When I'm to calculate the needed base resistor in order to insert the transistor into saturation (where its VCE is minimal), I'm using only one resistor, connected between the voltage-source and the base (R1 for example).

I saw in several applications that a second resistor is between the base and ground.

I'd like to ask please some questions:

1. About Configuration:
Is it aimed to be connected between base and ground or in parallel to the BE junction? (In case the Emitter isn't grounded).

2. About Purpose:
Why is it used?

3. About Calculation:
What are the guidelines for calculating each of the two resistors?


I'd appreciate if you could treat each question separately for clearance.
Thank you very much.

 

R2 is used when the input to R1 goes high to turn on the transistor then floats. Then R2 turns off the transistor. A low value for R2 turns off the transistor quicker but forms a voltage divider with R1 that reduces the base current when the transistor is turned on.

R2 is not used when the input to R1 goes down to zero volts.

The datasheet for every transistor shows its "Maximum Vce Saturation Voltage" and shows how much base current is needed at certain collector currents. Most little transistors have a base current of 1/10th the collector current to saturate well. This is a "forced beta" and has nothing to do with the hFE of the transistor which is used when it is not saturated.

 

Thanks a lot Audioguro!

When you say that the input to R1 floats, you mean when its on tri-state?

Whats the problem when the input floats?
How having the input floating could still turn on the transistor?

 

You forgot to show the source that feeds R1 so I don't know if it goes down to zero volts or if it floats. If it floats then leakage current in a hot transistor might keep it turned on or will make it turn off very slowly.

Look at the datasheet for a darlington transistor. Many have the base to emitter resistors built-in.

 

When I'm to calculate the needed base resistor in order to insert the transistor into saturation (where its VCE is minimal), I'm using only one resistor, connected between the voltage-source and the base (R1 for example).

As Audioguru stated, when you're forcing beta on small switching transistors, you use a base current of 1/10 the desired collector current.
The BE junction acts very much like a forward biased diode. With very low base current, you may measure a Vbe below 0.6v. As base current increases, Vbe can increase to over a volt. However, if you're driving the base from a uC, the most current you should consider sourcing is just under 20mA (or within the uC's specifications; consider both the datasheets' per pin I/O current and total I/O current); this will usually give a Vbe of around 0.7v

A rough way to approximate the base current limiting resistor is:
Rlimit = (uC's Vcc - 0.7v) / Ic/10
So, if your Vcc is 5v, and you have a load that requires a 100mA sink current:
Rlimit = (5v - 0.7v) / 100mA/10
Rlimit = 4.3 / .01 = 430 Ohms.

I saw in several applications that a second resistor is between the base and ground.

I'd like to ask please some questions:

1. About Configuration:
Is it aimed to be connected between base and ground or in parallel to the BE junction? (In case the Emitter isn't grounded).

Connecting it in parallel with the BE junction would suffice.

2. About Purpose:
Why is it used?

The idea is to lower Vbe to where the transistor goes into cutoff; usually below around 0.5v.

3. About Calculation:
What are the guidelines for calculating each of the two resistors?

The base current limiting resistor calculation is covered above.
For the pull-down resistor, just multiply the result by 5 to 10.
There won't be much current flow through it, as Vbe seldom gets above 0.7v when driven by a uC.

 

If it floats then leakage current in a hot transistor might keep it turned on or will make it turn off very slowly.

Could you please elaborate on this leakage current and how can it keep the transistor turned on?
(I'm only familiar with IS - the reverse saturation current of the base–emitter diode).

Regarding your question, the input is a digital output pin of a Uc that can either be 0V,3V or tristate.

The idea is to lower Vbe to where the transistor goes into cutoff; usually below around 0.5v.

Why would you want to lower it?
I mean, if the input is above 0.7V, then the transistor will be ON, and if the input is below 0.6V, then the transistor will be OFF (in most datasheets, VBE_ON_MIN = 0.6V).
So, i dont understand why would you want R2 there?

 

Could you please elaborate on this leakage current?
(I'm only familiar with IS - the reverse saturation current of the base–emitter diode).

For an example datasheet, go to http://www.national.com
Download the datasheet for a 2N3904.
Look at the plots on the bottom of the 3rd page;
"Base-Emitter Saturation Voltage vs Collector Current" and "Base-Emitter ON Voltage vs Collector Current". They're shown over three different temperature ranges. Note that the Vbe is much lower at 125° than at 25° or -40°.

Regarding your question, the input is a digital output pin of a Uc that can either be 0V,3V or tristate.

If you use the tristate mode, the output will then be floating; and you will need the pull-down resistor to ensure that the transistor goes into cutoff quickly.

Why would you want to lower it?
I mean, if the input is above 0.7V, then the transistor will be ON, and if the input is below 0.6V, then the transistor will be OFF (in most datasheets, VBE_ON_MIN = 0.6V).
So, i dont understand why would you want R2 there?

See the plots I mentioned above. The cutoff threshold for Vbe will vary over temperature. The solution for this is to choose values of resistance that will ensure saturation for the desired Ic, and to ensure positive cutoff in a reasonable amount of time.

[eta]
For more about Ib vs Ic, turn-on and turn-off times, have a look on page 4 of OnSemi's datasheet for the PN2222/PN2222A:
http://www.onsemi.com/pub_link/Collateral/PN2222-D.PDF
Figure 4, "Collector Saturation Region", shows why you want to use the Ic/10 for Ib. The lower the Vce, the lower the power dissipation will be in the transistor.

 

As I recall, the term for this resistor is the "base return resistor". As has already been stated in a number of different ways, the resistor is there to pull charge out of the base when no other means of doing so is present. If there is nothing to pull charge out of the base then the transistor may not turn off. Even if it does turn off it will not turn off as rapidly as it would with the base-return resistor present.

hgmjr

 

Also keeps the base at ground potential, so it is isolated from eratic signals.

I also found it is a good idea to put one at the gate of an SCR.
to keep it from switching on during spuratic voltages.

 

For an example datasheet, go to http://www.national.com
Download the datasheet for a 2N3904.
Look at the plots on the bottom of the 3rd page;
"Base-Emitter Saturation Voltage vs Collector Current" and "Base-Emitter ON Voltage vs Collector Current". They're shown over three different temperature ranges. Note that the Vbe is much lower at 125° than at 25° or -40°.



See the plots I mentioned above. The cutoff threshold for Vbe will vary over temperature. The solution for this is to choose values of resistance that will ensure saturation for the desired Ic, and to ensure positive cutoff in a reasonable amount of time.

Thank you very much for the detailed answer.
I understand from you and from the links you gave that the the cut-off threshhold varies over temperature.

I still dont understand where the base-to-ground resistor fits into all this?

 

As I recall, the term for this resistor is the "base return resistor". As has already been stated in a number of different ways, the resistor is there to pull charge out of the base when no other means of doing so is present. If there is nothing to pull charge out of the base then the transistor may not turn off. Even if it does turn off it will not turn off as rapidly as it would with the base-return resistor present.

hgmjr

Thank you.
Do you mean that the internal BE junction capacitor (Cbe) stores ~0.7V when the BE junction is ON?
And therefore, it keeps providing base current to the BE junction until it discharges to below ~0.6V? (VBE_OFF)

 

Basically that is what can happen. Keep in mind that this is only the case when the drive being delivered to the base of the subject transistor is not being driven to ground. In the case of a drive that is already going to ground, the series current limiting resistor is performing the function of this base return resistor so it is probably not needed.

hgmjr