LM723 Small Power Supply - psu as a beginner

[mike_mike]

Hello, I made a quick design of a schematic by reading some articles and schematics on the internet. Please find the schematic below.
I decided to use a LM317 power supply for supplying current to the 723 because the voltage from the main winding of the transformer was above 40V dc. Also, I found some time ago that using a regulated supply for 723, make the output voltage more stable (less ripple). So V3 (35V) is a LM317AHVT power supply, which is powered using the same transformer, but using another secondary winding which has 28Vac.

Regarding the value of C7, I established this value by experimenting. Without thic capacitor, there were some oscillations on the R11 (R shunt). Also, I made some quick tests, with the output voltage of the psu at max (30V approx.) and using 1n, 2n2, 4n7 was resulting in oscillations. Also, by connecting the 1n, 2n2 and 4n7 in parallel dampened the oscillation a little. During the tests, the value of C6 was always 100pF.

Screenshots:
1. Oscillations (with the probe of the scope on R11): 31_000, 31_002.
2. With 1n+2n2+4n7 in parallel: 32_000, 32_003.
3. With 10nF: 32_004, 32_006.
4. The simulation file.

Please have a look and tell me what you think.

[Simon]

schematic?

[mike_mike]

Plese find attached the schematic. It can also be found in the .asc file.
I am almost sure that the schematic need some improvements...

[blackdog]

Hi,
Some quick tips,

Place at least a 10nF capacitor over R14.

Remove Q4 and use the transitor here for the LM723.
Remove R11 and adjust R6 and R7 so you get the right value for your current limitation.

You can then use two resistors of say 100 Ohm from each emitter of your power transistors to the base of the current limit transitor in the LM723.

Another tip regarding stability, use faster power transistors!
This gives a larger phase margin.

Replace the 2N3055 with good 2SC5200 and then no China crap.
Q3 by e.g. a BD139.

Place a capacitor from the reference output to ground for less noise, 10uF is usually sufficient.

Use short wiring, do not use spaghetti wiring even when testing. 

Kind regards,
Bram

[mike_mike]

I found a problem in the circuit and I deleted the screenshots. I will try to solve the problem and I will come back.

[mike_mike]

I made the following modifications:
1. 10nF capacitor in parallel with R14 (8k2).
2. 10uF capacitor from pin 5 of 723 to gnd.

Screenshots:
1. With the probe on R_Shunt 37_003, 37_004.
2. On the PSU output 37_005.

[David Hess]

Place at least a 10nF capacitor over R14.

I agree, although I usually include space for a lower value capacitor and a series connected capacitor and resistor.  A little bit of phase lead in the feedback network helps although it is not necessary in lower performance design.

Another tip regarding stability, use faster power transistors!
This gives a larger phase margin.

I do not think there would be much advantage to using faster output transistors unless you want to minimize the output capacitance which is done in high performance designs.

Replace the 2N3055 with good 2SC5200 and then no China crap.
Q3 by e.g. a BD139.

The ubiquitous D44H11 ring emitter transistor is the common inexpensive output transistor.  Transistors intended for audio power amplifiers are also suitable if you want larger devices.  In addition to the BD139, another D44H11 could be used as the driver or the MJE182, MJE200, and MJE243 are also suitable.

Use short wiring, do not use spaghetti wiring even when testing.

Make sure the bulk input capacitor is connected close to the collector of the output transistor array.  Use a single point power ground.

Also, get rid of C4 or move it to the output terminals; all it does in its shown position is add high frequency phase lag an encourage oscillation of the output transistors, although this is unlikely with the resistors in series with the emitters of the output transistors.  This is an advantage of placing the current shunt in series with the output transistors.

A more advanced design could take AC feedback directly from the emitters of the output transistors before the current shunt, and this is what is done to allow a capacitor like C4 or a large ceramic output capacitor in place of C5, but it is hardly ever necessary.  By doing this, the ballast and current shunt resistance at the emitters of the output transistors is used to add phase lead.

[mike_mike]

Hi! I made the power supply from the attached schematic.
I tested it using the oscilloscope, using a PSU shorter, which is a circuit with LM555 that is generating pulses through a mosfet gate and is connecting and disconnecting a load. ln case of short circuit testing the load is 0 ohms, while testing with load, the load is 7.8 ohms. The set output voltage of the PSU is 25V. The rectifier is not present into schematic, but it is connected between the 30V ac transformer and filter capacitors.

The results are:
1. DS0018 = with 0 ohms at PSU shorter, and the yellow trace, CH1, on the output of the PSU, CH2, is the output of the PSU Shorter pulse generator.
2. DS0019 = with 0 ohms at PSU shorter, but with the probe on the R_Shunt resistor (on the 2 0.33R resistors).
3. DS0020, 21 = with 7.8 ohms at PSU shorter, and the scope probe on the output of the PSU (yellow trace) and the CH2 probe (blue trace) on the output of the PSU shorter pulse generator.

Please have a look at the screenshots and tell me what you think.

[David Hess]

The voltage spike when the output recovers is higher than it should be.  The problem is that charge is not being removed from the output and driver transistors quickly enough.  The 723 has no way to do this directly so the simple solution is to add low value resistors in parallel with the base-emitter junctions of the output and driver transistors to remove charge.  I usually make the resistors draw 1/10th of the base current but 1/20th or even lower works also.

If the output is 4 amps, so the base current of the output transistors is about 0.2 amps, then 1/10th would be 0.02 amps and with a Vbe of 1.0 volts, so a 50 ohm shunt resistor.  The resistor for the drive transistor could be about 40 times that from its current gain so 2200 ohms although I would make it lower.

None of these values are critical and transistor gain (hfe) varies widely anyway.

Darlington transistors have these base-emitter shunt resistors built in because they have the same problem; the drive transistor cannot remove charge from the output transistor.

[wizard69]

On your last schematic you have transistors labeled starting with a "T" and some with a "Q".   Is there a reason behind that.    I'm by no means a power supply engineer but I have to read a lot of schematics at work and things like this bug me.   I do like however the clearly labeled power input terminals and output terminals.

Best of luck with your power supply.