Search the Forum

(Advanced Search)


Bleeder Resistor question

Posts: 49
Threads: 5
Joined: Aug 2020
Location: Canada
Bio: I am really a novice builder relative to most out there and more as a hobby but I guess you can say a serious hobby. I have built the Ultimate amp starting from a JCM800 with power scaling, extra gain stage, aux power supply, and modified to take any octal power tube based on Kevin's works. This amp is really the latest and greatest for me and galactic grounding from TUT book and am extremely happy with the build. Ultra quiet and sounds awesome the way I like it.
Country:
Hi all.
I've always put in a bleeder resistor across the main power filter cap but not knowing how hot it gets. In a jcm800 style build I had a 27k 5W resistor across the main cap. Then I tested it with a 110K 7W resistor. It gets really hot to the touch. Not glowing hot but hot enough you wouldn't want your finger on it too long. Any good way to discharge caps? It's for safety but I'm worried the resistor might overheat and cause more danger.
I got these new higher resistor values based on the formula to calculate the resistance value as a function of the time and voltage.
I've heard of an active bleeder circuit which shuts off through a relay when the amp is on but connects back when you turn the amp off. Basically senses the mains voltage when on and disconnects from the cap. But this seems like too much of a project with the limited space I have in this amp build.
Any ideas?
Posts: 401
Threads: 40
Joined: Aug 2018
Location: CANADA
Country:
Hi Guys
Champ81, do you know Ohm's Law? That is what guides you here and there is no need for any active circuitry when it comes to discharging capacitors, or in assuring that they share voltage when wired in series.
Ohm's Law tells us the relationship between voltage (V in volts), current (I) in amperes, and resistance ® in ohms. When a voltage is impressed across a resistance, a current will flow through the resistance. The magnitude of current is inversely proportional to the resistance This gives us an equation:
I = V / R
You can think of V as pressure, I as flow and R as what limits the flow. What fllows here are electrons.
You can look at it another way, and think that a current through a resistance generates a voltage across the resistance, as:
V = I x R
If you know the voltage across the resistance and also know the current, then you can calculate the resistance:
R = V / I
So, if you have 400V across 100k, the current is:
I = V / R = 400V / 100,0000R = 0.004A
If we reduce R to 10k, there is a lot more current:
I = V / R + 400V / 10,000R = 0.04A
The Power equations tell us how much heat (W) in watts will be generated by the voltage (V) in volts, and current (I) in amperes, as:
P = V x I
which can be rearranged as :
P = I*2 x R
and
P = V*2 / R
The *2 means "squared".
So, with the 400V across 100k, the power is:
P = V x I = 400V x 0.004A = 1.6W
and with 10k:
P = V x I = 400V x 0.04A = 16W.
In each case, you need a resistor with a power rating in excess of the heat that is being dissipated, and preferably it is a flameproof type, either wirewound or metaloxide  NEVER carbon.
If you look at schematics, you see highvalue bleeders across filter caps, typically 220k or higher. If we use the equations above, a 220k resistor will dissipate 0.72W with 400V across it. We could use a 1W but it will get hot, so maybe use a 2W so it will be a bit physically larger and run cooler. There is no need to use a lowvalue resistor in this highvoltage position as it simply wastes power and heats up the inside of the chassis for no reason.
If the supply voltage were only 40V, a 10k resistor would dissipate 0.16W and you could use a halfwatt resistor without worry.
Our book Ready, Set, Go! (RSG) explains all the highschool science that is the foundation of understanding electronic quantities and basic circuits.
Posts: 49
Threads: 5
Joined: Aug 2020
Location: Canada
Bio: I am really a novice builder relative to most out there and more as a hobby but I guess you can say a serious hobby. I have built the Ultimate amp starting from a JCM800 with power scaling, extra gain stage, aux power supply, and modified to take any octal power tube based on Kevin's works. This amp is really the latest and greatest for me and galactic grounding from TUT book and am extremely happy with the build. Ultra quiet and sounds awesome the way I like it.
Country:
05032023, 12:46 AM
(This post was last modified: 05032023, 12:53 AM by Champ81.)
[quote="K O'Connor" pid="1106" dateline="1683084464"]
Hi Guys
Champ81, do you know Ohm's Law? That is what guides you here and there is no need for any active circuitry when it comes to discharging capacitors, or in assuring that they share voltage when wired in series.
