W
w0lfd0g
Guest
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What kind of voltages do you see at T1 and T5 with the ECC32? I'd imagine the current would need to be nudged down to lower the plate voltage.
You could punch a larger hole in the front of the chassis to install an octal socket, that would be the bigger challenge.
You could punch a larger hole in the front of the chassis to install an octal socket, that would be the bigger challenge.
To make it fairly simple, we will start by assuming that:
1. The high voltage supply available is 200V.
2. We will keep the 22.1K plate load resistors.
3. We will keep the 1.5V diodes for biasing.
200V and 22.1K allows us to draw the load line on the curves. It hits (200V, 0mA) and (0V, 9mA) by Ohm's law with the minimum current across the 22.1K giving us B+ on the plate, and maximum possible current drawn through the 22.1K resistor from 200V to be 9mA.
Where this load line intersects 1.5V of bias is where the tube will operate in the Crack. I have marked both the load line and the approximate location of this point on the attached image. As you can see, the tube will draw ~4mA and the plate voltage will be up around 120V.
For the Crack, I'd like to see 75V on the plate, and by using the same strategy, I would estimate that a 50K plate load would likely hit that mark.
As far as performance goes, I don't believe that the extra gain that this tube offers will be helpful to the circuit, and going to the 50K plate load will increase that gain a bit.
1. The high voltage supply available is 200V.
2. We will keep the 22.1K plate load resistors.
3. We will keep the 1.5V diodes for biasing.
200V and 22.1K allows us to draw the load line on the curves. It hits (200V, 0mA) and (0V, 9mA) by Ohm's law with the minimum current across the 22.1K giving us B+ on the plate, and maximum possible current drawn through the 22.1K resistor from 200V to be 9mA.
Where this load line intersects 1.5V of bias is where the tube will operate in the Crack. I have marked both the load line and the approximate location of this point on the attached image. As you can see, the tube will draw ~4mA and the plate voltage will be up around 120V.
For the Crack, I'd like to see 75V on the plate, and by using the same strategy, I would estimate that a 50K plate load would likely hit that mark.
As far as performance goes, I don't believe that the extra gain that this tube offers will be helpful to the circuit, and going to the 50K plate load will increase that gain a bit.
Attachments
Grainger49
New member
This is a common misconception. Regardless of the value of the pot or attenuator if your source puts out 1V, 1V appears across a 20k pot, a 50k pot and a 100k pot. Turning any of these to half volume will give you the same volume out. But as the resistance lowers the source has to provide more current into the lower resistance.
But the key here is to put a series resistor between the input and the "top" of the pot. The top is where the red and white wires from the RCA jacks to the pot. It is easy to add the resistors here. Try a 100k Ohm in series. If this doesn't give enough attenuation try a 200k Ohm in each channel.
But the key here is to put a series resistor between the input and the "top" of the pot. The top is where the red and white wires from the RCA jacks to the pot. It is easy to add the resistors here. Try a 100k Ohm in series. If this doesn't give enough attenuation try a 200k Ohm in each channel.
Grainger49
New member
Ok, Whoooo, boy!
Yes the voltage doesn't change as the resistance across it changes. But the current is higher through lower resistances. I.E. less resistance allows higher flow (current).
Good Old Ohm's law: V=IR or I=V/R or R=I/V Algebra applies here. (Finally found a use for it!)
A number of Bottleheads have removed the pot altogether. You put a 100k Ohm resistor across the RCA input jacks, from the center to ground. That provides a proper load for your source.
The input wires can be brought forward, keeping the original path, and landed on a terminal strip. Alternately make the wires 2 inches longer and bring them to the tube pins. You will need to tack down the wire to keep it away from the rest of the circuit. I use a hot glue gun. you could just as well use RTV/Silicone Seal.
The white wire (left) goes to the 12AU7 pin 2 and the red wire goes to the 12AU7 pin 7.
Gain is defined as 10Log10 Vin/Vout in dB. That is logarithmic. You probably don't want to mess with it.
See Ohm's Law for resistance, current and voltage relationships. But these don't linearly relate to output volume. Crack amplifies a constant amount. What you do is feed it different voltages and the gain (amplification factor) stays the same.
The 75V on the plate is where the tube is linear. Linearity says that the bass, midrange and treble gets amplified equally. (that should answer two questions) I would keep the biasing as designed to start with then post later if you want to change the operating point. However, I would trust Paul Birkeland's design. He is drawing a load line on top of those curves and I'm sure he selected a good point.
Yes the voltage doesn't change as the resistance across it changes. But the current is higher through lower resistances. I.E. less resistance allows higher flow (current).
