how to bias a transistor when base and collector have separate voltage sources?

Thread Starter

tpny

Joined May 6, 2012
216
My base is supplied by a 15V source, my collector is supplied by 5V source. Will this turn on the npn? Do I have to do something between the base and collector to prevent diode breakdown or something? I forgot what's going on between between the b and c junction..

Rich (BB code):
                       5v
                        | 
                   resistor
                        |
                        |
                       /
15v ---/\/\/\------|/ npn
                     |\
                       \
                       |
                       |
                     gnd
 
Last edited:

ScottWang

Joined Aug 23, 2012
7,080
It's all about the current, you could thinking about this way:

C and B used the same +5V:
+5V → Rc(1K) → C of Bjt → E(GND)
+5V → Rb(4.7K) → B of Bjt → E(GND)

C used +5V, and B used +15V:
+5V → Rc(1K) → C of Bjt → E(GND)
+15V → Rb(15K) → B of Bjt → E(GND)

Above just for explanation, you have to change the Rb to match your need.
 

Thread Starter

tpny

Joined May 6, 2012
216
OK. Thanks! I see b terminal is at 0.6 even though b is getting a 15V supply (resistor dropped 14.4 of it).. What if you had this situation and b reads 10V for example. Do I need to protect the bc junction somehow?

Rich (BB code):
                      5v
                        | 
                        |
                        |
                       /
15v ---/\/\/\------|/ npn
                     |\
                       \
                       |
                   resistor
                       |
                     gnd
 

ScottWang

Joined Aug 23, 2012
7,080
If in that situation, the whole state will as this:

0,15V → Rb → Vb(0,10V) → Ve(0,9.4V) → Vout 0V,9.4V
If the situation as above, then the C of Bjt is useless, so also the transistor is useless.
If the Rb don't use a resistor too small, then bc junction is OK.
If really like that, then you can use the diode to replace the job of the transistor.
I don't know what kind of the situation will use that way.
 

Ron H

Joined Apr 14, 2005
7,014
OK. Thanks! I see b terminal is at 0.6 even though b is getting a 15V supply (resistor dropped 14.4 of it).. What if you had this situation and b reads 10V for example. Do I need to protect the bc junction somehow?

Rich (BB code):
                      5v
                        | 
                        |
                        |
                       /
15v ---/\/\/\------|/ npn
                     |\
                       \
                       |
                   resistor
                       |
                     gnd
This scheme is useful if you want to saturate the NPN in your circuit. For example, If Re=5k and Rb=50k, then Vce will generally be less than 100mV, since the transistor will be saturated.
You do not have to protect the CB junction. It will be forward biased, just as is in the circuit in post #1.
 

ramancini8

Joined Jul 18, 2012
473
The current flow through the emitter resistor establishes the base voltage. If the base resistor is small the emitter current multiplied by the emitter resistor establishes the base voltage Vb = IeRe + Vbe when Rb is small. Then Vb rises until the base clamps at Vb = Vce + 5V. If Rc = 0 the BC junction will eventually burn out.
When Rb is large it limits Ie (unless beta is large) and the equation is Vb = 15 -IbRb. What are you trying to do?
 

Ron H

Joined Apr 14, 2005
7,014
The current flow through the emitter resistor establishes the base voltage. If the base resistor is small the emitter current multiplied by the emitter resistor establishes the base voltage Vb = IeRe + Vbe when Rb is small. Then Vb rises until the base clamps at Vb = Vce + 5V. If Rc = 0 the BC junction will eventually burn out.
No, the current through the base resistor establishes the base voltage. Vb is almost independent of Re as long as the transistor is saturated (Ic/Ib<<beta).

The base clamps at
Vbc+5V≈5.7V.
Ve=+5V-Vce(sat)
 
Last edited:

Thread Starter

tpny

Joined May 6, 2012
216
When I did this in a real circuit (Rb=150), Vb=19V, Ve=18.3V, Vc=18.3V.
Rich (BB code):
                      5v
                       | 
                     560
                       |
                      /
24v ---150------|/ npn
                    |\
                      \
                       |
                     5k
                       |
                     gnd


When I do this (Rb=10k), Vb=6V, Ve=5.3V, Vc=5.3V
Rich (BB code):
                      5v
                       | 
                     560
                       |
                      /
24v ---10k------|/ npn
                    |\
                      \
                       |
                     5k
                       |
                     gnd


I guess current is flowing from b to c as well when vbb is higher than vcc. Does this also happen when vbb = vcc?

I'm playing around with different configurations. But essentially, I'm trying to use the transistor as a switch where control voltage is higher than load voltage.
 

Ron H

Joined Apr 14, 2005
7,014
I guess current is flowing from b to c as well when vbb is higher than vcc. Does this also happen when vbb = vcc?
It can if you have emitter AND collector resistors, depending on the resistor values. If you don't have a collector resistor, then no.

I'm playing around with different configurations. But essentially, I'm trying to use the transistor as a switch where control voltage is higher than load voltage.
That was what I was describing in post #5. Choose Rb such that Ic/Ib≈10. This is called "forced beta", and Vce(sat) is specified with a forced beta of 10 in the datasheets of almost all transistors.
 

Thread Starter

tpny

Joined May 6, 2012
216
So if i really want the transistor to saturate as a switch and don't want to do the math, can I just match the Ib and Ic at unity like below?

Rich (BB code):
                      5v
                       | 
                     5k
                       |
                      /
5v ----5k--------|/ npn
                    |\
                      \
                       |
                       |
                       |
                     gnd
 

Ron H

Joined Apr 14, 2005
7,014
So if i really want the transistor to saturate as a switch and don't want to do the math, can I just match the Ib and Ic at unity like below?

Rich (BB code):
                      5v
                       | 
                     5k
                       |
                      /
5v ----5k--------|/ npn
                    |\
                      \
                       |
                       |
                       |
                     gnd
That is a huge waste of current. Set Ic/Ib≈10. The math is not rocket science.
In your example above, Rb≈50k.
 
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