How does this pwr scaling thing work then?
- Thread starter Swedish Chef
- Start date
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Kev
Well-known member
.... what ?
:roll:
you predict it chops in half ... ?
OR
you have made this circuit and it DOES halve the output voltage.
Are you trying to create power supply sag
or stop it
with this circuit ?
If you are trying to OVER create supply sage and want more than just the effect of a resistor in series
then
:roll:
can't find the right words here
If this circuit were an exponential current limiter ... then there is the obvious situation that the unit will just shut down.
...
if there isn't a limit ... to the limit
:sad:
err
:roll:
...
now I have confused my self
.....
I'm back
still confused
but
please tell me that this IS NOT for a ClassA amp.
I would expect that at a given current ... the voltage output to drop significantly ( err ... that is more than linear )
:roll:
you predict it chops in half ... ?
OR
you have made this circuit and it DOES halve the output voltage.
Are you trying to create power supply sag
or stop it
with this circuit ?
If you are trying to OVER create supply sage and want more than just the effect of a resistor in series
then
:roll:
can't find the right words here
If this circuit were an exponential current limiter ... then there is the obvious situation that the unit will just shut down.
...
if there isn't a limit ... to the limit
:sad:
err
:roll:
...
now I have confused my self
.....
I'm back
still confused
but
please tell me that this IS NOT for a ClassA amp.
I would expect that at a given current ... the voltage output to drop significantly ( err ... that is more than linear )
DrFrankencopter
Well-known member
Pretty sure this circuit is used to drop the B+ voltage for a tube guitar amp. Lower's headroom and volume. Similar to using a variac, except your heaters will stay at their proper voltage.
Not sure exactly how it works though.....
Kris
Not sure exactly how it works though.....
Kris
looks like the 1 meg pot sets the current going to the 12V zener diode.
The gate of a MOSFETs often needs 2 to 4 V to turn on at all and often have a max gate voltage of 20V, 12V is in the range .
The NPN transitor turns on hard with greater current flow, as it turns on it shunts current around the Zener and I would guess drops the gate voltage from the 12V
The scaling sets the current to the zener that the NPN has to shunt to cause a sag by reducing the mosfet gate voltage
I would need to do some math but that what I see looking at the circuit quickly
The gate of a MOSFETs often needs 2 to 4 V to turn on at all and often have a max gate voltage of 20V, 12V is in the range .
The NPN transitor turns on hard with greater current flow, as it turns on it shunts current around the Zener and I would guess drops the gate voltage from the 12V
The scaling sets the current to the zener that the NPN has to shunt to cause a sag by reducing the mosfet gate voltage
I would need to do some math but that what I see looking at the circuit quickly
The first function is a simple source follower. The "Power Scale" pot sets a voltage on the Gate from maximum to near zero. The MOSFET Source puts out (essentially) the same voltage, except at very low impedance instead of the high impedance of the pot.
The Zener just protects against gross abuse, won't normally do anything. You could leave it out and the circuit would work, until the dry winter day you shuffle across a wool rug and throw a static spark into the pot and thus to the Gate.
The 15 ohm resistor and BJT transistor form a soft current limiter. When the resistor voltage increases enough, Q14 turns on, pulling down the Gate voltage and reducing output voltage. Class AB guitar amps vary in sound as the increased supply current on loud sounds interacts with power supply sag: this allows the hard/harsh sound of a hi-fi amp or the soft sag of an early Fender.
D3 seems useless. D2 may be needed to save Q14 from the 1uFd output cap when you turn off the amp. The 15, 33, and 1 ohm resistors all carry the same current and should have power ratings proportional to resistance; it is odd they don't. For one pair 6L6GC, they could all be 2 watt.
The 953(???) resistor keeps Q14 Base from burning up, and could as well be 1K. The 100K looks like an attempt to skew the Sag pot center setting, but is too large to do much; would certainly work without that. The 33K at the bottom of the 1Meg pot sets the minimum output to about 20V; otherwise you'd have to turn the pot to 1 or 2 to get any output at all. Not essential, just nice.
