I've been fixing a Cambridge A60 - turned out to be nothing more than both
speaker fuses blown. But thought I might as well check the quiescent
current while I'm at it. And the method given in the manual is weird -
replace the PS DC fuses with 15 ohm resistors and measure the voltage
across them.

Not having any high watt 15 ohm resistors to hand is there a different way?

--
*Toilet stolen from police station. Cops have nothing to go on.

Dave Plowman London SW
To e-mail, change noise into sound.

On Sat, 11 Jun 2011 18:20:33 +0100, "Dave Plowman (News)"
wrote:

I've been fixing a Cambridge A60 - turned out to be nothing more than both
speaker fuses blown. But thought I might as well check the quiescent
current while I'm at it. And the method given in the manual is weird -
replace the PS DC fuses with 15 ohm resistors and measure the voltage
across them.

Not having any high watt 15 ohm resistors to hand is there a different way?

Take the fuse out and just put in an ammeter of suitable range. I
think that if it is a modern digital type, its internal resistance
will be low enough. Ohms law will translate volts across 15 ohms into
current.

d

"Dave Plowman (News)" wrote in message
...
I've been fixing a Cambridge A60 - turned out to be nothing more than both
speaker fuses blown. But thought I might as well check the quiescent
current while I'm at it. And the method given in the manual is weird -
replace the PS DC fuses with 15 ohm resistors and measure the voltage
across them.

Not having any high watt 15 ohm resistors to hand is there a different
way?


I don't know that those resistors have to have a lot of power rating. If
you are adjusting quiescent current, you're only talking a few dozen
milliamps. That doesn't drop a lot of voltage and therefore, the power
being dissipated is nominal.

In article ,
Arny Krueger wrote:

"Dave Plowman (News)" wrote in message
...
I've been fixing a Cambridge A60 - turned out to be nothing more than
both speaker fuses blown. But thought I might as well check the
quiescent current while I'm at it. And the method given in the manual
is weird - replace the PS DC fuses with 15 ohm resistors and measure
the voltage across them.

Not having any high watt 15 ohm resistors to hand is there a different
way?


I don't know that those resistors have to have a lot of power rating.
If you are adjusting quiescent current, you're only talking a few dozen
milliamps. That doesn't drop a lot of voltage and therefore, the power
being dissipated is nominal.

They say to use 5 watt ones.

--
*A backward poet writes inverse.*

Dave Plowman London SW
To e-mail, change noise into sound.

"Dave Plowman (News)"

I've been fixing a Cambridge A60 - turned out to be nothing more than both
speaker fuses blown. But thought I might as well check the quiescent
current while I'm at it. And the method given in the manual is weird -
replace the PS DC fuses with 15 ohm resistors and measure the voltage
across them.

Not having any high watt 15 ohm resistors to hand is there a different
way?


** Measure the DC voltage across the emitter resistors - easy done with
any DMM.

Then use Ohm's law.


..... Phil

In article , Don Pearce
wrote:
On Sat, 11 Jun 2011 18:20:33 +0100, "Dave Plowman (News)"
wrote:

I've been fixing a Cambridge A60 - turned out to be nothing more than
both speaker fuses blown. But thought I might as well check the
quiescent current while I'm at it. And the method given in the manual
is weird - replace the PS DC fuses with 15 ohm resistors and measure
the voltage across them.

Not having any high watt 15 ohm resistors to hand is there a different
way?

Take the fuse out and just put in an ammeter of suitable range. I think
that if it is a modern digital type, its internal resistance will be low
enough. Ohms law will translate volts across 15 ohms into current.

If you do this I recommend also having a shunt capacitor across the place
(fuseholder) where you connect the ammeter. (i.e. shunting the ammeter at
HF.)

Otherwise you may get amplifier instability due to the increased rail lead
inductance, etc. No idea if this *will* happen with the A60. But it can
occur for some designs/layouts. Use 0.1 microF or more. FWIW I always used
1 microF.

Slainte,

Jim

--
Please use the address on the audiomisc page if you wish to email me.
Electronics http://www.st-and.ac.uk/~www_pa/Scot...o/electron.htm
Armstrong Audio http://www.audiomisc.co.uk/Armstrong/armstrong.html
Audio Misc http://www.audiomisc.co.uk/index.html

On Sun, 12 Jun 2011 09:40:53 +0100, Jim Lesurf
wrote:

In article , Don Pearce
wrote:
On Sat, 11 Jun 2011 18:20:33 +0100, "Dave Plowman (News)"
wrote:

I've been fixing a Cambridge A60 - turned out to be nothing more than
both speaker fuses blown. But thought I might as well check the
quiescent current while I'm at it. And the method given in the manual
is weird - replace the PS DC fuses with 15 ohm resistors and measure
the voltage across them.

