current can disappear.

So let's say you have a B+ of 100V, and you two resistors in series between B+ and ground. Let's call them R1 and R2. R1 is 50k and R2 is 12k.

So we have a total of 62k of resistance between B+ and ground. Ohm's law says that means about 16mA of current flows thru that series. The same amount of current must flow thru both these resistors, since they are in series. That's pretty much the point of Kirchof's law.

From the point of view of the PSU (PT?) the whole of the circuit is just a bunch of stuff in series and in parallel between the PSU terminals. So equal amounts of current must flow in as flows out of the PSU. It's just a matter of parallel currents paths then.

 

[klnglim]

Indeed.

Equations used in my post were Ohm's law, dissipation, and simple proportional calculation.

Ohm's law in conjunction with Kirchof's law are the most important to understand, they will come up everywhere all the time. Luckily they are very simple and easy to understand.

Ohm's law is simply 1A of current flowing thru a 1 ohm resistor will cause 1 V of voltage to appear across the resistor.

Or, if you apply a 1 V voltage across a 1 ohm resistor, you will cause 1 A of current to flow thru the resistor.

Or, if you have 1 A flowing thru a resistor and you have 1 V of voltage across the resistor, the resistor must be 1 ohms.

These three variations will help you determine a great number of things in electronics. Memorize them.


Dissipation is simply voltage (in volts) times current (in amperes, so 1mA is 0.001 A), and the result is watts. If you have AC, you must first calculate RMS values. DC is most useful though, you rarely need to calculate AC dissipations.


Kirchof's law simply means that the same amount of current that leaves the power supply must return to the power supply.

A common application of this is that if you have a current path within a circuit, let's say you have two resistors in series between B+ and ground, they both must have the same amount of current thru them.

No extra current cannot appear in the middle of that resistor series, and no current can disappear.

So let's say you have a B+ of 100V, and you two resistors in series between B+ and ground. Let's call them R1 and R2. R1 is 50k and R2 is 12k.

So we have a total of 62k of resistance between B+ and ground. Ohm's law says that means about 16mA of current flows thru that series. The same amount of current must flow thru both these resistors, since they are in series. That's pretty much the point of Kirchof's law.

From the point of view of the PSU (PT?) the whole of the circuit is just a bunch of stuff in series and in parallel between the PSU terminals. So equal amounts of current must flow in as flows out of the PSU. It's just a matter of parallel currents paths then.

I understand that resistor in series = R1+R2, 50k+12K=62K, let say if power rating both resistors are 0.5W, so in series is 1W?

 

[MrCurwen]

I understand that resistor in series = R1+R2, 50k+12K=62K, let say if power rating both resistors are 0.5W, so in series is 1W?

Power rating is Pd + Pd like that only when you parallel two resistors of same resistance value.

Dissipation is always voltage times current.

So in my example you now know the current from using Ohm's law. Now you can again use Ohm's law to determine the voltage drop across each resistor.

There was a mistake in my current calculation, current is 1.6 mA not 16 mA. Anyway; the variation of Ohm's law we'll use next is this:

Ohm's law is simply 1A of current flowing thru a 1 ohm resistor will cause 1 V of voltage to appear across the resistor.

So we have 1.6 mA or 0.0016 A of current. Multiply this by 50 000 ohms, and we get a voltage drop of 80 volts for R1.

For R2 it's 0.0016 A times 12 000 ohms, 19.2 volts.

As we can see, the combined voltage drops equal (after taking rounding into consideration) 100 volts, so we know the calculations are correct since there is 100 volts between B+ and ground.

So now we know both voltage drops and current thru the resistors, and we can calculate dissipations.

1.6 mA times 80 volts means 0.128 W for R1.

1.6 mA times 20 volts means 0.032 W for R2.


This is how you calculate dissipations for components. Same process applies for tubes and FETs and all kinds of components that pass current.

 

[baronbeehive]

Anyone any idea where I could get an ammeter for the LD, one of the contacts is dead on one of mine, I believe a wire has come off inside it and it doesn't come apart to fix it.

 

[bloodhawk]

Anyone any idea where I could get an ammeter for the LD, one of the contacts is dead on one of mine, I believe a wire has come off inside it and it doesn't come apart to fix it.

I have 2 extra if you want. Just shoot me a shipping label and ill drop em off.