Good Old Ohm's law: V=IR or I=V/R or R=I/V Algebra applies here. (Finally found a use for it!)
A number of Bottleheads have removed the pot altogether. You put a 100k Ohm resistor across the RCA input jacks, from the center to ground. That provides a proper load for your source.
The input wires can be brought forward, keeping the original path, and landed on a terminal strip. Alternately make the wires 2 inches longer and bring them to the tube pins. You will need to tack down the wire to keep it away from the rest of the circuit. I use a hot glue gun. you could just as well use RTV/Silicone Seal.
The white wire (left) goes to the 12AU7 pin 2 and the red wire goes to the 12AU7 pin 7.
Gain is defined as 10Log10 Vin/Vout in dB. That is logarithmic. You probably don't want to mess with it.
See Ohm's Law for resistance, current and voltage relationships. But these don't linearly relate to output volume. Crack amplifies a constant amount. What you do is feed it different voltages and the gain (amplification factor) stays the same.
The 75V on the plate is where the tube is linear. Linearity says that the bass, midrange and treble gets amplified equally. (that should answer two questions) I would keep the biasing as designed to start with then post later if you want to change the operating point. However, I would trust Paul Birkeland's design. He is drawing a load line on top of those curves and I'm sure he selected a good point.
Grainger49
New member
Nathan,
The cathode heater voltage being higher doesn't enter into the design unless you change the cathode voltage. (reading through your first post)
Paul's post looks to me to be for the ECC32. He suggests keeping the HLMP-6000 diode with the ECC32. Reading through his post, I think he is suggesting a 50k Ohm plate load resistor.
The tube's output will be wiggling around the dot on the green line. (Paul, tell me if I got this wrong)
The voltage on the plate of the driver is also the DC bias to the output tube. This is why Paul is wanting 75V there.
The cathode heater voltage being higher doesn't enter into the design unless you change the cathode voltage. (reading through your first post)
Paul's post looks to me to be for the ECC32. He suggests keeping the HLMP-6000 diode with the ECC32. Reading through his post, I think he is suggesting a 50k Ohm plate load resistor.
The tube's output will be wiggling around the dot on the green line. (Paul, tell me if I got this wrong)
The voltage on the plate of the driver is also the DC bias to the output tube. This is why Paul is wanting 75V there.
Grainger49
New member
You have just exhausted my tube design knowledge. I will start reading Jones' Valve Amplifiers. I bought the book earlier this year and I need more reading in it.
Sorry
Sorry
Man, you have a whole thread worth of questions here, I'll try to do my best! The Submissive idea is perfect, your implementation is exactly why I wanted to put the submissive out in the first place. Just be sure to put a resistor from pin to shell on each RCA jack on the Crack. Any value between 100K and 500K should be OK.w0lfd0g said:
Gain can be defined in terms of voltage, current, power, etc. In our discussions on the forum, it is generally assumed that we are only talking about voltage gain. Grainger stumbled on to why, as when you feed 1V into a 100K pot and 1V into a 50K pot, there is a difference of power in those situations, and that difference is not useful to us. So, we define gain as 20log(Vin/Vout).
Impedance is defined as either a resistance, or as something that behaves like a resistance. For example, if we have a rectifier that drops 0.7V at 1A of current, then we can say that the rectifier has an impedance of 0.7 Ohms. (Even though our voltmeter wouldn't measure a resistance across it)
If you have 1V coming into the pot, and 1V coming out of the pot at full level, a 100K resistor between the pot wiper and the tube will not decrease the level. It will, however, have a negative impact on the high frequency response of the amplifier.w0lfd0g said:
Resistance divides voltage, change in voltage goes into the 20log(Vin/Vout) equation, then you get dB. Current isn't generally considered important in this particular discussion.w0lfd0g said:
It allows about half of the available B+ to appear from plate to cathode on the output tube. It also helps define the maximum amount of heat that can be generated by whatever is under the cathode.w0lfd0g said:
There are some other options, other LED's and other resistors. I would try to keep the bias voltage above 1V if possible.w0lfd0g said:
Yes, this is very important, but you need to pick the operating point *and* draw the load line.w0lfd0g said:
It actually looks very linear with a 50K load.w0lfd0g said:
If you reduce the bias voltage to 0.5V, you'll limit how much voltage you can feed into the amp.w0lfd0g said:
Yes, a lot of diodes have a voltage/current curve that you have to be careful of. You may find a diode with a 0.5V forward drop, but only at 1A. If you go down to a few mA, that voltage may wander all over the place. (Some parts may also be a tad noisy)w0lfd0g said:
I think you'll get lower distortion and better overall performance with the 50K plate load resistors.w0lfd0g said:
Grainger49
New member
Nope, I was dead serious. I have read a couple of books on tubes that didn't help. The Jones book is supposed to be much better.
