Supply for sinking current

;)

I want to use an amplifier circuit powered by a +/-15V supply (DC-DC converter IC with one or two Watts) to interface an ADC. I want to use clamping diodes to prevent the ADC input from leaving the voltage range 0 ... 3.3V. For that purpose I need an additional 3.3V rail generated by a supply that is also capable of sinking current. I already read on the internet that I cannot use LDOs (to step down from +15V to +3.3V) because they are usually not capable of sinking current. My simple question: Which supply topology can I use in order to source AND sink current? The datasheets only mention maximum values of current that shall not be exceeded, but not a hint about the direction of current flow. Can someone maybe provide further/additional information about supply topologies (ICs for board supply) and their properties so that I can read a bit about the topic?

Best regards!

EDIT: Most of the answers referred to voltage regulators, such as LDOs. But what about DC-DC converters? Can any of those converters sink current?

• Why not use a zener diode? – Ignacio Vazquez-Abrams Mar 19 '18 at 17:38
• I already thought about that but I did not find one that has a "sharp" turn-on characteristic at 3.3V. For example, the datasheet of this 3V3 protection Zener does not even include its turn-on graph, only the one of the 3.6V version. And the latter already starts drawing some mA at 2V. – Geralt Mar 19 '18 at 17:42

You could use a source/sink regulator such as the LT1118. These types of regulators are made for termination supplies which must source and sink current. You will probably have to put an additional regulator ahead of any termination regulator you use (perhaps a 7805?), because termination regulators are unlikely to support 15V input.

• Unfortunately, the LT1118 only has a 2.85V output, not 3.3V. – Geralt Mar 19 '18 at 18:41
• From the data sheet: The feedback pin is used to program the output voltage of the adjustable S8 part. The output voltage range that can be achieved is 2.1V to 6.5V. – crj11 Mar 19 '18 at 21:05

You can use a regular regulator if you pre-load it with your worst-case sink current.

simulate this circuit – Schematic created using CircuitLab

Figure 1. R2 pre-loads the voltage regulator. Excessive signal will power the circuit safely.

How it works:

• Signal maximum voltage is +15 V.
• With R1 at 1k the worst case current through D1 will be about $\frac {15 - 3.3}{1k} = 12 \ \mathrm {mA}$.
• Lets say the normal load on the 3.3 V supply is 5 mA.
• We will set R2 to draw a fixed 12 - 5 = 7 mA. $R_2 = \frac {3.3}{7m} = 0.5k$ or about 500 Ω. 470 Ω will do.
• Now, if the signal maximum is given then D1 will conduct, the circuit will draw 5 mA as normal, R2 will shunt 7 mA and the regulator current will fall to zero.

I would recommend setting the values so that even on maximum signal excursion the regulator is still supplying a milliamp or two for safety.

Limitations:

At 1k you shouldn't see any significant error on the ADC readings but you need to check a few details such as your sampling rate, etc, to make sure capacitance doesn't low-pass filter the signal.

The "wasted" 7 mA could be used to power the "on" LED.

The circuit may be unacceptable for a battery powered application due to permanent shunt load.

• Thanks for the suggestion! Currently, I have not planned to connect a resistor (R1 in your example) between OpAmp buffer and ADC input. The sampling frequency is around 100 kHz. – Geralt Mar 19 '18 at 18:19
• If you don't connect it then what will limit the current into your regulator? i.e., How much current can your amplifier dump into your ADC protection? – Transistor Mar 19 '18 at 18:22
• The only limit will be the buffer. The OpAmp datasheet states that the voltage drops at its maximum current of around 70 mA. – Geralt Mar 19 '18 at 19:55