Otherwise i got them from here -

https://www.ebay.com/itm/Backlight-...e=STRK:MEBIDX:IT&_trksid=p2060353.m2749.l2649

 

[baronbeehive]

I have 2 extra if you want. Just shoot me a shipping label and ill drop em off.

Otherwise i got them from here -

https://www.ebay.com/itm/Backlight-100mA-DC-Current-Panel-Meter-Amperemeter-34mm-Header-for-Amplifier-New/132327543572?ssPageName=STRK:MEBIDX:IT&_trksid=p2060353.m2749.l2649

Thanks bloodhawk, I'll have 'em both if you don't need them. Where are you UK or US?

 

[klnglim]

Anyone any idea where I could get an ammeter for the LD, one of the contacts is dead on one of mine, I believe a wire has come off inside it and it doesn't come apart to fix it.

You can get more choices from Alibaba.com
https://www.aliexpress.com/wholesale?catId=0&initiative_id=SB_20180508030118&SearchText=+vu+meter

 

[klnglim]

Power rating is Pd + Pd like that only when you parallel two resistors of same resistance value.

Dissipation is always voltage times current.

So in my example you now know the current from using Ohm's law. Now you can again use Ohm's law to determine the voltage drop across each resistor.

There was a mistake in my current calculation, current is 1.6 mA not 16 mA. Anyway; the variation of Ohm's law we'll use next is this:

Ohm's law is simply 1A of current flowing thru a 1 ohm resistor will cause 1 V of voltage to appear across the resistor.

So we have 1.6 mA or 0.0016 A of current. Multiply this by 50 000 ohms, and we get a voltage drop of 80 volts for R1.

For R2 it's 0.0016 A times 12 000 ohms, 19.2 volts.

As we can see, the combined voltage drops equal (after taking rounding into consideration) 100 volts, so we know the calculations are correct since there is 100 volts between B+ and ground.

So now we know both voltage drops and current thru the resistors, and we can calculate dissipations.

1.6 mA times 80 volts means 0.128 W for R1.

1.6 mA times 20 volts means 0.032 W for R2.


This is how you calculate dissipations for components. Same process applies for tubes and FETs and all kinds of components that pass current.

Now that's what I have learn, Thanks a lot!

 

[baronbeehive]

My amp uses 390k and 100k, not 68k like you say? 390k is for high gain and 390 parallell with 100 = 80k is for low gain. I have not measured these resistors when soldered on the board though.

If you have a connection problem like that you'd hear it.

Yes, it must have happened after trying the gain settings which I did for the first time recently. The switch could have been stuck or not working properly because it hasn't been used for 10 years or so! After that I haven't listened anymore because of the PSU upgrades so it could have happened without my knowing,

There is definately variation with resistor values. My 390K's are actually 330K, doesn't affect the parallel calculations much, and where you have 100K I have 68K, therefore the 68K I measured for high gain must be this resistor on only. Also there is 100K in the board somehow between the 2 switch terminals. I will not put the switch back now so that means permanent high gain from now on.

My board is earlier than yours with the "quality" Dales LOL!

BTW, I've changed the 5K1 to 2K55 as was specified in the update from LD, but there is one other 5K1 resistor nearby which I have kept at 5K1 as I believe others have done, hope that is OK.

 

[bloodhawk]

Thanks bloodhawk, I'll have 'em both if you don't need them. Where are you UK or US?

Yeah you can have em both.
Im in the US.

 

[SonicTrance]

therefore the 68K I measured for high gain must be this resistor on only.

No, your 68k are in parallell with your 330k for low gain. Again, high gain = high resistance feedback R (low NFB), low gain = low resistance R (high NFB). Your 330k is for high gain (low NFB)
Where exactly are you measuring from? Did you measure like I said? From input tube grid to the output?

 

[baronbeehive]

No, your 68k are in parallell with your 330k for low gain. Again, high gain = high resistance feedback R (low NFB), low gain = low resistance R (high NFB). Your 330k is for high gain (low NFB)
Where exactly are you measuring from? Did you measure like I said? From input tube grid to the output?

I'm still struggling with schematics, I'm finding it difficult to see what is in parallel to what frankly. I don't see a 390K there, could you elaborate?

I tried measuring from the points in red.

 

[baronbeehive]

Yeah you can have em both.
Im in the US.

Right. If you can PM me with the price inc postage I will let you know my details.