Constant current diodes (CCDs), frequently called current regulating diodes (CRDs) or current limiting diodes, are two-terminal semiconductor devices that clamp current to a precise value regardless of voltage fluctuations across them. If you need to drive an LED, bias a transistor, or trickle-charge a small battery without the overhead of a multi-pin IC, a CCD is the most elegant solution on the bench.
The safe default part numbers for through-hole designs are the Vishay 1N5283 through 1N5314 series (covering 220µA to 4.7mA at 100V max) and the Central Semiconductor CCLD series (e.g., CCLD050 to CCLD500). Expect to pay between $0.80 and $1.50 per unit for through-hole glass packages, with SMD variants commanding a premium. This guide covers how to select, bias, test, and deploy these components in real-world circuits.
Symbol, Pinout, and Internal JFET Structure
On a schematic, a constant current diode is drawn similarly to a standard rectifier diode, but with an additional line bending back across the cathode bar, or a vertical line intersecting the cathode bar. This signifies its regulating behavior rather than simple one-way conduction.
Physically, most through-hole CCDs (like the 1N5314) come in a DO-7 glass cartridge package with an axial lead. The cathode is marked by a black band on the glass body. The anode is the unmarked lead. Current must flow from Anode to Cathode for regulation to occur.
Internally, a CCD is not a PN-junction diode at all. It is an N-channel JFET (Junction Field Effect Transistor) with its gate terminal permanently shorted to its source terminal. When voltage is applied across the drain (anode) and source (cathode), the JFET conducts until the internal depletion region pinches off the channel. At that pinch-off voltage, the current flatlines at a specific value, known as the peak operating current ($I_p$). Understanding this JFET architecture is critical for troubleshooting, as it explains why CCDs behave differently than standard silicon diodes under test.
Operation Regions and Spec-Sheet Table
To use a CCD reliably, you must keep it operating in its saturation (active) region. Pushing it into the ohmic region ruins regulation, while pushing it into breakdown destroys the silicon. The table below maps the operating regions for a typical 4.7mA device like the 1N5314.
| Operation Region | Voltage Across CCD ($V_{AK}$) | Current Behavior | Power & Thermal Notes |
|---|---|---|---|
| Ohmic (Linear) | 0V to ~1.5V (Knee Voltage, $V_k$) | Current rises linearly with voltage. Acts like a low-value resistor. | Negligible dissipation. Regulation is lost here. |
| Saturation (Active) | 1.5V to 80V | Current is clamped flat at $I_p$ (e.g., 4.7mA ±10%). | Dissipation = $V_{AK} \times I_p$. Must stay under 600mW. |
| Breakdown | >100V (Max Rating) | Avalanche effect. Current spikes uncontrollably. | Destructive. Device will permanently short or open. |
Source reference: For deeper JFET physics and pinch-off mechanics, see the All About Circuits guide on constant current sources.
How to Select and Bias a Constant Current Diode
Biasing a CCD is simpler than biasing a BJT, but thermal math is where most hobbyists fail. Follow this numbered sequence to select and bias your part:
- Match the Load Current: Select a CCD with an $I_p$ rating that matches your load. If driving a standard 5mm LED (20mA), a single CCD won't work (max is ~4.7mA). You must either use a CCD to bias a pass transistor, or choose a low-current load like an optocoupler LED or a NiCd trickle charge circuit.
- Verify Headroom Voltage: Calculate the voltage dropped across the CCD: $V_{CCD} = V_{Supply} - V_{Load}$. This value must be greater than the knee voltage ($V_k$, typically 1.5V to 4V depending on the part). If $V_{CCD} < V_k$, the device is in the ohmic region and will not regulate.
- Calculate Power Dissipation: Multiply the voltage drop by the regulated current: $P_d = V_{CCD} \times I_p$. The standard DO-7 glass package has a maximum power dissipation of 600mW at 25°C ambient.
Bench Tip: Glass packages have terrible thermal conductivity. If your calculated $P_d$ exceeds 300mW, the CCD will be hot enough to burn your finger. Keep $P_d$ under 200mW for reliable, cool operation.
- Check Maximum Voltage: Ensure $V_{Supply}$ never exceeds the CCD's maximum rated voltage (usually 100V for the 1N5xxx series) under any transient or open-load condition.
Complete Application Circuit: 24V Optocoupler LED Driver
A common industrial requirement is driving an optocoupler LED from a 24V PLC (Programmable Logic Controller) output. Using a standard resistor wastes power and allows current to fluctuate if the 24V rail sags to 18V or spikes to 28V. A CCD provides rock-solid isolation current.
Target: Drive a PC817 optocoupler LED at exactly 4.7mA from a 24V DC industrial rail.
