Bottlehead C4S as a differential pair constant current sink?

[Mikey]

Hi Gang,

I'd like to try to use the Bottlehead C4S as a differential amplifier cathode load.
According to the C4S manual, this is possible by configuring the kit with MJE340's and 2N2222's.
The C4S would be used in place of an existing triode CCS (see attached file).

Is the replacement as simple as cutting out the triode and inserting the C4S in it's place?

Mike

[Paul Joppa]

Yes it is that simple. Be sure to check the operating limitations, 300 volts and 0.67 watts dissipation for the MJE340. (You can add an Aavid 5775 heat sink and get up to 1.67 watts dissipation.) Run at least 10% of the output current in the bias string; 20% is better but not always practical.

[Mikey]

Hi PJ, thanks for the reply!

So....just to document the retrofit for the forum archives:

1.  Remove the existing triode and it's bias components from the circuit
2.  attach C4S terminal A to the B- supply
3.  attach C4S terminal B to the cathodes of the vacuum tube
4.  attach C4S terminal C to GND

Is that correct?
If so, now I need to attach some values to resistors R1 and R2.

Measurements of the existing circuit show the following:

- B+ supply measures +250 VDC
- B- supply measures -250 VDC
- voltage drop across plate resistor R1 (49.9K) -> 125V drop = 2.5mA
- voltage drop across plate resistor R6 (49.9K) -> 125V drop = 2.5mA

My rudimentary math skills indicate that the existing CCS is set up to pass 5.0mA, no?
If so, how do I calculate the proper values for R1 and R2?  Do the same formulas apply
when using the C4S as a plate load?

If my logic is correct, let me know and I will try to calculate the proper R1 and R2 values
in a follow-up post...

Mike

[Paul Joppa]

R1 = 0.855/current, approximately. So 5mA = 0.005A, and R1 = 171 ohms

R2 passes 20% of 5mA = 1mA, and drops 250v, so R2 = 250K ohms.

R2 dissipation is 250v * 1mA, or 0.25 watts. I always downrate resistors by at least a factor of 3, so use a 1-watt resistor. Make sure it's rated for well over 250v!

The transistor drops 252v (assuming the differential pair biases at 2v) so the dissipation is 252 * 0.005, or 1.26 watts. It needs a heat sink.

[Mikey]

Hi Paul,

Let's assume for a moment that I didn't read your post above, and decided to calculate R1 and R2 from the procedure outlined in the C4S manual.
Let's also assume that I would make the calculations based upon the 'high current' application, and using 2mA (preferred?) LED bias.

Going strictly 'by-the-C4S-book', here is what I would have come up with:

R1 (from the chart on page 11) = 205 ohms
R2 (from the formula on page 15) = E/I = B-/.002A = 250V/0.002A = 125K ohms

Now that I've read your post, it seems you're OK with 1mA (20%) LED bias in this case.  That certainly explains the different R2 value that you calculated.

So, is the 'R1 = 0.855/current' formula a better way (rather than the chart on page 11) of determining the value of R1?

Mike

[Mikey]

One more quick question:

If two constant current sinks were laid out on a small circuit board,
could they share the bias components (LED's, R2) between them?

The CC sinks may or may not be running at the same current limit...

MIke

[Paul Joppa]

...
So, is the 'R1 = 0.855/current' formula a better way (rather than the chart on page 11) of determining the value of R1?
...
One more quick question: If two constant current sinks were laid out on a small circuit board,
could they share the bias components (LED's, R2) between them?...

The old manual was based on the old LEDs, which have not been in production for years. The current HLMP-6000 LEDs fit the 0.855/current estimate pretty well. The thing is, there are several different red LED formulations, each of which had a different voltage drop and different deviations from perfect linearity - if you change the LED, you have to re-calculate the resistances. If you have a stash of the old LEDs that Buddha originally specified (they had a clear package and radial round wire leads) then use his chart.

Yes, you can share the bias components - remember to double the bias current if you are feeding double the current sources. I don't think it buys you anything in performance unless it keeps the bias current above 1mA. Saves a 10-cent resistor and a couple 10-cent LEDs though.

[Mikey]

Thanks PJ,

I've got the HLMP-6000 LED's, so I'll stick with your formula.

I'll scrap the idea of a shared bias circuit...not worth the hassle.


Mike

[Mikey]

Yes it is that simple. Be sure to check the operating limitations, 300 volts and 0.67 watts dissipation for the MJE340. (You can add an Aavid 5775 heat sink and get up to 1.67 watts dissipation.) Run at least 10% of the output current in the bias string; 20% is better but not always practical.

Hi Paul,

Looking at the datasheet for the MJE340, it seems as thought this part can handle up to 20 watts dissipation.
Is this simply a matter of bolting on an even bigger heatsink?

Mike

[Paul Joppa]

Yes, you are correct. However, the full 20 watts requires a huge heatsink - think 4" by 5" by 2" deep - and good circulation of room-temperature air, not in the summer without air conditioning. I try to limit the internal chip temperature to 100 degrees C even though they are rated for 150C - this is based on various real-world failures. With the clip-on heat sinks we use, you are limited to the dissipation I mentioned. You can use TO-220 transistors - we use the MJE5731A in PNP alpplications - and get 1 watt bare, 2.5 watts with the clip-on we use for the Paramount soft-start upgrade, and 5.5 watts for the big extruded one used in Eros and the Speedball.

As usual, transistors are much touchier than tubes. A 20-watt tube will run, by itself, at 20 watts for its full life; practical transistors can be counted on for at most 1/20 of their rating alone, and 1/4 of the rating with generous heatsinks. Tubes can take transient plate voltages of at least twice their rated maximum, some of them 5 times for short intervals; transistors are damaged ijmmediately with the tiniest excess.

[Mikey]

Hi Paul,

Here's another thought:

A vacuum tube acts like a NPN transistor, correct?
So, could a rugged triode vacuum tube take the place of the
MJE340?  I'm guessing that it would have to bias a bit differently...

Would the calculations to derive R1 and R2 remain the same,
or would they need to be re-formulated?

Mike

[Paul Joppa]

Ought to work; make sure the tube bias voltage and current are within the limitations of the remaining small transistor.

[Mikey]

Is the replacement as simple as cutting out the triode and inserting the C4S in it's place?

Hi Paul,

I think I should have originally written:

Is the replacement as simple as cutting out the triode and it's cathode resistor, and inserting the C4S in it's place?

Would that be more accurate?

Mike

[Paul Joppa]

Sorry, I didn't see the attachment on the first post. Now I see it but I can't open it. I assumed there was more than just a resistor to operate the triode as a current source, so now I'm just confused!

[Mikey]

Hi Paul,

I've attached an earlier version of the Word document, hopefully you can open this one...

Mike