Gottlieb System 1 Power Supply Analysis
+5V regulator
Design Approach:
Utilize switching type regulator for minimum heat and maximum efficiency.
Current limiting set to 4 amps max
Adjustable output voltage
Minimum input voltage = 8 volts

Design Analysis:
Most items in this circuit (and most power supplies designed by others) are derived from a ‘cook book’ design, the design values are provided by the manufacturer of the regulator circuit (Texas Instruments). The only deviation from the cook book design is the fine tune voltage adjustment circuit plus a few minor improvements on efficiency.
The fine tune voltage circuit consists of a 1.2K (1%) fixed resistor in parallel with a 5K (10%) trimming resistor fed into the regulator’s feedback pin. The 1.2K fixed resistor provides a maximum limit on the trimming resistor capability. When paralleled with the 5K trimmer, the introduced resistance can be anywhere within the range of zero to 993 ohms. This allows the user to increase voltage from the regulator by up to 0.48V beyond the devices preprogrammed 5V output. In the unlikely event of a complete trimming resistor failure, the regulator is limited by the 1.2K resistor to a maximum output voltage of 5.59 volts.
Built in short circuit protection can limit the device’s output from 3 to 5 amps. For this application – a 4 amp limit was used. Isc = 37125 / R2 = 37125 / 9100 = 4.08 amps.

Another factor affecting the +5V regulator is the input circuit. Sufficient input current must be available for proper operation. To operate at the peak 4.08 amps, the input circuit must be capable of providing 2.25 amps. If output power = 4.08A x 5V = 20.4W and assuming worst case 80% efficiency, input power = 25.5W. Using rectified 12V input then current = 2.2 amps. A 1N5404 can provide 2.4 amps max (derated by 20% due to capacitive load) but this is more than 50% loaded (I prefer to keep less than 50% load) Using 6A rectifiers provide sufficient current and keep heat buildup minimal. I took a shortcut on input capacitance and just used the original value as the base number and sized it up from there. Multiple Low-ESR type caps were placed in parallel to reduce overall ESR value (in addition to using high current caps specifically for regulator’s current spikes).

Design Efficiency:
The typical efficiency for this device is 84% when using a 12V input and 5V output. To enhance this efficiency, high quality paralleled low-ESR capacitors are used along with a very efficient catch diode circuit (MBR1635). Also, a torroid type inductor was used to improve efficiency and minimize EMI noise. Axial type ferrite stick type inductors were ruled out – they cost less but tend to have higher EMI noise generation. After adding these improvements, measured power supply efficiency is at 96%.

High voltage regulator
Design Approach:
Utilizes traditional series pass regulator with foldback current limiting.
Key components of series pass regulator: Q1, R11, D10
Current limiting established by Q2 and resistors R12, R14 and R15

Typical rectified DC voltage at input of Q1 = 90VDC
Voltage set by zener diode D10 at base of Q1 = 62VDC (10% tolerance)
Desired maximum output current = 200mA (typical load < 20mA)

Design Analysis:
Normal operation (output not shorted and Q2 turned off):
Desired maximum output current = 200mA
Imax = maximum amount of current from regulator before it enters current limiting mode.
Imax = Q1Vbe[(R12 + R15)/(R15 * R14)] + Vout[R12 / (R15 * R14)]
Q1Vbe for TIP31C = 1.8V max, Vout = 62V
Imax = 1.8[(1K + 12K)/(12K * 33)] + 62[1K / (12K * 33)]
Imax = 0.059 + 0.156 = 0.215A or 215mA (close enough to desired Imax = 200mA)
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R1 max = (Vs - Vz) / (IL + IZK)
Vs = 90 (supply voltage)
Vz = 62 (zener voltage)
Using Imax of 215mA and Q1hfe of 25 min (at 1A) then IL ~= 8.6mA
IZK = Zener knee current of 1N5372 = 1mA minimum...using 4mA
R1 max = (90 - 62) / (0.0086 + 0.004) = 2.2K max
R1 min = (Vs - Vz) / (IL + IZmax)
IZmax for 1N5372A = 5W / 62V = 80mA
R1 min = (90 - 62) / (0.0086 + 0.08) = 320 min
R1 range = 320 to 2.2K
Too low of value and too much heat created by D1. Too high of value will not provide enough current for both zener knee and Q1IB
Using 1.8K, 10% resistor –> actual value = 1620 to 1980 ohms.

