1. Why this part survives

The ULN2003A was designed to let low-voltage logic drive relays, lamps, solenoids and small stepper motors. It does that with seven Darlington pairs, each rated to sink up to 500 mA at 50 V, plus a diode per channel for clamping inductive kickback.

Nothing about that function has changed. Microcontrollers still output 3.3 V, relays still need 12 V, and designers still want one part that handles seven loads with no external components. That is why the same 16-pin package appears on boards designed decades apart.

The result is a commodity part with a large second-source market and a handful of variants that look identical and behave differently. Most substitution mistakes come from those variants.

2. What the device actually does

Each channel is an inverting Darlington pair with an open-collector output. A logic high on the input turns the output transistor on, pulling the output pin to ground. The load is connected between the positive supply and the output pin.

That arrangement is convenient: the driver sinks current, so the load voltage can be higher than the logic supply, up to the 50 V rating. It also means the output cannot source current — a detail that matters when someone tries to drive an LED from the output to ground.

The internal diode connects each output to the COM pin. When an inductive load is switched off, the diode clamps the kickback to the COM rail instead of letting it destroy the transistor. The COM pin must be connected to the load supply for that to work.

Why the saturation voltage is high

A Darlington pair is two transistors in series. The first drives the base of the second, so the output cannot saturate below roughly two base-emitter drops — about 1.0–1.2 V at moderate current, and more at higher current.

That matters for two reasons. Electrically, the load sees the supply minus that voltage, so a 12 V relay coil actually receives about 10.8 V. Thermally, the drop multiplied by the current becomes heat inside the package: 200 mA at 1.1 V is 0.22 W per channel.

A single MOSFET would have a far lower drop, but it would also need a gate drive and a clamp diode. The Darlington trades efficiency for simplicity, and that trade is the reason this part survives.

3. Pinout and pin compatibility

The reverse ordering of the outputs is a classic source of wiring errors. Pin 10 is channel 7 and pin 16 is channel 1, not the other way round.

4. The variants that look identical

Two of these are true pin-compatible substitutes for the ULN2003A: the ULN2003 variants and the ULQ2003A. The ULN2004A is physically compatible but has a much larger input resistor, and the ULN2803A is a different package with a different pinout.

Resistor math for 1.8 V and 3.3 V logic

The input current is roughly (Vin − 1.4 V) / R, where the 1.4 V is the two base-emitter drops of the Darlington. With a 2.7 kΩ resistor, 3.3 V gives about 0.7 mA; 5 V gives about 1.3 mA. Both are enough to saturate most channels.

At 1.8 V, the same resistor gives only 0.15 mA, which is marginal for full saturation under load. A design that runs 1.8 V logic should either use a driver with a smaller input resistor or add a pre-driver.

The ULN2004A's 10.5 kΩ resistor is designed for 6–15 V logic, where the higher voltage compensates for the larger resistance. Using it at 3.3 V is the reverse of its intended application.

5. The input resistor decides the logic level

The input is a resistor in series with the Darlington base. With a 2.7 kΩ resistor, a 3.3 V logic high drives roughly 1 mA into the base, which is enough to saturate the channel at rated current. With a 10.5 kΩ resistor, the same 3.3 V produces only about 0.3 mA — enough to turn the output on, but not always enough to keep it fully saturated under load.

That is why the ULN2004A exists: it is intended for higher-voltage logic, where the larger resistor limits current. Using it with 3.3 V logic is not a direct swap, even though it fits the socket.

The practical test is simple: connect the intended load and measure the output saturation voltage. A channel that is not fully saturated runs hot and can fail.

6. Output limits and thermal reality

The 500 mA rating is per channel, and it assumes the others are not all at full current at the same time. The package thermal resistance is the real constraint.

A DIP-16 package with a modest copper area has a junction-to-ambient resistance in the range of 60–100 °C/W. At 1 W dissipation, the junction rises 60–100 °C above ambient. That is acceptable at 25 °C, marginal at 50 °C inside an enclosure, and impossible at 70 °C.

The usual fix is not a different driver — it is fewer channels on, lower current, or an external MOSFET for the heavy loads.

Stepper motors and the classic 28BYJ-48

The most common hobby application is a unipolar stepper such as the 28BYJ-48, driven directly from four channels. Each coil draws a few hundred milliamps, and the driver switches them in sequence.

Two details decide whether it works reliably. First, the coils are inductive, so the COM pin must connect to the motor supply. Second, the current is continuous while the motor holds position, so the package dissipation is not a transient — it is the steady-state operating point.

If the motor stalls, current rises to the winding limit and the driver heats quickly. A design that works on the bench but fails in a hot enclosure is usually a thermal design problem, not a driver defect.

When calculating, use the saturation voltage at your actual current rather than the datasheet's headline figure, which is often quoted at a lower current. At 500 mA the drop can be well above 1 V, and the dissipation scales with it during every switching event.