At minimum setting, a Fender Twin would be down to about 80dB SPL at "full" power, a level you could easily talk over.
But wait. This trick won't work right on Fender Twin or any fixed-biased amplifier. Turn down the G2 voltage without altering the G1 bias, and the amp will just cut-off. It will be unresponsive except at maximum input. There may be a musical style where this "works", but it may just sound like the kind of bad connection that only passes the loudest signals. It should work fine on any self-biased (cathode resistor) amplifier.
The MOSFET should be heat-sunk for about the amplifier's maximum output. Pair of 6L6GC with 500V, you need a 50 watt heatsink. This could be the heatsink from one channel of a 100 watt transistor amplifier. This will generally be a sizable addition and packaging problem.
Use the very best insulator on the MOSFET and remember that all these parts, especially the MOSFET case/tab, live at fatal voltage.
The Zener just protects against gross abuse, won't normally do anything. You could leave it out and the circuit would work, until the dry winter day you shuffle across a wool rug and throw a static spark into the pot and thus to the Gate.
The 15 ohm resistor and BJT transistor form a soft current limiter. When the resistor voltage increases enough, Q14 turns on, pulling down the Gate voltage and reducing output voltage. Class AB guitar amps vary in sound as the increased supply current on loud sounds interacts with power supply sag: this allows the hard/harsh sound of a hi-fi amp or the soft sag of an early Fender.
D3 seems useless. D2 may be needed to save Q14 from the 1uFd output cap when you turn off the amp. The 15, 33, and 1 ohm resistors all carry the same current and should have power ratings proportional to resistance; it is odd they don't. For one pair 6L6GC, they could all be 2 watt.
The 953(???) resistor keeps Q14 Base from burning up, and could as well be 1K. The 100K looks like an attempt to skew the Sag pot center setting, but is too large to do much; would certainly work without that. The 33K at the bottom of the 1Meg pot sets the minimum output to about 20V; otherwise you'd have to turn the pot to 1 or 2 to get any output at all. Not essential, just nice.
At minimum setting, a Fender Twin would be down to about 80dB SPL at "full" power, a level you could easily talk over.
But wait. This trick won't work right on Fender Twin or any fixed-biased amplifier. Turn down the G2 voltage without altering the G1 bias, and the amp will just cut-off. It will be unresponsive except at maximum input. There may be a musical style where this "works", but it may just sound like the kind of bad connection that only passes the loudest signals. It should work fine on any self-biased (cathode resistor) amplifier.
The MOSFET should be heat-sunk for about the amplifier's maximum output. Pair of 6L6GC with 500V, you need a 50 watt heatsink. This could be the heatsink from one channel of a 100 watt transistor amplifier. This will generally be a sizable addition and packaging problem.
Use the very best insulator on the MOSFET and remember that all these parts, especially the MOSFET case/tab, live at fatal voltage.
Swedish Chef
Well-known member
Thank you very much PRR! You realise that this is all steming from the Olde Radio guitar amp that you talked me through at the other place...!
Now how can I make this pup work on a single-ended 8W amp?
As it stands at the moment, there seems to be too much voltage drop across the 9K i.e. about 45V so approx. 5mA when set for max output.
Is there any way around this?
Cheers
chef
Now how can I make this pup work on a single-ended 8W amp?
As it stands at the moment, there seems to be too much voltage drop across the 9K i.e. about 45V so approx. 5mA when set for max output.
Is there any way around this?
Cheers
chef
Swedish Chef
Well-known member
Having gingerly stepped down through 5K to 1K I seem to have closer to what I expect at the max. V+ setting. The 1K does get nice and warm but then so did the 9K. I haven't investigated to see if this has effected the Sag Control yet but will have a look later... :thumb:
chef
chef
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