Why before base (of bjt) a resistor goes to ground?

[alphadog]

I was adviced by an engineer that in addition to a resistor one of its legs connected to Vsource and second to Base, to connect another resistor from Base to ground.

1. Why is it recommended?

2. Should the resistor be very large so the Vsource (MCU GPIO pin) wont source large current?

Thank you.

 

[Nigel Goodwin]

It's to help ensure the transistor turns off, values depend on the exact circumstances.

 

[alphadog]

How exectly does it ensure that the transistor is in cut-off?
The present Base resistor is 300ohm (The base-Vsource resistor).
Thanks.

 

[Nigel Goodwin]

In two ways, first it discharges the base/emitter capacitance, and it forms a potential divider with the series resistor - not all outputs feeding it may go low enough to switch it off, by adding a potential divider this ensures it will.

 

[alphadog]

In two ways, first it discharges the base/emitter capacitance, and it forms a potential divider with the series resistor - not all outputs feeding it may go low enough to switch it off, by adding a potential divider this ensures it will.

Thanks!
I very understand the C_BE discharging issue.
I dont quite understand the reisitve divider issue.
In cut-off, the BE resistance is very large, so connecting in paralel to BE (E goes to ground) another resistor, would only decrease the total resistance from Base to ground.
So therefore less voltage will drop on base?
I think i probably have a mistake on that, so i want to figure this issue out.

 

[Nigel Goodwin]

300 ohm resistor to base ONLY - logic feed to resistor is 4V when HIGH, and 1V when LOW - so transistor is always ON.

300 ohm resistor to base, 300 ohm resistor base to emitter, input voltage is divided by two, so transistor turns OFF when LOW is supplied.

 

[alphadog]

But the resistor-to-base and base-to-emitter resistors are not connected in series, so how can you say that when both used then base voltage is divided by two?

And what will happen when source is HIGH?
Aslo here the base voltage is divided by two?

 

[Nigel Goodwin]

But the resistor-to-base and base-to-emitter resistors are not connected in series, so how can you say that when both used then base voltage is divided by two?

Of course they are - both are connected together at the base.

And what will happen when source is HIGH?
Aslo here the base voltage is divided by two?

Yes, that's divided by two 'sort of' - but not really, because the Vbe of the transistor prevents more than 0.7V across the lower resistor. In reality it just bypasses some of the base current.

 

[alphadog]

Of course they are - both are connected together at the base.



Yes, that's divided by two 'sort of' - but not really, because the Vbe of the transistor prevents more than 0.7V across the lower resistor. In reality it just bypasses some of the base current.

How come there are connected in series?

the current that flows through base resistor, splits to two:
1. BE current
2. Base-GND current.

 

[Roff]

Below is a simplified view of when Rbe is required.

The fact that the base is connected to the junction of the two resistors does not mean the resistors are not in series. They are.

 

[alphadog]

In series = same current goes through them.
When HIGH and when LOW, IB is still not zero, so how can you say that both when HIGH and when LOW, they are connected in series?
What about IB?

By the way, adding this resistor means that when HIGH, the VBE will take its maximum value thats written on datasheet - 1.3V ? (becaues VCC/2 is 1.65, so VBE will aspire to be 1.65V?).

 

[Roff]

In series = same current goes through them.
When HIGH and when LOW, IB is still not zero, so how can you say that both when HIGH and when LOW, they are connected in series?
What about IB?

What about when Vbe<0.5V? Then, Ib is essentially zero. The resistors then just act as a voltage divider.

By the way, adding this resistor means that when HIGH, the VBE will take its maximum value thats written on datasheet - 1.3V ? (becaues VCC/2 is 1.65, so VBE will aspire to be 1.65V?).

What?????
No. And where did Vcc/2=1.65 come from?
EDIT: What datasheet are you referring to? Post a link, please.

 

[alphadog]

What about when Vbe<0.5V? Then, Ib is essentially zero. The resistors then just act as a voltage divider.

Thanks.
So when input is HIGH, therefore Vbe > 0.7, they are not connected in series, right?
And at this state (input HIGH) the voltage on the 'base-to-GND' resistor is forced to be the around 0.7V? (depend on the transistor)

What?????
No. And where did Vcc/2=1.65 come from?
EDIT: What datasheet are you referring to? Post a link, please.

Sorry for that, i meant to the input which its HIGH is 3.3V.
By the way, What set the V_Forward of B-E?
lets say that Vbe_on_max is 1V (accordint to datasheet of https://www.fairchildsemi.com/ds/PN/PN2222A.pdf ), then what happens if i apply 2V on B-E?

 

[ccurtis]

I was adviced by an engineer that in addition to a resistor one of its legs connected to Vsource and second to Base, to connect another resistor from Base to ground.

1. Why is it recommended?

2. Should the resistor be very large so the Vsource (MCU GPIO pin) wont source large current?

Thank you.

The need for a base resistor to ground (particularly a high valued resistor) might be considered if the source is disconnected at times, in which case the transistor input is floating and the gain of the transitor may result in some small amount of collector current to flow randomly and/or sporadically due to noise pick-up. Since the base is connected to an "MCU GPIO pin" it may at times be configured as an input pin rather than an output pin, in which case, the transistor input may be floating.

Otherwise, generally, a voltage divider is used for biasing of the transistor as in an amplifier, or as stated earlier, for turn-on/off level control as needed.