Converting from Tube to SS rectification

[Ed_Chambley]

Can't you simply wire a diode from pin 4 to 8 and another from 6 to 8 to convert to solid state rectification without removing the octal socket.  Will this make a full wave?  Adding a resistor like a 5 watt/100ohms between pin 4 and the diode going to pin 8 simulate sag, or simply drop voltage?  Of course the tube would need to be removed.

[firemedic]

That is often done as a failsafe, when using the rectifier tube. I don't know how well it would work with the 5v heater wires still attached. Good question, I've wondered that myself at times.

[sluckey]

The failsafe diodes are usually placed between the HT leads and the rectifier plates, ie, in series with the tube. I think Ed is asking for something different. The heater wire is not a problem. I'd do it like this...

[Ed_Chambley]

I am asking something different.  To convert a tube rectifier to solid state without removing the socket and not buying a copper cap or similar, can't you simply wire a diode from pin 4 to 8 and another from pin 6 to 8.  I liiked inside a copper cap and it looks as if that is all they have done.  They have resistors in place as well, but the one AES sells does not.   2 diodes and a little solder is much cheaper.  The resistors in the weber copper cap are supposed to simulate tube sag and it looks as if they have simply added a resistor between pin 4 and the diode going to pin 8. The difference in this resistor I assume will simulate different amounts of sag and also reduced voltage.  I am asking this because if a crazy build I am doing.  I may need a little additional vdc at the B+ and I think I can get all I need at the rectifier, but I want to be able to tube it if I decide to convert to 6v6 inplace of El-34's.  I am tnking I should get an additional 30+ vdc.  But thanks for the failsafe diagram.  I can use that as well.

Also, I do not see any reason to remove the 5vac heater wires as these solid state plugs do not require it.  This may be the same thing as the Falisafe.  Just an Idea I had looking inside a solidstate plug in I have.

[sluckey]

To convert a tube rectifier to solid state without removing the socket and not buying a copper cap or similar, can't you simply wire a diode from pin 4 to 8 and another from pin 6 to 8.

Yes you can.

The resistors in the weber copper cap are supposed to simulate tube sag and it looks as if they have simply added a resistor between pin 4 and the diode going to pin 8.

This is the part that doesn't sound right.


The diagram I posted above is not a failsafe diagram. It's what you asked for. The sag resistor inside the copper cap does not go between pin 4 and pin 8. It should go between the junction of the diode cathodes and pin 8. There are many ways to physically do this. I showed you how I would do it.

Another way to do it would be to connect the anodes of the diodes to pins 4 and 6, standing them upright. Twist the cathodes together. Now connect one end of the sag resistor to the junction of the cathodes and connect the other end of the sag resistor to pin 8. It's the same thing electrically. And one more way to do this... Bust an old dead octal tube, clean up the mess, and mount the diodes and resistor in the tube base. IOW, roll your own copper cap. Now you can really swap between tubes and SS quickly and easily.

Finally, just to feed more confusion, here's a pic of a typical "FAILSAFE" diode circuit for your tube circuit. Notice that the diodes are in series with the tube plates...

[Ed_Chambley]

To convert a tube rectifier to solid state without removing the socket and not buying a copper cap or similar, can't you simply wire a diode from pin 4 to 8 and another from pin 6 to 8.

Yes you can.

The resistors in the weber copper cap are supposed to simulate tube sag and it looks as if they have simply added a resistor between pin 4 and the diode going to pin 8.

This is the part that doesn't sound right.


The diagram I posted above is not a failsafe diagram. It's what you asked for. The sag resistor inside the copper cap does not go between pin 4 and pin 8. It should go between the junction of the diode cathodes and pin 8. There are many ways to physically do this. I showed you how I would do it.

Another way to do it would be to connect the anodes of the diodes to pins 4 and 6, standing them upright. Twist the cathodes together. Now connect one end of the sag resistor to the junction of the cathodes and connect the other end of the sag resistor to pin 8. It's the same thing electrically. And one more way to do this... Bust an old dead octal tube, clean up the mess, and mount the diodes and resistor in the tube base. IOW, roll your own copper cap. Now you can really swap between tubes and SS quickly and easily.

Finally, just to feed more confusion, here's a pic of a typical "FAILSAFE" diode circuit for your tube circuit. Notice that the diodes are in series with the tube plates...

Not confusing at all. The first diagram I thought was an answer to firemedic.  I prefer the first diagram.  If there were no resistor at all, would you have a full wave SS with no sag?  If so varying the resistors will vary the voltage.  In doing this I should get more vdc, correct?  I will have that 1.414 you mentioned instead of 1.2 as in 5U4 or 1.3 as in GZ34.  I know this is not exact, but it is a general idea.

[sluckey]

Yes you will get more voltage if you just use the diodes. Adding a series resistor will decrease the B+ voltage. A larger resistor will decrease the B+ voltage even more. You can even increase the value of the resistor such that you will have less B+ than you would have had with a tube.

