2
\$\begingroup\$

I am a little confused about power supply markings on ICs. For example, if I have an IC named U1 stating that Vcc should be 5V, can I connect it this way:

schematic

simulate this circuit – Schematic created using CircuitLab

Or this way:

schematic

simulate this circuit

Or this way:

schematic

simulate this circuit

etc.

Are these correct or am I missing a point?

Also, if these are correct, then the problem would be calculating the voltage drop on the parts located between + and Vcc AND GND and -, since we don't know the resistance of the IC right?

\$\endgroup\$
  • \$\begingroup\$ Stick with the first of these. Use a voltage regulator to get 5V from your other power supplies if necessary. \$\endgroup\$ – Brian Drummond Dec 25 '15 at 11:00
  • \$\begingroup\$ Usually, the voltage regulating part is placed in the Vcc line instead of GND. \$\endgroup\$ – JimmyB Dec 25 '15 at 14:29
  • \$\begingroup\$ Don't forget the decoupling capacitors! \$\endgroup\$ – pjc50 Dec 25 '15 at 19:04
2
\$\begingroup\$

You can indeed use a resistor to cause a voltage drop on the integrated circuit power line(s). Fact it's never done that way (at least not without clamping diodes, such as a reverse-biased Zener diode) because the resistor value depends on the current consumed by the IC, which varies and can vary a lot. Instead circuits are powered with a low impedance power supply that maintains its output voltage level regardless of the consumed current — let's forget about current limitation for a second for simplicity's sake.

So if you need, say, 5V to power your IC, then you'd use a 5V voltage regulator, which keeps its output to a steady 5V as long as the current it feeds is kept in the supported component's tolerance. For instance an LM7805 in a TO-220 case typically can bear 1A output current. In general you'd also power the whole circuitry (or board) with 5 volts, not just the IC.

The 7805 is a linear regulator because it simply causes a voltage drop using a linear behaviour. As a major drawback, it produces an increasing amount of heat (that must be dissipated with a heat sink) when the input voltage increases. That's one reason why it bears a maximum input voltage. That maximum input voltage is about 35V for a 7805.

\$\endgroup\$
  • \$\begingroup\$ It was (and may still be) common to add a small resistor on the Vdd pin of small CMOS devices. A fault on the power input is not always detectable as the devices could be powered from the signal inputs via the ESD diodes. \$\endgroup\$ – Peter Smith Dec 25 '15 at 11:03
  • \$\begingroup\$ You're right. There are also circuits that are powered directly from mains with a series resistor (e.g. dimmers) but they use at least clamping diodes. It's just that powering an IC using only a resistor and no protection normally is, well, never done. I'm updating my answer accordingly. \$\endgroup\$ – user59864 Dec 25 '15 at 11:22
  • \$\begingroup\$ @Nasha So in a nutshell, all 3 of these schematics are correct, had the voltage drop on the resistors not vary with current (which would vary but just for the sake of argument) \$\endgroup\$ – Utku Dec 25 '15 at 11:22
  • \$\begingroup\$ @Utku Yes, they're electrically correct. On a practical point of view they are just difficult to implement as is. As a means of protection, you'll instead use a clamping diode like a Zener and make the IC current negligible compared with the Zener reverse current to minimize voltage fluctuations due to variations in current consumed by the IC. \$\endgroup\$ – user59864 Dec 25 '15 at 11:27
4
\$\begingroup\$

IMHO the question, and hence answers, have a significant potential to be misleading.

It is necessary to consider how to power an IC (via Vcc and GND pins). However, it is not sufficient to ensure an entire circuit will work.

As well as power and ground, the vast majority of (Vcc/GND powered) ICs also have input and output signals. These signals must be considered too as part of a practical solution.

Typically, a signal will need either a shared voltage reference, and/or a common current path. Putting resistors into ground and power could significantly compromise the signals integrity.

For example, consider an output signal from the IC. The 'load' being driven may be sensitive to the voltage driving it, or the current flow. If its ground is tied directly to the negative side of the power supply it may input the (dangerously) wrong input voltage. If its 'ground' must be tied to the same ground as the IC, then its downstream inputs may also need to be tied to the same ground. However, this might be impossible, because it is separate equipment.

Further, that load may need current, disturbing the GND reference voltage. For multiple IC outputs this may be unacceptable.

The inputs to the IC will likely have requirements to, which may also be difficult to satisfy without common ground and voltage references.

So, IMHO, options two and three may be adequate to power the IC, however they are poor places to start when considering the whole circuit's signal requirements.

It may even be that option one is not appropriate, but the analysis and solution design will likely be easier starting from such a simple starting point.

Put another way; if I asked someone to design a way to power a circuit, and option two or three were presented when one was adequate, I would ask for a good analysis for all of the inputs and outputs to demonstrate the alternative was at least as good, and understood by us all. Similarly if they presented option two or three, even if it were adequate, without a good analysis covering inputs and outputs, I would ask for that analysis.

IMHO option two and three imply sufficient extra complexity that we'd need to have a documented analysis to identify and explain the assumptions and benefits of those approaches. If there are no clear benefits, we'd revert to option one. The common understanding of option one would help us develop, test, interface and evolve the circuit. If we departed from option one, we'd be clear why.

IMHO, it is unlikely that we would use option two or three to overcome practical problems caused by using option one.

\$\endgroup\$

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.