Zener-Transistor Constant Current Source - How to built constant current source

I think its been 10 days now that I started to learn more about negative feedback amplifier and how to use it. I covered the 3 types of negative feedback amplifier, voltage amplifier, transimpedance amplifier, transconductance amplifier and current amplifier. Among these, transimpedance amplifier(current to voltage converter) and the current amplifier(current to current converter) is current and so I started searching how to build a currrent source. In photodiode sensor with transimpedance blog post photo diode was used as current source. This note is about how to built Zener-Transistor Constant Current Source.

The following shows the circuit diagram of Zener diode and transistor based current source.

Zener-Transistor Constant Current Source circuit diagram

In the circuit a 5.1V zener diode 1N4372A is used to set the base voltage at 5.1V. This turns on the TIP31 power transistor. The resistor $R_E=44\Omega$ sets the voltage at the emitter voltage(4.4V in above circuit). The load is connected across the +12V and the collector.

Let's analyze the TIP31 + 1N4733A Zener current regulator circuit with simulation measurement and mathematically, compare theory with your measurements, and explain why you're seeing these voltages/currents.

Zener-Transistor Constant Current Source

Simulation Measurements:

Collector Voltage (Vc): 12V
Emitter Voltage (Ve): 4.54V
Current (Ie): 98.6mA

Theoretical Analysis

1. Zener Diode Operation

1N4733A Zener Voltage (Vz): 5.1V (nominal).
Zener Current (Iz):
$I_{z} = \frac{V_{c c} - V_{z}}{R_{B}} = \frac{12 V - 5.1 V}{1 k Ω} = 6.9 m A$
- This ensures the Zener stays in regulation (Iz > 1mA, typical minimum).

2. Transistor Base-Emitter Voltage (Vbe)

Base Voltage (Vb): Fixed by Zener = 5.1V.
Emitter Voltage (Ve):
$V_{e} = V_{b} - V_{b e} (where V_{b e} \approx 0.6 V for TIP31)$ $V_{e} = 5.1 V - 0.6 V = 4.5 V (matches your measured 4.54V)$

3. Emitter Current (Ie)

Theoretical Ie:
$I_{e} = \frac{V_{e}}{R_{e}} = \frac{4.5 V}{44 Ω} = 102.3 m A$
- Your measured Ie = 98.6mA (≈3.6% lower due to component tolerances).
- Possible reasons for discrepancy:
  - Actual $V_{b e}$ ≈ 0.56V (slightly lower than 0.6V).
  - Zener voltage slightly below 5.1V (e.g., 5.05V).
  - Resistor tolerance (44Ω may be 45Ω in reality).

4. Collector Voltage (Vc)

Expected Vc:
- If the load is disconnected (or very high impedance), the transistor cannot sink current, so:
  $V_{c} \approx V_{c c} = 12 V (matches the measurement)$
- If a load resistor (Rload) is connected:
  $V_{c} = V_{c c} - I_{c} \cdot R_{l o a d}$
  - Example: For $R_{l o a d} = 50 Ω$ ,
    $V_{c} = 12 V - (98.6 m A \times 50 Ω) = 7.07 V$

Comparison: Theory vs. Measurement

Parameter	Theoretical Value	Measured Value	Error	Explanation
Emitter Voltage (Ve)	4.5V	4.54V	+0.9%	Slightly higher $V_{z}$ or lower $V_{b e}$ .
Emitter Current (Ie)	102.3mA	98.6mA	-3.6%	Tolerance in $R_{e}$ , $V_{z}$ , or $V_{b e}$ .
Collector Voltage (Vc)	12V (no load)	12V	0%	No load → transistor in cutoff.

Why is Vc = 12V?

Your collector is at 12V because:
1. No load is connected (open circuit), so the transistor cannot sink current.
2. The TIP31 acts as a current sink—it only pulls current when a load provides a path to ground.
3. To see regulation, add a load resistor (Rload) between collector and ground:
  - For $I_{c} = 98.6 m A$ , choose $R_{l o a d} = \frac{12 V}{98.6 m A} \approx 122 Ω$

Power Dissipation Check

Transistor (TIP31):
$P_{T I P 31} = (V_{c} - V_{e}) \times I_{c} = (12 V - 4.54 V) \times 98.6 m A \approx 735 m W$
- Close to TIP31’s limit (TO-220 package: 2W max without heatsink).
- Solution: Reduce $V_{c c}$ or use a heatsink for long-term operation.
Zener Diode (1N4733A):
$P_{z e n e r} = V_{z} \times I_{z} = 5.1 V \times 6.9 m A \approx 35 m W (safe, << 1W rating)$

Key Takeaways

Your circuit works correctly as a 98.6mA current sink (close to theoretical 102.3mA).
Vc = 12V means no load is connected (expected behavior).
To use it as a current regulator:
- Add a load resistor (e.g., 50Ω–100Ω) to see $V_{c}$ drop.
- For precision, replace the Zener with a TL431 shunt regulator or LM4040.

Final Calculation Summary

\begin{aligned} V_{e} = V_{z} - V_{b e} = 5.1 V - 0.6 V = 4.5 V (Measured: 4.54V) \\ I_{e} = \frac{V_{e}}{R_{e}} = \frac{4.5 V}{44 Ω} = 102.3 m A (Measured: 98.6mA) \\ V_{c} = 12 V (No load → transistor inactive) \end{aligned}

In the example circuit below a load for testing is added and the current and voltage measurements are also shown.

The following video shows animation of how the Zener-Transistor Constant Current Source circuit works.

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