Basic problem with transistors

marcbarker said:

What is not realised with "legitimate current", is that if you as an engineer 'covered yourself' by setting each IC output driving a load for 24.99 mA, but there are 16 outputs, you'd have an extra 0.4 A supply current! There is a thing called 'maximum package power dissipation', and you've probably exceeded it by this extra power.

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Well, surely this depends at what point in the circuitry of the board the maximum current becomes an issue? I would imagine the 8mA limit on the parallel port bit outputs is set so low to avoid the effect you describe: if the total current leaving *all* of the terminals must be under 8mA, it's not exactly an incredibly useful board. There are 5 8-bit parallel ports, so it would make more sense that the total current through the circuit supplying these 40 terminals should be less than ~320mA, i.e. that each terminal can handle 8mA simultaneously. I assume this is what you were getting at.

It's certainly worth checking that the 8mA is what's allowed to flow through each parallel port bit terminal though.



Anyway, did you understand my explanation of using PA0 and PA1 as complementary outputs? Now that you know what I was trying to achieve, would my schematic achieve this, or would I need to make some adjustments?

Edit:

marcbarker said:

can you edit the previous posting and add a schematic

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Done.

As I mentioned before the maximum permitted supply current is 50mA. This means that you can have up to 10 outputs supplying 5mA or two outputs supplying 25mA

When its output current is 8mA the output voltage of your micro-processor will not be shorted to the 3.3V supply, it might be only 2.8V.

A 412 ohm resistor feeding a 2.0V LED from a supply of 3.3V limits the current to only (3.3V - 2.0V)/412 ohms= 3.2mA, not 8mA. If the output voltage is actually only 2.7V then the current in the 412 ohm resistor and 2V LED is only 1.7mA, not 8mA.

Hero999 said:

As I mentioned before the maximum permitted supply current is 50mA. This means that you can have up to 10 outputs supplying 5mA or two outputs supplying 25mA

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Sorry, I didn't see you mention that. Where did you find this information?

audioguru said:

When its output current is 8mA the output voltage of your micro-processor will not be shorted to the 3.3V supply, it might be only 2.8V.

A 412 ohm resistor feeding a 2.0V LED from a supply of 3.3V limits the current to only (3.3V - 2.0V)/412 ohms= 3.2mA, not 8mA. If the output voltage is actually only 2.7V then the current in the 412 ohm resistor and 2V LED is only 1.7mA, not 8mA.

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Sorry, could you quote the messages which you are responding to in future messages? This thread is becoming so long it's getting difficult to figure out what's being referred to in each message. The mention of an LED makes me assume you were referring to my 'pedestrian crossing' test though. Anyway, I wasn't suggesting using a 412ohm resistor, just trying to ensure that the circuit has a minimum of 412 ohms resistance in total: in this case, the LED would contribute to this, and the value for a resistor required could be calculated in the way you described given the voltage drop across the LED, correct?

However comparing this to my H-Bridge circuit, I believe I read that the transistor should have a resistance of < 1 ohm when fully on, though perhaps that was collector-emitter rather than base-emitter, and the motor resistance fluctuated when tested with a multimeter, but appeared to generally be < 5 ohms (though 200mA flowed through the motor when connected to a 3V supply, suggesting 15ohms, so perhaps the motor is not quite 'ohmic'. Either way these values are more or less negligible, and a resistor close to 412ohm should produce a current close to the maximum allowed by the board, I think. Please correct me if any of what I just said was incorrect.

From the 74HC series datasheet.

audioguru said:

When its output current is 8mA the output voltage of your micro-processor will not be shorted to the 3.3V supply, it might be only 2.8V..

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Please elaborate on this information So that it can be related to the text that appeared earlier (below). Please also quote the source of your micro-processor information if poss.

If it's a Line Driver IC (possibly good for a h-bridge too), the resistance is going to be low, 10 ohms, but if it's a "4000 series" IC running of a small battery, this resistance is something like 1000 ohm. The uC we've been talking about I'll guess is about 100 ohm. When you connect a logic IC output to ground and make it try assert a logic H, a current flows, and the IC starts to get hot. In practice it doesn't blow up, but it can burn off its printed number. But having said that another IC, say a 4000 series I mentioned earlier, can drive an LED directly without bothering with a 'current limiting resistor'.

Hero999 said:

From the 74HC series datasheet.

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74HC series...? Sorry, maybe you misread something. We were talking about the allowed output of a 4000-series RabbitCore Microprocessor prototyping board. Is 74HC series a type of PIC? As I mentioned in a previous post, I've never used PICs before and while I plan to in future, I don't currently own one.

giftiger_wunsch said:

74HC series...? Sorry, maybe you misread something. We were talking about the allowed output of a 4000-series RabbitCore Microprocessor prototyping board. Is 74HC series a type of PIC?

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Most microcontrollers are based on the 74HC logic series but yours might be different, either way we'll never know unless you post the datasheet.

Hero999 said:

Most microcontrollers are based on the 74HC logic series but yours might be different, either way we'll never know unless you post the datasheet.

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http://www.rabbit.com/products/rab4000/rab4000.pdf

You didn't post the datasheet of the micro-processor. Instead you posted its brief sales sheet without the details we need.
Datasheet Archive dot com has never heard of it.
EDIT: I went to RabbitSemi and they don't have a detailed datasheet!

