DC elevation of heaters

[EFK]

I'm learning here so please bear with me! Fixed bias amp with filament CT. I lift the CT from ground and tie it to DC line out of one of the filter caps - this DC comes through a resistor of (unknown value - how do I determine this?) and goes to ground through a second resistor of (unknown value). The second resistor is bypassed with an approximate 10uf cap, and the line to the CT is tied into the junction of the cap and two resistors. Is this really all there is to it? If so, I guess my questions are how to determine the values of the resistors, and exactly where to tie into a DC line. One of the feeds to the preamp after dropping resistors? Or right off one of the filter caps? Or...??? Found a very good explanation of it on the 'Valve Wizard' site (clearest explanation I've thus far found) but I'm unclear on some of these details. Any help appreciated! Thanks.

[Chuck H]

I guess my questions are how to determine the values of the resistors, and exactly where to tie into a DC line.

A series resistor of 100k to 220k would seem appropriate. The dropping resistor makes the circuit a voltage divider. So the ratio between your series and dropping resistor is the same as the ratio between your starting voltage and your desired output voltage for the circuit, ie: "I have a 100V source and I'd like to get 10V". A 90k resistor with a 10k dropper will get you there.

I've seen this circuit tied to the main filter "bleeder" resistors. You know 'em. Those resistors that parallel the main filter or series filters in most amps. If you have series filters and two resistors (one across each filter) you should create your voltage divider across the cap that is tied to ground and the total resistance of your two resistors should be about the same as the resistor that you are replacing. It seems to me that placing it here should relieve the need to bypass the dropping resistor with a 10uf cap. If your amp doesn't have a bleeder resistor/s this would be a good time to add this along with the elevated heater circuit.

Chuck

[tubeswell]

I did this on a bassmanish build a few months ago.

In this case the B+ was 430VDC. The bleeder network was 220k (or equivalent) for each filter cap at the reservoir position.

The voltage divider is basically 39k/(39k + 180k + 220k) = 39k/439k = .089 x 430 (or 8.9% of 430VDC) = 38VDC: I call it 40V (the actual resistors weren't quite accurate). 2CW

[Gaz]

A series resistor of 100k to 220k would seem appropriate. The dropping resistor makes the circuit a voltage divider. So the ratio between your series and dropping resistor is the same as the ratio between your starting voltage and your desired output voltage for the circuit, ie: "I have a 100V source and I'd like to get 10V". A 90k resistor with a 10k dropper will get you there.

I was just wondering how to approximate the voltage drop in that circuit! I'm at the point where I'm trying to learn some simple math to help me with my designs, rather than just go by other's schematics and experiment, but often get confused.

So, with your example in mind: I have approximately a 385vdc source, a 200K series resistor, and a 47k dropping resistor to ground. How would I use the ratios to find the answer?

A hand holding through this would be greatly appreciated - I went to liberal arts school!!

[tubeswell]

I was just wondering how to approximate the voltage drop in that circuit!

... example in mind: approximately a 385vdc source, a 200K series resistor, and a 47k dropping resistor to ground. How would I use the ratios to find the answer?

385= voltage in.

47k/(47k + 200k) = 47k/247k = 0.19

385 x 0.19 = 73V out at the knee

run a 10uf - 47uF 100V cap in parallel with the 47k (+ve end at the knee) to stop unwanted oscillations wreaking sonic havoc with your heater-to-cathode voltage

[Gaz]

Tubeswell, thanks so much. I feel like I'm back in Algebra I (which wasn't that long ago).

[Chuck H]

run a 10uf - 47uF 100V cap in parallel with the 47k (+ve end at the knee) to stop unwanted oscillations wreaking sonic havoc with your heater-to-cathode voltage

Tubes, did you try the circuit without the 22uf cap across the dropping resistor? It seems to me that with the main filter on top of the circuit that decoupling is already achieved. But I have never built this circuit, I just know about it. What is your insight?

Chuck

[tubeswell]

Tubes, did you try the circuit without the 22uf cap across the dropping resistor? It seems to me that with the main filter on top of the circuit that decoupling is already achieved. But I have never built this circuit, I just know about it. What is your insight?

Chuck

If neither of the two resistors in the sub-divider are bypassed, there is the possibility of micro-rises in potential causing an oscillation. I didn't bother trying it without the extra bypass cap. loudthud also endorsed this approach, and the cap was nicks in terms of cost, so I'm convinced.

[txstrat]

I elevated the heaters in my latest build and used a 10nF disc cap parallel to the dropping resistor (the value of 10nF is suggested in TUT3). I couldn't make out a difference with cap or no cap. It's silent, though.
BTW
What is the difference of using higher values of the resistors (I use a 470k and a 100K dropping resistor)?
I mean the ratio can be achieved by many different values but I think a minimum value must be needed to keep the B+ from being bled off.

[tubeswell]

I
What is the difference of using higher values of the resistors (I use a 470k and a 100K dropping resistor)?
I mean the ratio can be achieved by many different values but I think a minimum value must be needed to keep the B+ from being bled off.

I think its a matter of optimising it to achieve the function of bleeding the filter caps in a reasonably short time after switchoff on one hand, and keeping the B+ from shorting to ground on the other.

Having said that, you don't need really high resistance to stop the B+ shorting to ground - 220k per cap seems to be a tried and true value (making 440k from B+ to ground where the caps are in series).

The value you end up selecting will also depend on whether there are any other resistive paths from the B+ to the ground return in the (pre-)amp - such as a resistive divider in a DC coupled pair for example - that could be in parallel to any bleeder cap network/elevated DC heater supply, that you are configuring.

[Steve Conner]

I like to use a 39V zener in place of the bottom resistor.

The reason: There are scenarios where a fair proportion of B+ can get onto the heater line. (say a power tube shorts, and a fuse in its cathode blows.) The zener at least has a chance of failing as a short circuit and saving all of the other tubes from heater-cathode breakdown.

And it is good practice to bypass that elevated heater line to ground with a decent-sized cap. The main filter cap prevents noise from the power rail getting down the divider, but it doesn't stop noise sneaking from one tube cathode to another along the heater wires, which the bypass cap will.

[hasserl]

Cool thread, this is the type of thread that makes this board the best on the net IMO.

For the guys looking for how to determine the resistor values, there is a formula for voltage dividers. Vout = Vin*R2/(R1+R2)

If you already some of the numbers, such as Vin and the desired Vout, and in the case of Tubeswell's example above you know the sum of R1 + R2 (220k), then it's just a matter of calculating the rest of the numbers.

[EFK]

Thank you guys so much! This has been very helpful and essentially answered all of the questions I had. I'm going to give it a try. Currently this amp does not have bleeders but perhaps I should look into that as well.

As an aside, does elevating the heaters draw any more current or put any additional stress on the PT in any way?

[Albert Kreuzer]

As an aside, does elevating the heaters draw any more current or put any additional stress on the PT in any way?

Marginally. Assuming 440V B+ and 440k (2 x 220k) bleeders there will be an additional current draw of 1mA.
Most PTs should be ok with that.

Cheers,
Albert

[EFK]

One more question - I have a 32/32 can cap that has an unused side for an extra 12ax7 which is no longer being used in the amp. It's fed by a 27K power resistor so the voltage on the line out is only 305v (was the plate feed for the now-removed triode). I can use this line for my DC feed to the filament CT, but I'm unsure as to the watt rating of the resistors to form the voltage divider. Are 1W sufficient or do I need to go higher? Any thoughts?