Voltage compliance and load resistor range for BJT current source

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Thank you for any help that you can offer.

Did I answer the questions?

Homework Statement



What is the voltage compliance range of the current source?
Between what range of RL values will the circuit function as a current source?

https://scontent-a.xx.fbcdn.net/hphotos-ash3/t1.0-9/1488711_10151945891295919_440459508_n.jpg

VB = 1.6V
VE = 1.0V
IE = 1.0mA
IL = 1.0mA

Homework Equations



VLimit = VCC - ILRE
VCE < V < VLimit

............. ...
Minimum staturation voltage:

VCE (Sat) = 0.1V
............. .....
compliance range is:

VCE < V < VLimit
............. .........
VCE < V < VCC - ILRE
.........
0.1V < V < 10V - 1V
.........
0.1V < V < 9V
.........

Between what range of RL values will the circuit function as a current source?

The range of load resistors can be found by:

iOUT = Vin/R
............
VCE/iE < R < (VCC - ILRE)/iL
...............
0.1V/1.0mA < R < (10V - 1V)/1.0mA
............
100(ohm) < R < 9000(ohm)
Are answers? Are there any errors located?

Thanks again for your help.
 
Last edited:
Can you tell me why you think that for RL<100R this current source won't work?
Also calculate Vce for RL = 9K.
 
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If you assume that transistor is in saturation region for Vce<0.1V. In this case the active region (Vce range) is indeed between 0.1V and 9V. But still you have incorrect resistor range.
 
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Jony130 can you look at what I did and tell me where I went wrong?
 
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  • Handwritten working for a BJT current source: VCE < V < Vlimit, VCE = 0.1 V, Vlimit = VCC - IL RE, VCC = 10 V, RE = 1 kΩ, IL = 1 mA, Vlimit = 10 - 1x10^3 x 1x10^-3 = 9 V, so 0.1 < V < 9 V, R = V/I, Rmin = 0.1/(1x10^-3) = 100 Ω, Rmax = 9/(1x10^-3) = 9000 Ω = 9 kΩ, load range 100 Ω to 9 kΩ.
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Vce voltage is not the voltage that "appears" across RL resistor.
V_RL = Vcc - Vce - Ve. So for Vce = 0.1V --->V_RL = (10V - 0.1V - 1V) = 8.9V.
And if Vce = 9V --->V_RL = ??
 
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I'm lost .

I knew my answers were incorrect.

V(limit) was said to be the maximum before saturation

so the compliance voltage range is okay.

the load resistance range is not

so what do I do? Cause I don't get this now
 
Do you see the difference between Vce voltage and voltage across RL resistor?
 
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Vce is a junction voltage, The voltage across RL is a branch voltage,

But I don't understand this range business.

What does range mean with respect to Vce and RL?
 
Q1. What is the voltage across V_RL if Vce = 0.1V ?
Q2. What is the voltage across V_RL if Vce = 9V ?
 
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See I confused because I wasn't given an RL value,

Ic = Ie only during the active region so I have no currents,

I'm stuck like a duck.
 
You don't need to know RL value. First we need to know V_RL voltage range. Next we can use this range to find RL value. The only think you need is to find V_RL range by using KVL and find V_RL voltage.
 
Vcc-(ILxRL) - Vce = 0

10V-(IL x RL) -0.1 = 0

9.9V -(ILxRL) = 0

9.9V = ILxRL

9.9V/IL = RL

9.9V/1.0mA = RL

9.9 k(ohm) = RL
 
You forget about Ve voltage.
 
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Jony130 said:
You forget about Ve voltage.

Vcc-(ILxRL) - Vce -Ve = 0

10V-(IL x RL) -0.1 - 1V= 0

8.9V -(ILxRL) = 0

8.9V = ILxRL

8.9V/IL = RL

8.9V/1.0mA = RL

8.9 k(ohm) = RL

So is this the lower or the upper limit?
 
Duave said:
Vcc-(ILxRL) - Vce -Ve = 0

10V-(IL x RL) -0.1 - 1V= 0

8.9V -(ILxRL) = 0

8.9V = ILxRL

8.9V/IL = RL

8.9V/1.0mA = RL

8.9 k(ohm) = RL
Finally, some good work.
So is this the lower or the upper limit?
What do you think ?