Superimpose DC output with Sine wave

Fred,

Starting with schematic on data sheet page 6 (top).
Add a resistor between AC input and IN- pin. This will modulate the feedback voltage and the output will be modulated to keep IN- constant.

This will work well at 50Hz, but I am unsure how much bandwidth can be achieved.

Hello Ron,

Not sure that I understand what do you mean by  "Add a resistor between AC input and IN- pin".

I have COTS linear PS that uses transformer. Primary is connected to AC 110V , secondary is connected to the traditional linear regulator controlled by uA723.

Which AC input do you have in mind in your response?

Do you think that it is realistic expectation to achieve 20kHz bandwidth?

Thank you,

Fred Kleytman

Fred,

The AC input is your pulse generator sine wave. I did not mean the 50/60Hz wall outlet.

I added the resistor and a DC blocking capacitor. The new resistor is the same value as R1.
R1 and R2 are chosen for a VOUT of 14V.

Here is the attachment.

Ronald,

Any suggestion on AC amplitude level I have to apply? 1V p-p does not make any effect on the Vout.

Fred Kleytman

Freq,

That is odd. I expected the output sine Vpp to match the sine wave Vpp.

What values did you use for R1, C1, Rnew, and Cnew.
Also what is VIN and do you have an output capacitor?

Ronald,

R! = 5.3k; R2 = 3.9k Rnew = 5.3k Cnew = 47uF

Any way to attach word document that I can show to you PS schematic and what I am trying to achieve?

Or give me your email please.

Thank you,

Fred Kleytman

Hello Ronald,

Do you have any recommendations for me? 

Thank you,

Fred Kleytman

Fred,

Yes, there is an insert file button.

Hi Ronald,

Attached is what I am trying to achieve. Please review if it is possible.

Thank you,

Fred

Fred,

The values and connections point look good. The input cross over frequency is 0.6Hz (also good)

The uA723 and output transistors can increase the output voltage and the load can decrease the output voltage.
The rate that this can happen is a function of C6, current limit, and load current.

I suggest starting with a low frequency minimum amplitude signal, 10Hz 0Vpp. Measure VOUT; it will be a DC voltage.
Now increase the 10Hz amplitude slowly an watch the output. If the DC output voltage increases but there is no 10Hz AC then the problem is the output capacitor is holding too much charge. Pin 10 should have a 10Hz signal, but I suspect it won't be a sine wave.