Possibly stupid question.

when a digital output is in the low state, can it act as a ground?

Yes. When a digital output is low it is connected to ground. See, for instance, how a LED can be connected between two output pins.

However, there a limit to how much current each output can sink (about 20mA or something close). A resistor may be necessary to limit current, so as to avoid damaging the Picaxe.

Not a stupid question at all. One that needs to be answered properly on a regular basis in electronic design.
_

More precisely, the output is connected to ground "through" a low impedance semiconductor junction. It is an "approximate" ground.

If via bipolar transistor there is a voltage drop of about .7V that the connection will remain above actual ground. If via a CMOS device the connection will display a certain resistance based on the characteristics of the device.

There is never a direct connection to ground, and thus there is the need to know the actual characteristics of the connection being made in order to ensure it is being made within the limits of the device being used to approximate ground.

Absolutely, and difficult to know exactly how to pitch the reply.
A 'ground' to switch a little LED is one thing but a ground for RF or an ADC is another thing altogether.
It is very much Horses for Courses.

JW has provided a good description, so I'll provide a picture.

This is from a PIC data sheet and is merely a 'typical' arrangement.
As a slight aside you can also see the clamping diodes.. you may need to know that one day so keep it in mind.

You'll see how the CMOS output drivers are arranged schematically - inside the green circle.
See the things that have a generalised FET symbol and marked "P" and "N"?

Then if you investigate the PIC DC characteristics tables in the PIC Data Sheet you can see the figures for "Output Low Voltage".

This is one reason why it is important to know the base PIC used or be supplied with detailed specs.

If you need detailed nitty-gritty information you'll be best off wrestling with Microchip Data Sheets and any other info on the Microship website.

John West said:

If via bipolar transistor there is a voltage drop of about .7V that the connection will remain above actual ground.

Click to expand...

This is a common misconception. The Vce(sat) of most common transistors is less than the bias voltage required on the base to create that saturation.

Examples:
BC546-548 Vce(sat) = 80 to 200mV @ Ic = 10mA with Vbe = 700mV @ 0.5mA
BC546-548 Vce(sat) = 200 to 600mV @ Ic = 100mA with Vbe = 900mV @ 5mA
2N2222 Vce(sat) = 400mV @ Ic = 150mA with Vbe = 600 to 1200mV @ Ib = 15mA
2N2222 Vce(sat) = 1000mV @ Ic = 500mA with Vbe = 2000mV @ Ib = 50mA

I concur with IWP.

For small signal transistors such as the BC547/548/549 when used for signal purposes, I usually use a value of 0.2 Volts for Vce(sat).

Certainly as the Collector current (and required base current) draw increases then Vce(sat) also increases. For example, with a 3 Amp TIP31 transistor at Ic = 3 Amps and Ib=0.375 Amps, the Vce(sat) can be as high as 1.2 Volts.

Suppose it depends on what you mean by ‘ground’.

A low output is essentially at the same voltage as ground, so any ground voltage references can also be taken from the low output. But that’s only true so long as you treat it as a signal voltage (don’t load it). Once you begin to draw any appreciable current from it (sink any appreciable current through it), the pin voltage will rise, like it’s connected to ground through a resistor.

Most Picaxe outputs are basically symmetrical. They can source or sink around 20mA before becoming too heavily loaded.

Try it for yourself. Set an unconnected pin high and measure it’s voltage. Same as VDD yes? Now put a 220 ohm resistor from the pin to ground. Voltage has dropped a little yes?

Same applies in reverse. Set an unconnected pin low and measure the voltage. Zero yes?. Now connect it to VDD through a 220 ohm resistor. Voltage on the pin has risen a little yes?

westaust55 said:

I concur with IWP.

For small signal transistors such as the BC547/548/549 when used for signal purposes, I usually use a value of 0.2 Volts for Vce(sat).

Certainly as the Collector current (and required base current) draw increases then Vce(sat) also increases. For example, with a 3 Amp TIP31 transistor at Ic = 3 Amps and Ib=0.375 Amps, the Vce(sat) can be as high as 1.2 Volts.

Click to expand...

We're getting off topic here (since PICAXE outputs aren't fitted with internal TIP31s) but you left out an important parameter:

When the TIP31's Ib = 0.375A, Vbe = 1.8v, so the rule still holds.

and . . .

more in line with the PIC datasheets,

The datasheet (eg for 28X1 and 40X1 others are similar) indicate that the max voltage on a pin defined as an output when set to the low state (VOL) is 0.6 Volts when drawing a current of 8.5 mA and Vdd is 4.5 Volts.

Same applies for the 28X2 and 40X2 5V versions and the humble 08M.

....that's the data that matters

HaHa, of course we are going off-topic.
That's one of the qualities of this Forum.

Yes, Vce is much lower than originally said and no it has nothing to do with CMOS drivers and yes, the Vdrop is related to current and yes, it is with CMOS based drivers too for different reasons and yes, aren't we all clever and yes, we all want the last word and yes, some people will argue black is white and yes, this thread was really finished by Post#4

But, boy, that's part of the fun! Keep it going