Power dissipated by R1 (PR1):
VR1 = 90 - 62 = 28VDC
IR1 = 28V / 1800 = 0.016A
PR1 = (28 * 0.016) = 0.45W
R1 rated at 5W

Power dissipated by D1 (PD1):
Current thru D1 (ID1) ~= IR1 - Q1Ib (negligible) ~= Current thru R1.
PD1 = VD1 * ID1 = 62 * 0.016 = 0.992W
D1 rated at 5W
Thermal resistance ~= 35C/W. Worst case temperature rise = 35C

Power dissipated by current sensing resistor R14 (PR14):
PR14(typ) = R * I^2, R = 33, I = 20mA (typical)
PR14(typ) = 33 * 0.02^2 = 0.013W (typical)
PR14(max) = R * I^2, R = 33, I = 215mA (maximum)
PR14(max) = 33 * 0.215^2 = 1.53W (maximum)
R14 rated at 2W (R14 will not be subject to 1.53W during normal operation)

Power dissipated by R12 (PR12):
VR12 = (90 * 1K) / (1K + 12K) = 7V
PR12 = 7^2 / 1K = 0.049W
R12 rated at 0.5W

Power dissipated by R15 (PR15):
VR15 = (90 * 12K) / (1K + 12K) = 83V
PR15 = 83^2 / 12K = 0.574W
R15 rated at 1W

Power being dissipated by Q1 (PQ1):
Ice = 0.020A (typical load current)
Vce = 90 - 62 = 28V
PQ1 = Ice * Vce = 0.02 * 28V = 1.24W (typical)
Ice = 0.215A (maximum load current)
Vce = 90 - 62 = 28V
PQ1 = Ice * Vce = 0.215A * 28V = 6W (maximum)
NOTE: These power dissipation values are the same regardless of type of linear regulator – could be a series pass transistor (this design) or one of the integrated linear regulators such as the LM317AHVT or TL783CKC.

Q1 Thermal Resistance (worst case)
TIP31 thermal resistance, Junction to Case = 3.125C/W
Heatsink pad thermal resistance, Case to Sink = 0.35C/W
Heatsink thermal resistance, Sink to Ambient = 11.4C/W
Q1 accumulative thermal resistance = 3.125 + 0.35 + 11.4 = 14.875C/W

Q1 Temp rise normal operation with typical load:
Q1 temperature rise = 14.875C/W * 1.24W ~= 19C
Assume ambient temp = 25C
Q1 temperature at ambient room temp = 25 + 19 = 44C

Q1 Temp rise normal operation with maximum load:
Q1 temperature rise = 14.875C/W * 6W = 89.25C (don't touch!)
Assume ambient temp = 25C
Q1 temperature at ambient room temp = 89.25C + 25C = 114.25C
Q1 maximum allowed operating temperature = 150C
If Q1 did not have heatsink ...
Q1 thermal resistance, Junction to Ambient = 62.5C/W
Q1 Temp rise = 62.5 * 6W > 375C!
Regulator and heat sink as used by ’somebody else’…
Regulator thermal resistance, Junction to Ambient = 31.975C/W
Regulator Temp rise = 31.975 * 6W = 192C. Nearly 200C!
Maximum temp of device = 150C.
200C is hot enough to release the ‘magic smoke’.

Power being dissipated by Q2 (PQ2):
Q2 turned off in non-shorted configuration - power dissipation by Q2 = 0.

Shorted supply operation (Q2 turned on):
Ishort = Q1Vbe[(R12 + R15)/(R15 * R14)] + Vout[R12 / (R15 * R14)]
Q1Vbe for TIP31C = 1.8V max, Vout when shorted = 0V
Ishort = 1.8[(1K + 12K)/(12K * 33)]
Ishort = 0.059A or 59mA
When this power supply is shorted – the regulator will continue to try to supply power to this circuit but at a reduced 59mA current level. If the short circuit persists, R1 will absorb this supplied current.

Power dissipated by R1 (PR1) during short circuit condition:
VR1 = 90 - 0 = 90VDC (Q2 saturation voltage negligible)
PR1 = (90^2 / 1800) = 4.5W

R1 rated at 5W – resistor will get hot if supply left shorted but still operating at lower than maximum limits.

Power dissipated by D1 (PD1):
Current consumed by D1: ID1 ~= IR1 = [4.5W / 1800] = 0.0025A
Power dissipated by D1 worst case = (VD1 * ID1) = 62V * 0.0025A = 0.155W
D1 power rating = 1W

Power being dissipated by Q1 (PQ1):
Ice ~= Ishort = 0.059A
Vce = 90 - 0 = 90V
PQ1 = Ice * Vce = 0.059A * 90V = 5.3W

Q1 Temp rise during short circuit:
Q1 temperature rise = 14.875C/W * 5.3W = 78.84C
Assume ambient temp = 25C
Q1 temperature at ambient room temp = 78.84C + 25C = 103.84C
Q1 maximum allowed temperature = 150C
Again, this transistor will get quite hot during a short circuit but will still be operating at lower than maximum limits.