Why is this PNP conducting?

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vlad777

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Hi.

I want to make H bridge and I made a half of it just to test the idea.

**broken link removed**

It's something like this but inputs are at NPN s.

I have the problem that if I test if PNP is open ,it conducts.

Any ideas?

Thanks.
 

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What do you mean the PNP "conducts".

What does "polarize the pnp base" mean.

The circuit you show will always have the PNP off, whether the switch is open or closed. With the switch open, you may measure some voltage at the collector due to the leakage current from the off PNP.
 
conducts - ON (conducts from e to c)
polarize - directly polarize ,in this case, negative voltage to b and positive to e.

I think when switch is on PNP base gets negative voltage in respect to e which turns it ON.

Edit:
Can someone explain to me what bias or biasing (forward bias ) means.
Judgeing from something I read on the net I think it has something to do with this I call polarization.
I can't find anything definitive on Wikipedia.
 
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What's not to like:
 

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Can someone explain to me what bias or biasing (forward bias ) means.
Judgeing from something I read on the net I think it has something to do with this I call polarization.
I can't find anything definitive on Wikipedia.
Forward bias for a diode or bipolar transistor base means to apply voltage in the direction that it conducts. Thus for a diode, it's positive voltage on the anode with respect to the cathode. For a transistor, its positive base-emitter voltage for an NPN, and negative base-emitter voltage for a PNP.

The term "polarization" is not normally used when discussing the operation of transistors. Polarization can refer to the permanent electric field that some materials can achieve.
 
I would like to control it with ttl 5v signal and drive 12V bipolar step motor. (I'll use two hs)

So I just reverse the power and exchange PNPs and NPNs ?

How much current does your stepper draw?

The circuit is not likely suitable for operation at 12V, or at your current level. You would have to carefully simulate it to determine that there are not huge "shoot-through" currents during switching. Also, reconfiguring it so that it can be driven with TTL levels is not trivial.
 
I guess it draws about 500 mA.
(It is 2 inch wide and 1 inch high and it's coil measures 28 Ohm,
so is my guess about right or is it 5V motor?)

I have to provide enough current from TTL signal and reduce the voltage for about 0.8V .

What is this "(switching) shoot-through current" ?
 
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