Not having any high watt 15 ohm resistors to hand is there a different
way?

Take the fuse out and just put in an ammeter of suitable range. I think
that if it is a modern digital type, its internal resistance will be low
enough. Ohms law will translate volts across 15 ohms into current.

If you do this I recommend also having a shunt capacitor across the place
(fuseholder) where you connect the ammeter. (i.e. shunting the ammeter at
HF.)

Otherwise you may get amplifier instability due to the increased rail lead
inductance, etc. No idea if this *will* happen with the A60. But it can
occur for some designs/layouts. Use 0.1 microF or more. FWIW I always used
1 microF.

Slainte,

Jim

Shouldn't be a problem from what I remember of the Cambridge design,
but it never hurts to be safe. Somewhere in a junk box I still have a
modified fuse holder for the panel mounted bayonet fixture. It has a
couple of fat leads with banana plugs on the end for a meter for
current measurement.

One thing that could be a gotcha, though, is an autoscaling meter.
This could just induce oscillation in combination with the amp. A
fixed scale meter would be best.

d

"Dave Plowman (News)" wrote in message
...
In article ,
Arny Krueger wrote:

"Dave Plowman (News)" wrote in message
...
I've been fixing a Cambridge A60 - turned out to be nothing more than
both speaker fuses blown. But thought I might as well check the
quiescent current while I'm at it. And the method given in the manual
is weird - replace the PS DC fuses with 15 ohm resistors and measure
the voltage across them.

Not having any high watt 15 ohm resistors to hand is there a different
way?


I don't know that those resistors have to have a lot of power rating.
If you are adjusting quiescent current, you're only talking a few dozen
milliamps. That doesn't drop a lot of voltage and therefore, the power
being dissipated is nominal.

They say to use 5 watt ones.


Not exactly high power in my book. My high power resistors are rated at
250-300 watts.

On Sun, 12 Jun 2011 07:11:48 -0400, "Arny Krueger"
wrote:


"Dave Plowman (News)" wrote in message
...
In article ,
Arny Krueger wrote:

"Dave Plowman (News)" wrote in message
...
I've been fixing a Cambridge A60 - turned out to be nothing more than
both speaker fuses blown. But thought I might as well check the
quiescent current while I'm at it. And the method given in the manual
is weird - replace the PS DC fuses with 15 ohm resistors and measure
the voltage across them.

Not having any high watt 15 ohm resistors to hand is there a different
way?


I don't know that those resistors have to have a lot of power rating.
If you are adjusting quiescent current, you're only talking a few dozen
milliamps. That doesn't drop a lot of voltage and therefore, the power
being dissipated is nominal.

They say to use 5 watt ones.


Not exactly high power in my book. My high power resistors are rated at
250-300 watts.


The 5 watt rationale is obvious. Still physically small and wire ended
yet big enough not to go pop if the current is a bit higher than
expected on firing up.

And of course if you are expecting a dissipation of about a milliwatt,
5W is an extremely high power resistor.

d

"Don Pearce" wrote in message
...
On Sun, 12 Jun 2011 07:11:48 -0400, "Arny Krueger"
wrote:


"Dave Plowman (News)" wrote in message
...
In article ,
Arny Krueger wrote:

"Dave Plowman (News)" wrote in message
...
I've been fixing a Cambridge A60 - turned out to be nothing more than
both speaker fuses blown. But thought I might as well check the
quiescent current while I'm at it. And the method given in the manual
is weird - replace the PS DC fuses with 15 ohm resistors and measure
the voltage across them.

Not having any high watt 15 ohm resistors to hand is there a
different
way?


I don't know that those resistors have to have a lot of power rating.
If you are adjusting quiescent current, you're only talking a few dozen
milliamps. That doesn't drop a lot of voltage and therefore, the power
being dissipated is nominal.

They say to use 5 watt ones.

Not exactly high power in my book. My high power resistors are rated at
250-300 watts.

The 5 watt rationale is obvious. Still physically small and wire ended
yet big enough not to go pop if the current is a bit higher than
expected on firing up.

I suspect that the 5 watt rationale is based on making the part finger-tip
safe. Or it makes for a durable part with heavy leads.

If someone makes a mistake with a bias pot, both they and the equipment are
probably better off with the low value part - it will act like a fuse.

And of course if you are expecting a dissipation of about a milliwatt,
5W is an extremely high power resistor.

If you do the math, you can easily see that 5 watts is vast overkill.