I am sorry. I thought you were driving LEDs but you are actually driving motors from transistors. The 5 ohm motors might actually be 4.5 ohms (subtracting the resistance of the leads of your meter) and draw 667mA from 3V when starting or when stalled.

Then the base current of the transistors should be 66.7mA which your micro-processor cannot supply.

If your supply is 3.3V then the output voltage of the micro-processor will be about 2.8V and the base voltage of the transistor will be about 0.8V. Then your 412 ohm resistor will provide a base current of only (2.8V - 0.8V)/412 ohms= 4.9mA which is too low.

audioguru said:

You didn't post the datasheet of the micro-processor. Instead you posted its brief sales sheet without the details we need.
Datasheet Archive dot com has never heard of it.

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The link on the website which pointed to that PDF document was entitled "Rabbit 4000 Microprocessor Data Sheet". The only other document is the manual, which is 358 pages, or you could try the quick reference poster. PDF versions of both can be found at Rabbit® 4000 Microprocessor | Documentation | Rabbit

audioguru said:

If your supply is 3.3V then the output voltage of the micro-processor will be about 2.8V

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I don't understand where the 2.8V comes from... the output voltage of the microprocessor's parallel ports is 3.3V.

Yes, the output voltage of the output port is 3.3V when it does not provide any current.
But you want it to provide at least 8mA. Then its output will have a voltage loss.
If it has an output resistance of 100 ohms then with a 3.3V supply and an 8 mA load its output will be 2.5V.

I am not going to look through 358 pages of this unknown IC.

audioguru said:

I am not going to look through 358 pages of this unknown IC.

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Indeed not.

It helps if you use the search function.

I did a case sensitive search for mA, see the table on page 301.

Output drive
(TXD+, TXDD+, TXD-, TXDD-) 24mA
All other I/O 8mA
http://www.rabbit.com/documentation/docs/manuals/Rabbit4000/UsersManual/R4000UM.pdf

I was considering doing that but I got too caught up in cursing Rabbit for hiding everything in a 358-page document.


Anyway I assume that means my assumption was correct, each I/O port can have up to 8mA current? Or at least, the average current of all the I/O ports should be a max of 8mA.

You can't do much with only 8mA and a 3.3V supply.
I wonder if it has an output of 8mA with a 5V supply but only 2mA or less with a 3.3V supply?

Since MOS-FETs are voltage-operated, they could be controlled with a current as low as 8mA though, right? Providing I can find some MOS-FETs which are 'fully on' at 3.3V gate voltage.

Also, I believe the IC suggested by blueroomelectronics involves logic-level MOS-FETs, so that's another reason why I need to make time to take a look at the data sheet for that device...

audioguru said:

I wonder if it has an output of 8mA with a 5V supply but only 2mA or less with a 3.3V supply?

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I don't understand what you mean. The specifications seem to state that 8mA is the max allowed current through the parallel port bits, and you said yourself that it doesn't limit the current itself, it simply suggests a maximum current which should be used. Whether the 5V or 3.3V supply is used shouldn't make a difference to the permissible current, surely?

**broken link removed**

**broken link removed**

cyan and magenta colours are referenced to the Controller ground

red and black are referenced to a theorectical ground I have placed at the centre tap of the battery Don't confuse these grounds.

REMEMBER: think currents, not voltages


diodes: BAT85
fets: vgs(th) 1.5 V 'automotive' type

*** Edit: Using data from Table 28.1 in datasheet:

"Vol= 0.4 V @ 8 mA" = equivalent source impedance '50 Ohms' (not 100 as drawn)
"Voh=2.4 V @ 8 mA" = e. s. i. '110 Ohms'

Means it seems to be even better than I first calculated.

It might be time to actually build this and see if it really works!

The upper transistors cannot have an emitter voltage (+2.2V) that is higher than the collector voltage (+1.5V). The emitter will be about 0.4V less than the collector voltage. about +1.0V. Then the motor will have a total of 2.4V.

Where will you find a Mosfet that is completely turned on with a gate to source voltage of only 4.3V (a logic level one) but does not conduct with a gate to source voltage of +1.5V? Maybe if you test hundreds you might find one.

The annotations are in different coloUrs for a reason!! There are two independent circuits, each having it's own theorectical ground which voltage annotations are referenced to. I intentionally chose to reference the motor circuit at the centRE tap of the battery.

1.

The upper transistors cannot have an emitter voltage (+2.2V) that is higher than the collector voltage (+1.5V).

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I don't think you've interpreted the data correctly. The 2 voltages you're referring to are referenced to two entirely separate circuits, and to compare them with oneanother is invalid analysis.

2.

The emitter will be about 0.4V less than the collector voltage. about +1.0V. Then the motor will have a total of 2.4V.

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How do you calculate that figure? Where did 0.4 V come from please?

3.

Where will you find a Mosfet that is completely turned on with a gate to source voltage of only 4.3V (a logic level one)

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What do you mean by 'completely turned on' ? There's more design considerations in operating fets than "completely turning it on"!! Don't you mean "biased into it's ohmic region + a margin"? So, how do you figure it out that 4.3 V is not sufficient?

4.

but does not conduct with a gate to source voltage of +1.5V?

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Where did you get "1.5V" from ? Possibly you confused again the interpretation as you did with 1.

5.

Maybe if you test hundreds you might find one.

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Maybe you didn't notice. This isn't a practical 'to-be-built' circuit, it's a response to a specific question that the OP has been trying to ask since he started the thread, but still no one had answered properly yet!.