Bill of Materials:
- 1x 1N5314 Constant Current Diode (4.7mA, 100V max)
- 1x PC817 Optocoupler (LED $V_f$ = 1.2V typical, max continuous current 50mA)
- 1x 100Ω 1/4W Resistor (Transient protection only)
Wiring Steps:
- Connect the 24V PLC positive output to the Anode (unbanded lead) of the 1N5314.
- Connect the Cathode (banded lead) of the 1N5314 to one end of the 100Ω resistor.
- Connect the other end of the 100Ω resistor to Pin 1 (Anode) of the PC817 optocoupler.
- Connect Pin 2 (Cathode) of the PC817 to the PLC ground (0V).
Circuit Verification:
The voltage dropped by the optocoupler LED and resistor is roughly 1.7V. The voltage across the 1N5314 is $24V - 1.7V = 22.3V$. The power dissipated by the CCD is $22.3V \times 0.0047A = 104.8mW$. This is well below the 600mW limit, ensuring the glass package stays cool. The 100Ω resistor drops about 0.47V and limits transient fault currents if the PLC output accidentally shorts to a higher voltage bus, protecting the CCD from instantaneous avalanche.
Failure Modes and Multimeter Testing
Testing a constant current diode with a standard digital multimeter (DMM) is a frequent point of confusion. If you put a DMM in 'Diode Test' mode across a 1N5314, the meter will likely display 'OL' (Over Limit) or a very high voltage drop, even when forward-biased. The diode is not broken.
Because the internal JFET pinches off and limits current to 4.7mA, it cannot supply the 1mA to 2mA test current required to forward-bias the DMM's internal test circuitry properly, nor does it behave like a standard 0.7V PN junction. To properly test a CCD, you must build a quick bench jig:
- Set a bench power supply to 12.0V DC.
- Connect the CCD Anode to the positive terminal.
- Connect a 1kΩ, 1% precision resistor between the CCD Cathode and the power supply ground.
- Set your DMM to measure DC voltage across the 1kΩ resistor.
- Apply Ohm's Law: $I = V / R$. If you measure 4.70V across the resistor, the current is exactly 4.7mA. The CCD is healthy.
How They Fail:
CCDs rarely fail from old age. They fail from thermal runaway or overvoltage. If the power dissipation exceeds the DO-7 package limit (600mW), the silicon overheats. Unlike some power transistors that fail open, overheated JFET structures tend to melt internally and fail short-circuit. If subjected to voltage spikes exceeding 100V, the depletion region avalanches, usually blowing the internal wire bond and causing an open-circuit failure. Always use a transient voltage suppressor (TVS) diode in parallel with the CCD if the supply line is prone to inductive kicks.
Frequently Asked Questions
Can I use a constant current diode in AC circuits?
Yes, but with a major caveat. A CCD is a unipolar device; it will only regulate current during the positive half-cycle of the AC waveform. During the negative half-cycle, the internal JFET's gate-source diode becomes forward-biased, and it will conduct heavily like a standard shorted diode, potentially destroying itself if current isn't limited. To use a CCD for AC regulation, you must place it inside the DC terminals of a full-wave diode bridge rectifier, with the AC load connected to the bridge's AC terminals.
What is the difference between a constant current diode and an LM317?
The LM317 is a three-terminal linear voltage regulator that can be configured as a constant current source using an external sense resistor. The LM317 can handle up to 1.5A of current, but it requires a minimum dropout voltage of about 3V to regulate properly, and it dissipates massive heat at high voltages. A CCD is a two-terminal device requiring no external resistors, has a much lower dropout voltage (knee voltage ~1.5V), but is strictly limited to low currents (under 5mA). Choose the LM317 for power LED strings; choose a CCD for precision biasing, sensor excitation, or low-current indicator LEDs.
Why is my constant current diode getting too hot to touch?
The DO-7 glass package has a very high thermal resistance to ambient air (often >100°C/W). If your CCD is dissipating 400mW, the silicon junction will be roughly 40°C to 50°C hotter than the room temperature, making the glass envelope hot enough to burn skin. Recalculate your $V_{CCD} \times I_p$. If the dissipation is too high, you must either lower the supply voltage, or use the CCD to drive the base of a high-voltage BJT pass transistor, shifting the thermal burden to a TO-220 package with a heatsink.
Are there SMD equivalents for the 1N5283 series?
Yes. Manufacturers like Central Semiconductor produce SMD current regulating diodes in SOD-123, SOT-23, and SOT-89 packages (e.g., the CCLD and CMCD series). However, SMD variants are significantly more expensive (often $1.50 to $3.00 each in low quantities) and have lower maximum power dissipation ratings due to the smaller silicon die and PCB thermal coupling. For high-volume automated assembly, they are excellent; for bench prototyping, stick to the axial DO-7 parts.
