Tom VerbeureThe Scenic Route to Repairing a Self-Destructing SRS DG535 Digital Delay Generator
https://tomverbeure.github.io/2025/12/24/Repair-of-SRS-DG535.html
Philip Bragg@tom_verbeure I thought the boost resistors were horror until I got to the variable pin one geometry. SRS, why are you like this?!
Henning Paul DC4HP@synx508 @tom_verbeure I still don't understand why they didn't use series resistors in front of the regulators instead.
@hennichodernich @synx508 That would be a series resistor in front of both the 7805 and the current boost resistor, right?
@tom_verbeure @synx508 Yes, but no current boost resistor. You just have to make sure that the regulator sees enough voltage drop at maximum expected current.
So the regulator has to carry all the current but doesn't see the full voltage drop (the resistor dissipates the rest).
Here the corner case is when the current is higher than expected: In this case, the regulator doesn't see enough input voltage to properly regulate and its output voltage will drop below the regular value.
So the regulator has to carry all the current but doesn't see the full voltage drop (the resistor dissipates the rest).
Here the corner case is when the current is higher than expected: In this case, the regulator doesn't see enough input voltage to properly regulate and its output voltage will drop below the regular value.
@hennichodernich @synx508 That works as long as you don’t exceed that maximum output current of the 7805. The datasheet has a peak value of 2.4A. I didn’t measure the sustained current of the 5V rail, but I doubt that it’s this high.
@tom_verbeure @synx508 The series resistor should make it hard to reach maximum current (1A continuous) if it has been dimensioned correctly.
For example: 0.5A max. expected output current, 9V in. Minimum input voltage at the 7805 at 0.5A is 7V, so choose e.g. 3.3 ohms. This will result in 1.65V drop at 0.5A, therefore 7.35V at the input of the 7805. 2.35V*0.5A=1.175W in the regulator, 1.65V*0.5A=0.825W in the resistor.
For example: 0.5A max. expected output current, 9V in. Minimum input voltage at the 7805 at 0.5A is 7V, so choose e.g. 3.3 ohms. This will result in 1.65V drop at 0.5A, therefore 7.35V at the input of the 7805. 2.35V*0.5A=1.175W in the regulator, 1.65V*0.5A=0.825W in the resistor.
@tom_verbeure @synx508 If load current hypothetically increased to 1A, only 5.7V would be at the input of the regulator, so only 3.something volts at its output, and most likely the circuit behind it wouldn't draw 1A if the voltage dropped that far.
@synx508 In the eevblog forum thread, someone who has 2 DG535 to repair managed to get one working, then still got stuck in a boot loop with the second one because he made the same exact mistake with the LCD connector.
Gary @N8DMT@tom_verbeure Thanks for sharing your journey.
Loved this quote:
“We buy test equipment NOT because we need it, but because one day we might need it.”
#TestEquipment #ItsDeadJim
Loved this quote:
“We buy test equipment NOT because we need it, but because one day we might need it.”
#TestEquipment #ItsDeadJim
@n8dmt I thought of that a day before I found a use for my variac. That was satisfying. 😀