I'm an EE but didn't study tubes in school.
I'm an EE but didn't study tubes in school.
Yeah, I think you'll be happy with that operating point.
Generally, a good rule of thumb is to keep a plate load around 3x Rp to keep distortion reasonable, and 50K should get you there.
Generally, a good rule of thumb is to keep a plate load around 3x Rp to keep distortion reasonable, and 50K should get you there.
JamieMcC
New member
Just a quick question about the Submissive implementation would it not be possible if building a new amp as Nathan is doing to make the Submissive components an integral part of the Crack build from the outset so as a one box solution giving the Crack the extra inputs and stepped control in place of fitting the stock pot and associated mods of resistors if adding the submissive as a passive second box after? I can see a larger enclosure would also be required.
Yes, Rp is plate resistance.
You could indeed built the Submissive into a larger chassis and also build the Crack into that.
As already pointed out, having the Submissive separated allows you to use it with all sorts of different components. (Crack, Stereomour, SEX, Smack, etc.)
You could indeed built the Submissive into a larger chassis and also build the Crack into that.
As already pointed out, having the Submissive separated allows you to use it with all sorts of different components. (Crack, Stereomour, SEX, Smack, etc.)
JamieMcC
New member
Kris said:
Kris, I can say its almost a 99.9% certainty I will build the mainline later on when both funds and spare time allow. I was more just curious that I wasn't missing some obvious reason why it could not be implemented like that.
I also have had a lot of fun building and modding the Crack which was my introduction into diy electronics, having never soldered or used a multi-meter before. It would be neat to start a fresh and build another one now my skill level has improved and I could implement some of the new ideas, skills and knowledge I have learned along the way. There would also be no rush to complete it so could take my time and hopefully do a pucker job on the inside as well as the outside.
The 0.95A will not bother the power transformer of the Crack.
The switch for plate loads is possible, but not super advisable, as high voltage DC is pretty hard on switch contacts. If you can be 100% sure that you'll never switch it while the amp is powered on, then it will be OK.
-PB
The switch for plate loads is possible, but not super advisable, as high voltage DC is pretty hard on switch contacts. If you can be 100% sure that you'll never switch it while the amp is powered on, then it will be OK.
-PB
Paul Joppa
Moderator
If you are switching DC, you should in principle check the DC voltage rating, not the AC rating. You won't like what you see, most switches are rated for a maximum around 30 volts DC.
Many people pretend they didn't see those limitations, and get away with it for quite a while. What you can get away with turns out to be very complicated - but the bottom line is that if you exceed ratings, your heirs will not be able to win any wrongful-death lawsuit. :^)
Many people pretend they didn't see those limitations, and get away with it for quite a while. What you can get away with turns out to be very complicated - but the bottom line is that if you exceed ratings, your heirs will not be able to win any wrongful-death lawsuit. :^)
2wo
Active member
Placing the switch underneath is the way go, better insures that no one touches it while the amp is on.
You are looking to switch between 50k and 21k. Get two resistors of 47k, or if you want to get fancy 50k and 39k. Wire the 50k in as the plate resistor and parallel it with the 39k and switch...John
You are looking to switch between 50k and 21k. Get two resistors of 47k, or if you want to get fancy 50k and 39k. Wire the 50k in as the plate resistor and parallel it with the 39k and switch...John
While you *could* do this, you'll have to switch the 120V AC power through this switch, and signal carrying connections.
We get away with this in the Quickie because the battery power isn't a source of noise.
We get away with this in the Quickie because the battery power isn't a source of noise.
The R1 resistors on the two small PC boards should be changed to ~450 Ohm resistors. (470 Ohm should be close enough)
Paul Joppa
Moderator
This is obsolete now but I'm going to give a technical answer anyway.w0lfd0g said:
1. is a fine idea.
2. reed relays have a gas inside; the way to do this is with a vacuum relay. They are hard to find and expensive.
3. It's the distance of the contacts that must meet and part to make the connection. Any switch that works will have infinitely close distance at some point in time.
4. The usual solution is to be aware of the arcing (and ionization - there are two mechanisms to worry about) and choose a switch whose contacts are so large that their erosion takes many many cycles. Anyone who remembers changing the points on a distributor is familiar with this!
Side story - way back when we were dating, my wife had a Dodge with a high-compression V-8 and a 6-volt battery, with dual paralleled points. It needed new points about every 750 miles (and a quart of oil every 1500). She used to pull the distributor and do the operation on the kitchen table, replacing the distributor the next morning to go to work.
