linear-reg & reference questions

Question 1

Drop-out vs. head-room
Scenario: 5 V, 500 mA rail from a 6 V battery; prove it stays in regulation down to 5.3 V.

Answer 1

Required head-room = 5.3 V – 5.0 V = 0.3 V
Pick TPS7A4700: Vdrop(max) = 0.2 V @ 0.5 A
0.3 V > 0.2 V → pass
PSRR 45 dB @ 100 kHz → 20 mV in becomes 0.2 mV out

Question 2

Worst-case output tolerance stack
Part: REF5030A (3.0 V) over –55 °C → +125 °C.

Answer 2

Temp coef: 180 °C × 3 ppm = 540 ppm = 0.054 %
RSS: √(0.05² + 0.1² + 0.1² + 0.054²) = 0.156 %
Vref_min = 3.0 (1 – 0.00156) = 2.995 V
Vref_max = 3.0 (1 + 0.00156) = 3.005 V

Question 3

Thermal shutdown math
Conditions: 1 A load, 4 V dropout, Tamb = 85 °C, no heatsink, θJA = 40 °C/W (SOT-223).

Answer 3

P = 1 A × 4 V = 4 W
Tj = 85 + 4 × 40 = 245 °C > 150 °C → fail
Choose DDPAK θJA = 8 °C/W → Tj = 85 + 4 × 8 = 117 °C → pass

Question 4

PSRR measurement setup
Goal: 1 Hz – 1 MHz PSRR on the bench.

Answer 4

Network analyzer → injection transformer → 50 mV AC on DC output
CH-A = input AC, CH-B = output AC
PSRR(f) = 20 log(|VIN_AC| / |VOUT_AC|)
Calibrate cables, keep injection ≤ 50 mV

Question 5

Load-transient response
Spec: 0 ↔ 500 mA, 1 A/µs, max ΔV < 50 mV.

Answer 5

ΔVESR = 0.5 A × 2 mΩ = 1 mV
ΔVloop = 0.5 / (2 π × 1 MHz × 22 µF) ≈ 3.6 mV
Total 4.6 mV ≪ 50 mV → pass

Question 6

Noise density hand-calc
Part: REF5050, 10 kHz BW.

Answer 6

What is in the datasheet is:
Table
Copy
Device 0.1 Hz – 10 Hz noise (typ)
REF5050 15 µVpp
So, for a real design you would write:
15 µVpp ÷ 6.6 ≈ 2.3 µVRMS (1/f region)
and then add the broadband term separately.
Always quote the actual number from the datasheet of the part you are using.

The “broad-band” (white-noise) term for the REF5050 is taken straight from its TI datasheet, Figure 7-14 and the Electrical Characteristics table:
Spectral density (white region, >1 kHz) is 0.9 µVrms / V
For the 5 V output that gives 0.9 µVrms/V × 5 V = 4.5 µVrms over any 1 Hz-wide slice .
To get the RMS noise in your bandwidth, simply multiply by √BW:
Broadband RMS = 4.5 µVrms × √(BW / 1 Hz)
Example – 10 kHz bandwidth:
4.5 µVrms × √10 000 = 45 µVrms

Total RMS = √( (15 µVpp / 6.6)² + (4.5 µVrms × √BW)² )
BW = 10 kHz → 45 µV
Total = √(2.3² + 45²) ≈ 45 µVrms

Question 7

Crow-bar protection
Threat: MCU shorts 5 V ref to 15 V bus.

Answer 7

P-ch MOSFET, 5.6 V zener gate-source
If Vload > 5.6 V → zener ON → gate rises → MOSFET OFF → ref safe
Zener power < 100 mW → no heatsink

Question 8

Cress factor

Answer 8

The 6.6 is a crest-factor shortcut engineers use to convert peak-to-peak noise into RMS for a band-limited 1/f (pink) noise waveform that is roughly Gaussian.
For Gaussian noise the theoretical crest-factor (peak / RMS) is about 3.3 on each side.
Peak-to-peak therefore spans ±3.3 σ → 6.6 σ.
So
RMS ≈ Vpp ÷ 6.6
That’s where the 5 µVpp ÷ 6.6 = 0.76 µVRMS comes from in the REF5050 datasheet example.