Ohm's Law tells us the relationship between voltage (V in volts), current (I) in amperes, and resistance ® in ohms. When a voltage is impressed across a resistance, a current will flow through the resistance. The magnitude of current is inversely proportional to the resistance This gives us an equation:
I = V / R
You can think of V as pressure, I as flow and R as what limits the flow. What fllows here are electrons.
You can look at it another way, and think that a current through a resistance generates a voltage across the resistance, as:
V = I x R
If you know the voltage across the resistance and also know the current, then you can calculate the resistance:
R = V / I
So, if you have 400V across 100k, the current is:
I = V / R = 400V / 100,0000R = 0.004A
If we reduce R to 10k, there is a lot more current:
I = V / R + 400V / 10,000R = 0.04A
The Power equations tell us how much heat (W) in watts will be generated by the voltage (V) in volts, and current (I) in amperes, as:
P = V x I
which can be rearranged as :
P = I*2 x R
and
P = V*2 / R
The *2 means "squared".
So, with the 400V across 100k, the power is:
P = V x I = 400V x 0.004A = 1.6W
and with 10k:
P = V x I = 400V x 0.04A = 16W.
In each case, you need a resistor with a power rating in excess of the heat that is being dissipated, and preferably it is a flameproof type, either wirewound or metaloxide  NEVER carbon.
If you look at schematics, you see highvalue bleeders across filter caps, typically 220k or higher. If we use the equations above, a 220k resistor will dissipate 0.72W with 400V across it. We could use a 1W but it will get hot, so maybe use a 2W so it will be a bit physically larger and run cooler. There is no need to use a lowvalue resistor in this highvoltage position as it simply wastes power and heats up the inside of the chassis for no reason.
If the supply voltage were only 40V, a 10k resistor would dissipate 0.16W and you could use a halfwatt resistor without worry.
Our book Ready, Set, Go! (RSG) explains all the highschool science that is the foundation of understanding electronic quantities and basic circuits.
Thank you for this Kevin. I have used the ohms law to calculate a new resistor value that I in fact need around a 220k or higher resistor across the main cap. This was the ohms law reiterated as a function of time and voltage.
I used a value that is too low. A tried a 110k resistor with a rating of 7w which got too hot. A 200k resistor was warm. From digikey I was able to find all flame proof resistors. I did try to measure the cap discharge without a resistor it discharged at about the same rate. I figure if I went higher it wouldn't make a difference as at that point the discharge rate would be about the same at 200kohm. I chose to just disconnect the resistor and have no bleeder. Would you say there are other reasons to have one connected?
Posts: 49
Threads: 5
Joined: Aug 2020
Location: Canada
Bio: I am really a novice builder relative to most out there and more as a hobby but I guess you can say a serious hobby. I have built the Ultimate amp starting from a JCM800 with power scaling, extra gain stage, aux power supply, and modified to take any octal power tube based on Kevin's works. This amp is really the latest and greatest for me and galactic grounding from TUT book and am extremely happy with the build. Ultra quiet and sounds awesome the way I like it.
Country:
(05032023, 12:27 AM)K O'Connor Wrote: Hi Guys
Champ81, do you know Ohm's Law? That is what guides you here and there is no need for any active circuitry when it comes to discharging capacitors, or in assuring that they share voltage when wired in series.
Ohm's Law tells us the relationship between voltage (V in volts), current (I) in amperes, and resistance ® in ohms. When a voltage is impressed across a resistance, a current will flow through the resistance. The magnitude of current is inversely proportional to the resistance This gives us an equation:
I = V / R
You can think of V as pressure, I as flow and R as what limits the flow. What fllows here are electrons.
You can look at it another way, and think that a current through a resistance generates a voltage across the resistance, as:
V = I x R
If you know the voltage across the resistance and also know the current, then you can calculate the resistance:
R = V / I
So, if you have 400V across 100k, the current is:
I = V / R = 400V / 100,0000R = 0.004A
If we reduce R to 10k, there is a lot more current:
I = V / R + 400V / 10,000R = 0.04A
The Power equations tell us how much heat (W) in watts will be generated by the voltage (V) in volts, and current (I) in amperes, as:
P = V x I
which can be rearranged as :
P = I*2 x R
and
P = V*2 / R
The *2 means "squared".