[Ed_Chambley]

how much will that 100Ω sag resistor drop the B+?  The load needed to calculate is derived form where?  The OT?  If so the primary is 8 ohms and the secondary is 4900ohms.  If this is not the load, where is the load information?

[Willabe]

The load for the B+ is the current draw from the tubes. The load for the output tubes is the 4K9 (with this OT) on the OT primary.

how much will that 100Ω sag resistor drop the B+?

The more the current draw, the more the B+ voltage will drop.

Here's a tube list that might help. It was put together by tubenit and loose change, look at page 2 for current draw of common PP pairs.


                  Brad    

[Willabe]

If so the primary is 8 ohms and the secondary is 4900ohms. 

You have the primary and secondary backwards.


                    Brad   

[Ed_Chambley]

If so the primary is 8 ohms and the secondary is 4900ohms. 

You have the primary and secondary backwards.


                    Brad   

Yep, I am kind of like that myself.

[Ed_Chambley]

The load for the B+ is the current draw from the tubes. The load for the output tubes is the 4K9 (with this OT) on the OT primary.

how much will that 100Ω sag resistor drop the B+?

The more the current draw, the more the B+ voltage will drop.

Here's a tube list that might help. It was put together by tubenit and loose change, look at page 2 for current draw of common PP pairs.


                  Brad    

OK, lets say a pair of EL34's in use and they have a current draw of 120-160ma according to the voltage of 385-475v respectively.  How will knowing this tell me what the voltage drop will be if using a 100ohm resistor?

EDIT... sluckey... untangled quotes

[Willabe]

Yep, I am kind of like that myself.

Me too Ed, me too.


                    Brad         

[Ed_Chambley]

Yep, I am kind of like that myself.

Me too Ed, me too.


                    Brad         

Brad, the deal is I am trying to understand the math involved.  I can always put a jumper in place and measure voltage  Through trial and error eventually I come up with the voltage I need.  I have read and read and read, but I can't seem to get it.   It is probably much simpler than I am making it.  This is just single resistors I am trying to understand.  Wait until there are multiple if you think I am driving you crazy now.

[Willabe]

I think I understand what your looking for. Sluckey or one of the other guys know the math, sorry to say I don't.

They'll help you out.


                    Brad     

[PRR]

> 120-160ma .... using a 100ohm resistor?

V = I*R

[Ed_Chambley]

> 120-160ma .... using a 100ohm resistor?

V = I*R

v=.12 (low side of EL-34) * 100ohms
V=12

On the high side would be v=16.  This would be the voltage drop.  So if I begin with 385 vac I should drop from 385 to 373.

Is this how it works?

[sluckey]

On the high side would be v=16.  This would be the voltage drop.  So if I begin with 385 vac I should drop from 385 to 373.

Is this how it works?

Not quite. The 385-475v is B+, not PT secondary voltage. Since the 100Ω resistor is in the B+ line, it will drop 16vdc. So, the 16 volts dropped across the 100Ω will subtract form the B+ value.

In review, you must know the total B+ current that will flow through the 100Ω sag resistor. Then it simply becomes E=IR to calculate the voltage dropped by the sag resistor. The total B+ current consists of currents drawn by the power tubes, PI, and preamp tubes. The power tubes draw the lions share of the total current so usually just ignore the PI and preamp currents. If you already know the current you'll bias your output tubes, use that figure for total B+ current. Alternately, put a 1Ω resistor under the output tubes cathode and measure the current. And if you must use the total current for the entire amp, put a 1Ω resistor between the HT center tap and ground. The voltage dropped by that resistor will directly represent the total B+ current.

[Ed_Chambley]

On the high side would be v=16.  This would be the voltage drop.  So if I begin with 385 vac I should drop from 385 to 373.

Is this how it works?

Not quite. The 385-475v is B+, not PT secondary voltage. Since the 100Ω resistor is in the B+ line, it will drop 16vdc. So, the 16 volts dropped across the 100Ω will subtract form the B+ value.

In review, you must know the total B+ current that will flow through the 100Ω sag resistor. Then it simply becomes E=IR to calculate the voltage dropped by the sag resistor. The total B+ current consists of currents drawn by the power tubes, PI, and preamp tubes. The power tubes draw the lions share of the total current so usually just ignore the PI and preamp currents. If you already know the current you'll bias your output tubes, use that figure for total B+ current. Alternately, put a 1Ω resistor under the output tubes cathode and measure the current. And if you must use the total current for the entire amp, put a 1Ω resistor between the HT center tap and ground. The voltage dropped by that resistor will directly represent the total B+ current.

Sorry, I meant to write 385 vdc since the resistor is post rectification which makes it the beginning of the B+.  I ask this so I can understand and know how to use dropping resistors with so much trial and error.  Thank you for the information.  I now have a few more building blocks.

[Jack1962]

I have does this a number of times Ed , it will work but it will increase the DC Voltage by 40 or 55 volts DC . I have always used a 100 ohm resistor directly attached to pin 8 of the socket , works great for me and everyone I have done this for.