[quote=“largon”]The included manual is:
1)
8 digital open collector ouput switches (max 50v/100mA)
The diagram is written 5-30V.
What is the actual maximum voltage?
[color=#00BF00]Is the current 100mA for each of the digital outputs (8 * 100mA = 0.8A)? or 100mA for all 8 digital outputs?[/color]
[/quote]
Velleman uses conservative numbers, as is good engineering practice and to prevent people with limited knowledge in electronics from frying their boards.
For reference here is the ULN2803 datasheet: semicon.toshiba.co.jp/info/d … =datasheet
The maximum output voltage is specified as 50V.
The maximum output current is a little more tricky. The datasheet specifies 500mA per channel. However, the entire IC has a maximum Power dissipation of 1.47W (at 25 degrees Celsius ambient temperature). The created heat in the circuit is not a constant. It is a function of VCE(sat) times current. VCE(sat) of a transistor (the ULN2803 is a transistor array) is the voltage, measured across the collector and emitter, when the transistor is fully ON (saturated). At 200mA, the forward voltage of this thing would be 1.3V (max), which would be 0.26W per channel. All 8 channels driving 200mA each would therefore be 2.08W, which clearly exceeds the maximum 1.47W and will melt the chip. So while you can have a 500mA load on a single channel, you cannot have 8 x 500mA simultaneously. You can’t even have 2 x 500mA at the same time. But you can have 8 x 100mA, because at 100mA VCE(sat) is only 1.1V and that results in a total of 8 x 1.1 x 0.1 = 0.88W. The chip will still get warm because that power must go somewhere.
[quote]2)
2 analogue output
0 to 5V, output resistance 1K5
Hmm, I = 5V/1k5 = 3mA
Does this mean that the load resistance can be any (even 0 ohms)?[/quote]
You got that right. Any load resistance would be in series with R14 or R16, so your total load resistance is always >= 1K5, even if you short circuit the output.
[quote] [color=#00BF00]PWM 0 to 100% open collector max load 100mA/40V
100mA? We know that the current drawn from the USB is only 70mA!
Is PWM also uses an external power supply with CLAMP input?[/color]
The diagram in manual is not clear.[/quote]
The PWM outputs are also open collector, like the digital outputs. There is no “CLAMP input”. An NPN open collector circuit (like the digital and PWM outputs of the K8055) acts like a switch/relay to ground.
The CLAMP connector is not to “supply” any power. It is only for the 8 digital outputs and is used with the clamping diodes inside the ULN2803 for switching inductive loads like a motor or a relay (anything with a coil). This is to protect the transistors inside the array from the back-EMF. When the power to a coil is turned off, the collapsing magnetic field creates a reverse voltage spike, which could destroy the transistor. Those tiny diodes, you see drawn inside the IC symbol on the schematic diagram, prevent that by shorting that reverse spike.
As for the maximum load on the PWM output, the BC337 is actually good for 50V, has a VCE(sat) of 0.7V at 500mA and a maximum of 625mW dissipation.
Hope this helps,
Jan