So, with the 400V across 100k, the power is:
P = V x I = 400V x 0.004A = 1.6W
and with 10k:
P = V x I = 400V x 0.04A = 16W.
In each case, you need a resistor with a power rating in excess of the heat that is being dissipated, and preferably it is a flameproof type, either wirewound or metaloxide  NEVER carbon.
If you look at schematics, you see highvalue bleeders across filter caps, typically 220k or higher. If we use the equations above, a 220k resistor will dissipate 0.72W with 400V across it. We could use a 1W but it will get hot, so maybe use a 2W so it will be a bit physically larger and run cooler. There is no need to use a lowvalue resistor in this highvoltage position as it simply wastes power and heats up the inside of the chassis for no reason.
If the supply voltage were only 40V, a 10k resistor would dissipate 0.16W and you could use a halfwatt resistor without worry.
Our book Ready, Set, Go! (RSG) explains all the highschool science that is the foundation of understanding electronic quantities and basic circuits.
I currently have an amp at 489V. Rounding up to 500V I used a resistor at 7w 110k. According to the equation it should not be heating up so much. Meaning the power rating is sufficient at that resistance.
W=V^2/R
= 500V^2/110,000ohm
= 2.273W
The resistor was rated a 7W which is much more than sufficient. At this number a 3W should be plenty.
What I mean is it wasn't glowing hot but touching the middle I had to take my finger off it.
Posts: 401
Threads: 40
Joined: Aug 2018
Location: CANADA
Country:
Hi Champ81
Absolutely leave the bleeder resistor in place.
Capacitors can retain charge for an extremely long time and with the voltages inside a tube amp you can have a nasty surprise. For example, a customer brought his Marshall to me for a repair. When I took the chassis out of the amp sleeve and was setting it up on my bench I got a good zap from the main filter which still had a voltage of over 400V on it after at least an hour of disuse.
It is a good practise to install bleeder resistors in every kind of amp and PSU. You only "see their absence" in cheaply designed products.
Your skin temperature is 24C and you can usually touch things that approach 40C  beyond that you definitely cannot hold your finger in contact with the hot device. You can gauge temperature by "proximity" without touching the device, which we naturally do anyway.
As suggested above, using a component with a larger body will allow the part to operate at a lower temperature for a given amount of heat. The large part has more surface area and its thermal resistance should be lower than for a smaller part. It is best to keep the surface temp of components to less than 50C and otherwise as low as you can within reason.
"Laser" thermometers are quite inexpensive now and are very handy when you want to probe a circuit and measure component body temperatures. Every device has a thermal resistance rating, or its data sheet shows a graph of temp / dissipation. The meter will read the surface temperature and you can use the thermal resistance to calculate the internal temperature, which will have a specified maximum.
Any resistor or component that dissipates heat  which is all of them except capacitors  must be mounted in a manner that allows that heat to escape and so its heating won't damage or effect adjacent components. This is why all of our kit notes specify to elevate power resistors off the PCB. In the case of your hot resistor, you do not want PVCinsulated wires in contact with it as the insulation will melt.
Just use the 'quick reply' instead of quoting the entire reply each time.
Posts: 49
Threads: 5
Joined: Aug 2020
Location: Canada
Bio: I am really a novice builder relative to most out there and more as a hobby but I guess you can say a serious hobby. I have built the Ultimate amp starting from a JCM800 with power scaling, extra gain stage, aux power supply, and modified to take any octal power tube based on Kevin's works. This amp is really the latest and greatest for me and galactic grounding from TUT book and am extremely happy with the build. Ultra quiet and sounds awesome the way I like it.
Country:
Thanks Kevin.
The lazer thermometer sounds like what I need.
I have ordered a few 270k resistor rated at 3W for three amps based on my calculations of ohms law. I took out the 27k resistors I had in there rated at 5W (white cement wirewound) started showing some discoloration (brown marks) which is not good. Based in P=V^2/R that resistor should at least have a rating of 10W. Some add a safety factor of 2. So with that 20W. Haha.
I was only perplexed because a resistor rated at 7W should have been sufficient at 110k and shouldn have gotten that hot. But.. temperature could be subjective in it was my own perception of what was hot. I'm going to look up some thermometers in the meantime.
Posts: 401
Threads: 40
Joined: Aug 2018
Location: CANADA
Country:
Hi Guys
When selecting power resistors, look at the dimensions for the prospective devices, assuming you are doing a search at Mouser or Digikey.
