10
\$\begingroup\$

I have been reading that NPN transistors are sinking and PNP are sourcing devices. I do not really understand this concept. It says current source device connects load to Vcc and current sinking device connects to ground (low voltage).

So does connecting a load at emitter of NPN transistor make it sourcing?

\$\endgroup\$
10
\$\begingroup\$

JYelton is right, and probably this is what whoever said "NPN transistors are sinking and PNP are sourcing devices" had in mind. But, that's not the only way to use a transistor. For example:

schematic

simulate this circuit – Schematic created using CircuitLab

This configuration is called common collector or emitter follower. Now the NPN is sourcing, and the PNP is sinking.

So, sourcing or sinking doesn't really have much to do with the type of transistor, but rather what it's doing. Is it pushing current from the positive supply rail (sourcing), or is it sucking current from ground (sinking)?

\$\endgroup\$
9
\$\begingroup\$

In a very simple form, relative to VCC, think of the transistor as either coming before or after the device.

If the transistor is connected between VCC and the device, it is sourcing current.

If the transistor is connected between the device and ground, it is sinking current.

Sink vs Source

(Image from CircuitsToday.com)

Some articles that describe things in more detail:

\$\endgroup\$
  • \$\begingroup\$ I Have a multiplexed 7 Segment led(common anode) anI want to drive it using transistors. I am currently using pnp transistors with base at control voltage ,emitter connected to Vcc and load(7segment) at collector.Is it possible to use npn transistors and still drive the 7-segement. \$\endgroup\$ – Abhishek Jul 2 '13 at 17:04
1
\$\begingroup\$

Think of it like this. SINKING = provides a path to ground. SOURCING = Provides a path to V+

Be careful because most electrical manuals etc refer to conventional current flow.(+ to -) We (electronics) tend to refer to actual electron flow (- to +)

Yes you could configure them the way you described, but its standard in industrial electronics to do NPN = Sinking and PNP = Sourcing..

\$\endgroup\$
0
\$\begingroup\$

When considering an output from an integrated circuit, source vs sink is very simply a matter of whether the current comes out of the pin (source) or goes into it (sinks). You often find NPN devices connected emitter to ground, and indeed configured to sink current when producing a logic low. Similarly PNP devices are often found with their emitters to the positive rail, and to source current to produce a logic high. But generally outputs don't have to be BJTs, and even BJTs don't strictly have to be used the way I just described.

So, bottom line, I would say yes. If you connect an NPN collector to the positive rail and expect to run your output from your emitter, that transistor will source current.

\$\endgroup\$
0
\$\begingroup\$

I second the final part of the Phil Frost's answer.

"Sourcing/sinking" is a property of an electrical source (power supply) - it sources a current by its positive terminal and, at the same time, sinks a current by its negative terminal... the source is both sourcing and sinking. Thus looking at the source terminals, we see that a current exits its positive terminal and a current enters its negative terminal.

When connecting some elements (transistors) to the source terminals, the currents flow through them and we see that a current exits the element connected to (after) the positive terminal and a current enters the element connected to (before) the negative terminal. Then we assign the sourcing/sinking attribute to these elements... and say that the first element sources, and the second - sinks current.

Simply speaking, if the current exits a device terminal (output or input), it is sourcing; if it enters the device terminal, it is sinking. It seems strange but some inputs can source current (e.g., TTL inputs).

\$\endgroup\$
0
\$\begingroup\$

This perspective from a boots on the ground non engineer electrical maintenance man who replaces pnp/npn proximity sensors a lot, in layman terms;

Sourcing switches the positive [high] side. Think of the lights in your house. 120 V goes through switch to energize bulb.

Sinking switches the negative [low] side. 120 V goes straight to the bulb and you put the switch on the neutral leg.

\$\endgroup\$
-2
\$\begingroup\$

I know this is an old post, but you can think of sourcing as how much a component can provide. For example, an arbitrary op amp might source 50 mA out of the output, and if you put small feedback resistors on it, the op amp might not be able to source enough current for stable feedback.

Sinking is the opposite. How much current can a component take? For example, some n channel mosfet operating at some Vgs that lets the drain current be 50 mA, with the source connected to ground. You can only sink 50 mA through the fet. If you have more current, it will need another place to sink.

Yes, in normal IC’s, P devices usually source current while N devices sink current (the current flows down from VCC to VEE, or VDD to VSS). This is also convenient because N devices conduct more (they can sink all of what a P device sources) which can establish a direct path to ground aka virtual ground with correct biasing for analog devices.

\$\endgroup\$
  • 1
    \$\begingroup\$ A vague answer at best. Confusing details of how mosfets supposedly operate or source/sink current. As long as polarity and bias are done correctly, a P-channel can source or sink current just as an N-channel can. OP was referring to BJT transistors. \$\endgroup\$ – Sparky256 Jan 19 '18 at 4:38
  • \$\begingroup\$ @Sparky256 The other users already covered how you can sink or source either way. Feel free to correct it. \$\endgroup\$ – Fillups Jan 19 '18 at 5:21
  • 1
    \$\begingroup\$ My comment was to you so you can write a clean and sensible answer. That way you can get up votes instead of down voted. Regardless of the other answers you should stay within the question(s) the OP asked. \$\endgroup\$ – Sparky256 Jan 19 '18 at 6:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.