Rockford Fosgate Punch 400a4

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Perry Babin

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Do Q403 and Q404 now match?

Basically, everything in the amplifier section from Q401 on determines the output voltage. Q401 and Q402 are only 'satisfied' when the voltage on their individual bases match. When they don't match, they drive the other components to correct for the error in voltage.

If you still have DC after replacing Q404, Lift Q405 and Q406. Does the DC go away?
 
seankane

seankane

Member
Joined 2010
Perry Babin said:
Do Q403 and Q404 now match?

Basically, everything in the amplifier section from Q401 on determines the output voltage. Q401 and Q402 are only 'satisfied' when the voltage on their individual bases match. When they don't match, they drive the other components to correct for the error in voltage.

If you still have DC after replacing Q404, Lift Q405 and Q406. Does the DC go away?
Click to expand...

Q403 and Q404 have been replaced and match, terminal still has DC.

Lifted Q405 and Q406 terminal still has DC but the IRF540 no longer gets hot.
 
seankane

seankane

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Joined 2010
Perry Babin said:
If you have 16v on the gate, you should see that same voltage across R452 (confirm). If you have 16v across a 560 ohm resistor, it should get too hot to touch within seconds. If it's not getting hot, it's either open or out of tolerance.

Post the DC voltage measured directly across R430, R431 and R432.
Click to expand...

R452: -16.01 vdc

R430: -8.41 vdc

R431: 0.000 vdc

R432: -0.0012 vdc

The 560 ohm resisitor is not getting hot at all. Maybe I should change it?
 
seankane

seankane

Member
Joined 2010
Perry Babin said:
R452 is likely open. Nearly 1/2w of dissipation in a 0805 resistor should create intense heat.

Remove Q408 after replacing it (and before applying power) and confirm that the voltage from the gate-source for the 9540 is 0v. Leave Q408 out of the circuit for now.
Click to expand...

R452 was not open but not sure it was within tolerance. The resistor I removed read 555 ohm and the new one i put in read 561 ohm, quite a difference.

Q408 is removed.

Powered up amp and gate-source voltage on 9540 is 0.000vdc

Q405 and Q406 are still out of circuit. Should I put them back in or leave them out? Do these being out affect my readings I just gave you?
 
seankane

seankane

Member
Joined 2010
Perry Babin said:
Confirm that you have 0v DC on the speaker terminals.

Is Q408 leaky?

Post the voltages on those three resistors again. They're all likely 0v.

What is the DC voltage on the solder pad for the base of Q408 (black probe on ground terminal)?
Click to expand...

LR+ terminal reads: -0.015 vdc

R430: 0.000 vdc
R431: 0.000 vdc
R432: 0.000 vdc

Q408 base voltage 0.542 vdc

Q408 had lots of resistance in ohms so if its suppose to read OL in all configurations, then YES it was leaky.

Still wondering when or if I should put Q405 and Q406 back in?
 
seankane

seankane

Member
Joined 2010
I compared it to a new one and its virtually the same. The only way I get any resistance is red probe on base and black probe on collector or emitter. I get like 15.50 m ohms. On the old one same configuration I get 14.45 m ohms. I get OL in all other probe configurations.

So it must not be leaky?
 
seankane

seankane

Member
Joined 2010
Perry Babin said:
That's a fraction of a volt and insignificant.

Isn't this significantly different than before you replaced the resistor?

Didn't you have full rail on the output with Q405/6 out and Q408 in?
Click to expand...

More than significant I had over 29 vdc before.

Yes I did have full rail with 405/406 out and 408 in. So that little resistor being slightly out of tolerance caused all this, or am I not out of the clear yet?
 
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