Note that both suppliers offer free shipping for orders over C$100.
When using their search engines, first select the type of part you are looking for: 'passive', then select 'resistors' or 'throughhole resistors', then select 'in stock'. Sometimes the system will not highlight the 'apply' button unless two parameters are chosen, but it is best to select only one parameter at a time for filtering. Trying to narrow the search too quickly often leads to 'no parts found'
Another detail regarding resistors is that they have voltage ratings.
Most 1W metaloxide resistors are rated for 350V yet they generally work without issue in typical tube amps with 500V supplies. It is better to select the larger body (approx. 15mm x 4mm) instead of the smaller, newer ones (7mm x 3mm) as the larger has a higher voltage rating. In positions where the voltage across the resistor will be low even if both ends are at highvoltage, you can get away with the smaller body provided the device is elevated.
In other places, I have written about thermal management for diodes, zeners and transistors inasmuch as all devices rely on their leads for their main heat transfer path. Therefore, it is advisable to retain as much lead length as is reasonable for the application.
Posts: 49
Threads: 5
Joined: Aug 2020
Location: Canada
Bio: I am really a novice builder relative to most out there and more as a hobby but I guess you can say a serious hobby. I have built the Ultimate amp starting from a JCM800 with power scaling, extra gain stage, aux power supply, and modified to take any octal power tube based on Kevin's works. This amp is really the latest and greatest for me and galactic grounding from TUT book and am extremely happy with the build. Ultra quiet and sounds awesome the way I like it.
Country:
I purchased these resistors from digikey.
https://www.digikey.ca/en/products/detai...0K/2390406
The length is 16mm with a 5.5mm diameter. Rated at 3W.
The last one I purchased were these 110k
https://www.digikey.ca/en/products/detai...0K/9926365
Rated at 7W 32mm length and 8.5mm diameter. These gor hot. The max working voltage according to the data sheet is 750V.
The one that was "warm" were these 200k
https://www.digikey.ca/en/products/detai...0K/9926548
54mm length and 8.5mm diameter.
Posts: 49
Threads: 5
Joined: Aug 2020
Location: Canada
Bio: I am really a novice builder relative to most out there and more as a hobby but I guess you can say a serious hobby. I have built the Ultimate amp starting from a JCM800 with power scaling, extra gain stage, aux power supply, and modified to take any octal power tube based on Kevin's works. This amp is really the latest and greatest for me and galactic grounding from TUT book and am extremely happy with the build. Ultra quiet and sounds awesome the way I like it.
Country:
Noticed something I never really payed much attention to. I was seeing the voltage rate drop of a 160k resistor with multimeter clipped to the main cap. It steadily decreased all the way down to mV range (safe to work on).
I then wanted to see the voltage drops with no bleeding resistor in place. As soon as I turned the amp off it dropped fast to about 4V I think. Then it steadily started to rise. Up to about 7.25V and just stayed there. It was interesting to see. I was wondering why it does this. I would think it would not rise back up. This confirmed that yes I do need a bleed resistor in place. I never noticed that it actually can go back up like that to a steady 7V.
Posts: 401
Threads: 40
Joined: Aug 2018
Location: CANADA
Country:
Hi Guys
What you see there is one of the effects of the imperfections of capacitors, especially of electrolytics which are pretty far from ideal as you get while still having a "decent" capacitor.
People think that a capacitor will hold a charge forever once the power is removed and there is no load. This is not the case. Maybe it should be, but real world assemblies are imperfect. Charge gets "trapped" in electrolytic dielectrics and redistributes itself after power is removed exhibiting the roller coaster voltages you saw.
"Super caps" with around a Farad of capacitance have optimised imperfection allowing them to be used as an alternative to a backup battery for digital circuit. Typically the drain is at the microamp level when in standby, so the device, say a VCR, will "remember" the setting used last time.


Come in where it's warm!

A warm welcome to tube amp modding fans and those interested in hifi audio! Readers of Kevin O'Connor's The Ultimate Tone (TUT) book series form a part of our population. Kevin O'Connor is the creator of the popular Power Scaling methodology for amplifiers. 
Please remember these three principles: respect, sharing, community. 
Not familiar with The Ultimate Tone book series? See discussion topics, or click here to visit London Power/Power Press Publishing. 
