Deep cur-off

I have a BC337 transistor

I followed the formulas to get the on-off mode but I do not see 0volts when I apply 5 volts in the base, how it cab be possible to get a deep cut-off.

IB>IC/HFE
HFE=100
IC=14.15mA

I double the result
IB=2*Ic/HFE =.3mA

Vin=5votls
then
Rx=Vin/IB =16.666K
I am using a 15 Kohms resistor

with this value I have a 30mV at the colector pin

Thanks in advanced

You can Never get Zero Volts. I'm Surprize you got down to 30mv.

Collector to Emitter voltage is typically 0.6 volts.

have a resistor pull the tiny extra voltage to ground maybe?

sardineta said:

I have a BC337 transistor

I followed the formulas to get the on-off mode but I do not see 0volts when I apply 5 volts in the base, how it cab be possible to get a deep cut-off.

IB>IC/HFE
HFE=100
IC=14.15mA

I double the result
IB=2*Ic/HFE =.3mA

Vin=5votls
then
Rx=Vin/IB =16.666K
I am using a 15 Kohms resistor

with this value I have a 30mV at the colector pin

Thanks in advanced

Click to expand...

with 5V at the base your transistor is fully on..

Ok, yes when I apply 5 volts at the base, through the resistor of 15k it is fully on, so the relay must work fine?, I just tried with a resistive load, because I haven't fount the deivce I must use. it is from FINDERNET series 34 at 12VC

about the 0.6 volts, does i have to ensure this value?

I tried with another resistors values: 3k 5k and the voltage values at collector pin dropped at 11mV. If I have to ensure .6V at the collector pin then I have to increase the resistor, isn't it?

chemelec said:

You can Never get Zero Volts. I'm Surprize you got down to 30mv.

Collector to Emitter voltage is typically 0.6 volts.

Click to expand...

Not true for almost any bipolar transistor. see the curves for BC337 below.

According to the diagram, if I require a current of 10mA on Ic

Ic=10mA (needed)

Then, I need a base current of 100uA, it means that the load will work?

what does saturation region mean? I think it is when the NPN let current flow. this is the on state?

I do not understant figure 5
I am confused?

i agree sardineta .. the Y axis on fig 5 is not labeled .. i assume it is Vce..

Sardineta, saturation is when the transistor is fully ON. In saturation, the application of more base current will not result in appreciably more collector current, because the voltage from collector to emitter is nearly zero.
Fig.4 says that, with 10ma of collector current, you might not reach saturation with only 100ua of base current. Standard engineering practice is to set Ic/Ib=10 if you want to guarantee saturation, although, with most transistors, you can use a larger ratio and still be OK I personally wouldn't go higher than Ic/Ib=20. Beta (Hfe) specifications on most transistors are misleading, because the spec is usually at Vce=5v or 10v, but beta is much lower near saturation.

Willi, the y axis of Fig.5 is labeled - it's volts. There are three different curves on that graph. The lower one is collector to emitter voltage when Ic/Ib=10 (so labeled). The upper two are also labeled. Think about it for awhile.

Fig.4 says that, with 10ma of collector current, you might not reach saturation with only 100ua of base current. Standard engineering practice is to set Ic/Ib=10 if you want to guarantee saturation,

Click to expand...

Ok, I think the consideration you use I will use, but even if I do this, I haven't see the 0.6 volts at Vce

when base is 0volts Vc is 12 volts, the voltage I apply to the load
when base is 5 volts Vc is almos 0 as you say, less than 50mV depending on the resistor on the base.

when off
Rb=7.5K
Vin at base 0 volts

Load: 848 ohms at 14.15mA

Vce = 16.4mV

when on
Vce=Vsupply of the load (12volts)


I think all is ok

Sardineta,
Indeed, all is OK. 0.6v is not a goal, and, as I said, is not a normal value for Vce(sat). Vce(sat) for low currents (like your application) is generally below 100mv.

Thank you! :lol:

plot said:

have a resistor pull the tiny extra voltage to ground maybe?

Click to expand...

A resistor will not help. It will only bypass a small amount of current away from the transistor when on, and conduct a lot of current when the transistor is off (thus leaving the circuit half on).

If the vce you have is too high, you can certainly replace it with a N-channel MOSFET. When vgs is high enough, it acts only as a very low ohms resistor (often down into the milliohms depending on the part). It draws no current off the 5v supply either.

Realistically, you show a 12V relay coil being driven with a 12V supply. That's unlikely to be affected by a small voltage drop and your setup sounds fine.