7. The COM pin and inductive loads

For relays, solenoids and motors, the COM pin must connect to the positive supply of the load, not to logic ground. The internal diode then clamps the inductive kickback to that rail.

If COM is left floating, the diodes do nothing and the outputs see the full kickback voltage. The device may survive a few events and then fail — often in a way that looks like a random channel death.

For loads above the 50 V rating or currents above the internal diode rating, add external clamp diodes. The internal ones are convenient, not unlimited.

The choice between a Darlington array and a MOSFET driver comes down to three numbers: voltage drop, channel count and board space. A seven-channel Darlington in a 16-pin package is hard to beat when the loads are modest and the design values simplicity.

When 500 mA is not enough

Above roughly half an amp per channel, the Darlington array stops being the right answer. The options are discrete MOSFETs, a dedicated low-side driver, or a relay driver with a higher rating.

A MOSFET reduces the voltage drop to tens of millivolts and eliminates most of the thermal problem, at the cost of gate drive and an external clamp diode. For a single heavy load, that is usually cheaper than paralleling channels.

Paralleling channels is possible but not free: the two Darlington pairs do not share current perfectly, and one carries more than the other. A small emitter resistor per channel can force sharing, which defeats the simplicity of the part.

The 50 V rating is an absolute maximum for the output, not a recommended operating point for continuous use. A 24 V relay with a 36 V transient is fine; a 48 V solenoid with a 60 V spike is not.

Clamp diode ratings in practice

The internal diode is rated for the same voltage class as the output, but its current capability is not the same as the transistor's. A relay coil that draws 100 mA steady can produce a brief current of similar magnitude when switched off; a solenoid with a large inductance can produce more.

For loads with high inductance or frequent switching, an external diode with a higher current rating is cheap insurance. Place it across the load, not across the driver, so the loop is short.

8. Substitution checklist

Layout and grounding

The driver carries load current, not logic current. Keep the return path for the loads separate from the logic ground until a single star point, or the switching current will lift the logic reference and cause false triggering.

Place the COM connection close to the package and give it a short, wide trace. The clamp current flows through that pin during every switch-off event, and a thin trace adds inductance that defeats the diode.

For DIP packages, a socket is convenient for prototypes and a liability in production. Sockets add contact resistance and inductance exactly where the clamp current flows.

Common failure modes

One channel dead. Usually a missing COM connection or an unclamped inductive load. Check the load side first, then the driver.

All channels weak. Usually the input drive: 3.3 V logic into a 10.5 kΩ part, or a series resistor added by mistake. Measure the input voltage at the pin, not at the microcontroller.

Intermittent switching. Usually grounding: the load return shares a trace with the logic return, and switching current shifts the reference.

9. Test plan

Second-source differences show up in saturation voltage at low base current and in the clamp diode forward drop. Both are specified with minimum and maximum values; compare the maximums, not the typicals, when the design runs close to a limit.

Automotive and industrial notes

The Q-grade variant is specified for automotive use and carries the corresponding qualification. It is not simply a temperature-rated version of the standard part: the qualification documentation is part of the product.

For industrial designs, the relevant questions are the ambient temperature inside the enclosure and the number of channels that can be on simultaneously. Both are thermal questions, and both are answered by measurement rather than by the datasheet's per-channel rating.

10. Cost and supply

This is one of the cheapest driver ICs on the market, so the price difference between sources is measured in fractions of a cent. Availability, package and second-source consistency matter more.

Second sources are common and generally acceptable for non-automotive use. For automotive, use the Q-grade variant and confirm the qualification documents, because the standard part is not a substitute for an AEC-Q100 device.

Reel quantities and date codes still matter: a cheap reel of old stock that has been stored badly can cost more in rework than the part saved.

The practical rule for a substitution: match the channel count and input resistor, confirm the clamp diodes are internal and COM is wired, then measure the saturation voltage at your load current and check package temperature with every channel that will be on at once. Everything else is secondary.

11. FAQ

Is ULN2004A a drop-in for ULN2003A?

It fits the same socket but has a much larger input resistor. With 3.3 V logic the output may not saturate fully. Check the input current and saturation voltage.

Can I use ULN2803A instead?

No. It has eight channels and a different package and pinout. It is a functional upgrade, not a substitute.

Why does my relay channel fail after a few weeks?

Usually the COM pin is not connected, so the internal clamp diode does nothing and the kickback degrades the output transistor.

Can I drive an LED from the output to ground?

No. The output sinks current to ground. Connect the LED from the supply through a resistor to the output.

How many channels can run at 500 mA at once?

Only one, in practice, because of package dissipation. Derate total power rather than counting channels, and measure the package temperature with the enclosure closed.

Do I need external diodes for a 24 V relay?

Not for the voltage alone; the internal diodes are rated for the 50 V class. Add external diodes if the load current or surge exceeds the internal rating.

12. Sources

• TI ULN2003A product page: ti.com
• ST ULN2003 documentation: st.com
• onsemi driver portfolio: onsemi.com