Power supply voltage is creeping

[SY]

I don't quite understand this. How can an opamp drive current it's not pulling from the supply?

Generally, supply current is spec'd at a particular load (or no signal, i.e., idle current). If you draw current into the load, the total supply current is the idle current plus the load current.

 

[AndrewT]

Hi,
you can use a pre-regulator followed by the adjustable regulator.

BUT,
you must protect both regulators from excessive voltage drop, particularly at start-up and during output short accident.

 

[veracohr]

Hi,
you can use a pre-regulator followed by the adjustable regulator.

What's the purpose of that?

 

[DF96]

Spread the voltage drop and power dissipation between two of them, so neither is near its limits. However, you would need to watch power-up and power-down, as AndrewT said.

 

[veracohr]

Well, I'm already committed to getting a more suitable transformer anyway.

 

[tomchr]

Again, the datasheet gives no maximum input voltage rating. It only gives a maximum input-output differential, which is 40V.

My bad... The LM317 is a floating regulator, hence, only the input-to-output differential voltage matters. At 1.25 V output, you can have 41.25 V on the input and still meet the 40 V max.

Now, maybe you guys can help me understand how to calculate the input voltage.

There are two issues here. 1) Max input voltage. 2) Ripple voltage.

The max theoretical input voltage is VAC(RMS) * sqrt(2) = VAC(RMS)*1.41. However, this assumes that the rectifying diodes conduct infinite current for zero time (zero conduction angle), which of course is not the case in real life. Typically, you'll get about VAC(RMS) * 1.3 as the voltage across the reservoir cap.

The ripple voltage and, thereby, the minimum input voltage depends on the reservoir cap and the load current. When calculating the capacitance needed, I use the following equation: C = (i * t)/V, where C is the capacitance in Farad, i is the load current in Ampere, t is time in seconds, and V is the ripple voltage in Volt. I generally design for worst case, hence, I use t = 0.01 second. This corresponds to operation at 50 Hz with a full-wave rectifier. V needs to be limited such that the input voltage to the regulator never drops below the minimum worst case voltage required by the regulator to maintain regulation. Hence, V < Vin_min - (Vout + Vdrop-out), where Vout is the output voltage, Vin_min is the worst case (lowest) input voltage the regulator will see (not accounting for ripple), and Vdrop-out is the worst case drop-out voltage for the regulator (2.5 V for the LM317 if I recall correctly).

Unfortunately when I was trying to replace the burned out components, a couple parts of the copper traces peeled away from the board. I'm going to have to etch a new board and rebuild the damned thing.

Not necessarily. You may be able to hack it with a few pieces of wire. Once you get the circuit working you can either live with it or etch a new board.

The trick to avoid this trace peeling issue is to make the pads and traces oversize. I use 80x100 mil oblong pads for IC pins and at least 100 mil pads for resistors, etc. On a 100-mil grid, I'd use 80 mil pads. Use wide traces. 20 mil minimum -- 15 mil if you have to go between two pads. Also, if the traces carry any significant current, use about 1 mm trace width per Ampere current. 1 mm = 40 mil (give or take).

~Tom