Question about Agilent E3634 design details

[prasimix]

I found on the net a schematic of Agilent E3634 power supply and using it to compare some design details with my current working on programmable power supply. There is a lots of thing that is still a mystery for me but two of them catch my attention and I would like to share that with other members of the forum. An edited screenshot is attached. There is a two things:

1. P-ch mosfet section looks to me as a "constant current" source that has a more favourable characteristic then resistor. I believe that is added as small "always on" load to achieve better regulation (loop stability). If this is correct I don't understand what is happened when Vout rise about 15V since this model deliver up to 50V.

2. AD706 DC precision opamp is used for current monitoring. It is possible that this opamp can withstand 50V common mode voltage? In datasheet is not stated any absolute maximum rating for common mode voltage only Differential Input voltage which should not be larger than 0.7V but even that is internally clamped with protection diodes.

[just_fib_it]

In these E36xx power supplies the positive output acts as the ground reference for the control circuitry. So when Vout is high, so are the +/-15V bias supplies.

[prasimix]

Thanks for quick answer. I'd like some more clarification. Does it mean that OUT- is actually more positive for control circuitry then OUT+? What is a purpose of diode CR40?

[just_fib_it]

From the point of view of the control circuitry, Vout+ is 0V (*) and Vout- is a negative voltage.

*) Nitpick: the ground point (A1 on the schematic) is actually before the current shunt, so Vout+ is also slightly negative relative to the ground of the control circuitry.

CR40 prevents the +15V from showing up at the output when Vout is below 15V.

[prasimix]

From the point of view of the control circuitry, Vout+ is 0V (*) and Vout- is a negative voltage.

*) Nitpick: the ground point (A1 on the schematic) is actually before the current shunt, so Vout+ is also slightly negative relative to the ground of the control circuitry.

CR40 prevents the +15V from showing up at the output when Vout is below 15V.

It starts now to make sense. Presuming that you have a whole schematics I'd like to ask you where is the point where +out becomes "negative"? How do they achieve that? Is it enough that bias power supplies used for control logic are "floating" in accordance to "main power loop"?

[void_error]

Take a look at the schematic of the 3610A, it uses a floating ground for the control circuitry and it's easier to understand than the 3634.

[David Hess]

1. P-ch mosfet section looks to me as a "constant current" source that has a more favourable characteristic then resistor. I believe that is added as small "always on" load to achieve better regulation (loop stability). If this is correct I don't understand what is happened when Vout rise about 15V since this model deliver up to 50V.

The +15 volt supply is isolated from the power ground so I think something else is going on here.

It looks to me like this connection is to prevent the +15 volt isolated supply and everything it is connected to from drifting away from the positive output.  If the +15 volt supply is high, then the p-channel MOSFET pulls it down through the resistor.  If it is lower, then the diode pulls it up sharply.

[prasimix]

Take a look at the schematic of the 3610A, it uses a floating ground for the control circuitry and it's easier to understand than the 3634.

Thanks for recommendation. Yes, this model is much simpler. What is a benefit of such connection? To invert control logic levels in reference to main power loop?

[prasimix]

1. P-ch mosfet section looks to me as a "constant current" source that has a more favourable characteristic then resistor. I believe that is added as small "always on" load to achieve better regulation (loop stability). If this is correct I don't understand what is happened when Vout rise about 15V since this model deliver up to 50V.

The +15 volt supply is isolated from the power ground so I think something else is going on here.

It looks to me like this connection is to prevent the +15 volt isolated supply and everything it is connected to from drifting away from the positive output.  If the +15 volt supply is high, then the p-channel MOSFET pulls it down through the resistor.  If it is lower, then the diode pulls it up sharply.

Does it mean that is a kind of "trade-off" resulted by connection of Vout+ to the ground of bias power supply used for the control circuitry?

[David Hess]

1. P-ch mosfet section looks to me as a "constant current" source that has a more favourable characteristic then resistor. I believe that is added as small "always on" load to achieve better regulation (loop stability). If this is correct I don't understand what is happened when Vout rise about 15V since this model deliver up to 50V.

The +15 volt supply is isolated from the power ground so I think something else is going on here.

It looks to me like this connection is to prevent the +15 volt isolated supply and everything it is connected to from drifting away from the positive output.  If the +15 volt supply is high, then the p-channel MOSFET pulls it down through the resistor.  If it is lower, then the diode pulls it up sharply.

Does it mean that is a kind of "trade-off" resulted by connection of Vout+ to the ground of bias power supply used for the control circuitry?

I do not know why they did it and the theory section in the service manual does not say anything about it.  Further study might reveal the reason.

My guess is that there was some overload condition like shorting the output which could result in failure if the bias supply got out of range.

[C]

What is the simple -* 15 volt supply
a center tapped winding
a full wave bridge
two caps

think more on the cap's a second. two conductive cylinder with an insulator, A cap to what is around it!!
all of those parts have some Capacitance to what is around them.

So if you change the world around them then you start charging or discharging that Capacitance.

Now think on that Op-amp that gets it's power from that floating supply.
The Op-amp inputs functions best when the inputs are in the center between the supply voltages and stays good until the signals get close to supply rails.

The Op-amp outputs will pull on the supply rails to produce their outputs. If the  Op-amp output is at what should be the floating supply center then the Op-amp  will pull the floating supply so that it is centered.

So with careful design that floating supply will float right where is it suppose to be. But remember that  Capacitance, for a fast change, it will be slow to charge/discharge and may need help for the very fast changes to where it should be floating.

CR40 takes care of the +Out increasing faster then the normal float adjust that the +-15 volt supply can handle.

I would guess that Q7 handles the problem that the +-15 volt supply wants to float up higher than wanted for static operation. And also handles the -Out dropping very fast.

You have here a +-15 volt supply. What is the dynamic range that the floating supply can be off center and the control system can still regulate? Add that the greater the change from the static float voltage center, the greater the pull on the supply rails to get back to that static float center.

So why float the +-15 Volt supply instead of just connecting the center of this supply to +Out?
There are probably many reasons.
One could be to keep some of the control noise from being transferred to +Out. 


The big problem with floating electronics is not that it is floating, it is that some repair tech will damage the circuit by thinking it is not floating.


C

[prasimix]

So why float the +-15 Volt supply instead of just connecting the center of this supply to +Out?

In this case Agilent made connection of bias common ground to +Out. Is it possible that they did that to insure bias voltage for serial mosfet gate (HEXFET) without using an additional voltage (and aditional transformer winding) that is few volts above drain potential?