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Transistor as IC input

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patroclus

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I'm studing transistor right now, but I came out with some practical questions. Some of you may know the JDM pic programmer.

https://www.jdm.homepage.dk/newpics.htm

My question are about Q2. Base is conected to 0V across R1, and colector is attached to a RS-232 signal (DTR, aprox +/- 12V) across R2. Chip is fed at 0V and grounded at -5V.

The theory says : when DTR is high (+12V), the transistor works as an emitter follower, with emitter voltage at Vb- 0,7... First problem here. Why? What is really connected to the emitter?? I drawed a diagram cricuit attached. If it is fine, Ve should be 0V, and Vb shoud be 0,7 (common emitter, base biased), and it is not an emiter follower... What is wrong in my diagram??

Many thanks
(also thanks to Nigel, who's been helping me with this so far)
 

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Nigel Goodwin

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patroclus said:
I drawed a diagram cricuit attached. If it is fine, Ve should be 0V, and Vb shoud be 0,7 (common emitter, base biased), and it is not an emiter follower... What is wrong in my diagram??

I can't see anything wrong, it's a straight forward common emitter switch circuit. Ve will be 0V, as it connects to the 0V line, likewise Vb will be roughly 0.7V, it can't be anything else - the collector voltage will be fairly low, depending on the current gain of the transistor.


An emitter follower by the way is a common collector configuration.
 

patroclus

New Member
That's exactly what I think.. but the transistor in the JDM circuit doesn't work like that. When collector current is high, +12V, Q2 works as an emiter follower, that is, common collector. I really CAN'T see this..
this is how it works in the circuit, as the author says, and is how a high byte is sent to the serial data output...
 

Nigel Goodwin

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Most Helpful Member
patroclus said:
That's exactly what I think.. but the transistor in the JDM circuit doesn't work like that. When collector current is high, +12V, Q2 works as an emiter follower, that is, common collector. I really CAN'T see this..

That's the problem with the JDM programmer, it works in very obscure ways (as a result of minimising the component count). You don't need to know, as long as it works don't worry - and you shouldn't really be designing something along those lines anyway.

this is how it works in the circuit, as the author says, and is how a high byte is sent to the serial data output...

Do you mean reading or writing to the PIC?. If it's reading the PIC, it works in common base - if it's writing to the chip it doesn't 'really' work in any mode, but it's essentially common collector.

Here are the three basic transistor configurations, notice that they are exactly the same - except for how you connect to them.
 

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patroclus

New Member
Yes I understand that, thanks :)
But, what I don't understand is why when writing to the PIC it is in common colector, as emitter has no resistor, and base does have one. Vb should be Ve + 0,7V, depending on Ve. and it seems more a common emiter.

But what really happends is all the way round, Ve depends on Vb as any emitter follower. But, where's the emitter resistor?
 

Nigel Goodwin

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Most Helpful Member
patroclus said:
Yes I understand that, thanks :)
But, what I don't understand is why when writing to the PIC it is in common colector, as emitter has no resistor, and base does have one. Vb should be Ve + 0,7V, depending on Ve. and it seems more a common emiter.

As I mentioned before it's not 'really' working in any configuration, but when DTR is high the transistor switches on, the base will be 0.7V below the base, and the collector will be slightly above that - R2 1.5K provides the current for that. When DTR goes low (-12V) Q2 stops working as a transistor at all, as the collector is 12V negative of the base the base/collector junction will act as a forward biased diode, so the voltage at the base will be that set by the ratio of the potential divider R1 and R2, this will allow the pin of the PIC to go low, with the base/emitter junction acting as a reverse biased diode to prevent it going negative.

But what really happends is all the way round, Ve depends on Vb as any emitter follower. But, where's the emitter resistor?

The input resistance of the PIC itself.
 

patroclus

New Member
Ah! It is that. the input resistance of the PIC was not included in my diagram. So the emitter really is conected to a resistor (the input resistance of
PIC), isn't it? This way yes, Ve will be Vb - 0,7V. But appart from the resistance, it is grounded as in my diagram, isn't it?
Anyway, as the base has a 10k resistor, there will be a voltage drop across it... input resistance of PIC should be much higher for it to still be a high pulse.

Damn, this programmer is really tricky.. but good to learn new things ;)
About that "inverted mode", current flows from base to colector, but what happends to the emitter current, flows from emiter to base? or viceversa?
 

Nigel Goodwin

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patroclus said:
Damn, this programmer is really tricky.. but good to learn new things ;)
About that "inverted mode", current flows from base to colector, but what happends to the emitter current, flows from emiter to base? or viceversa?

You're thinking too complicated!. The transistor is NPN, so it's essentially two diodes, NP and PN, when DTR is low the emitter/base diode is reverse biased and turned off.
 

patroclus

New Member
Ups, you're right.. Fine, all clear now. It is that I'm studying transistors now and this kind of working is not what I've been hearing about... thanks ;)
 

Nigel Goodwin

Super Moderator
Most Helpful Member
patroclus said:
Ups, you're right.. Fine, all clear now. It is that I'm studying transistors now and this kind of working is not what I've been hearing about... thanks ;)

It's a totally obscure way of working, it's not a good circuit to study as it's too unusual.
 

patroclus

New Member
Ok, I see :)

JDM appart, I just would like to know how can I simulate a transistor circuit as an Integrated Circuit input. Imagine any circuit in which a transistors is atached to a IC input. Imagine I use a program like Electronic Workbench simulator. How should I represent the IC input? as a resistor?
( this was indeed, my original question ;) )
 

Nigel Goodwin

Super Moderator
Most Helpful Member
patroclus said:
Ok, I see :)

JDM appart, I just would like to know how can I simulate a transistor circuit as an Integrated Circuit input. Imagine any circuit in which a transistors is atached to a IC input. Imagine I use a program like Electronic Workbench simulator. How should I represent the IC input? as a resistor?
( this was indeed, my original question ;) )

The input circuit of individual chips will vary considerably, if you wanted to do it with any degree of accuracy you would need to simulate the internal circuitry of the chip.

Personally I'm not in favour of simulating designs, from what I've seen in these forums it's often necessary to modify designs to make the simulator work - this seems to make such simulations totally useless!.

Obviously any simulation is only an approximation of what will (or may) actually happen, and is highly dependent on however wrote the original simulator software.

If you've been trying to simulate the JDM programmer, that could explain a lot :lol:
 

patroclus

New Member
Yes, jajaj :lol:
At least I made the JDM as my first PCB project and it works great ;)

By the way, I just got Electronic workbench simulator, as I said. You think it is useless to use software like this? Do you recommend anything?
 

Nigel Goodwin

Super Moderator
Most Helpful Member
patroclus said:
By the way, I just got Electronic workbench simulator, as I said. You think it is useless to use software like this? Do you recommend anything?

I don't recommend anything, it's easier (and far more accurate) to simply build the design. Many people happily use such software, and no doubt will suggest the one they like - but I don't see as it's accurate enough, if you look back through these forums you will see lots of advice on how to modify your circuits to make them simulate correctly - my point of view is that this makes the simulations useless.

If a simulator can't take any circuit and simulate it perfectly, without any changes, it's just a waste of space.
 

mozikluv

New Member
simulations

:D hi,

have always believe that simulation will only give you ideas of how the circuit will function, likewise with theories, once you put it down in actual circuit function the result most of the time will not be the same although the circuit is working. that's becoz of parts tolerances. :D
 
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