/sci/ - Science & Math

[Anonymous No.5559876]

Hey /sci/

I am having trouble understanding how to analyze a Widlar current source, and my professor is out of town for family issues. I grabbed a circuit diagram from Wikipedia, but could you guys maybe explain how one would analyze a circuit like this? perhaps hearing from a different source will help it stick.

Thanks!

[Anonymous No.5560028]

1) Vce = Vbe for Q1
2) Ib1 << IR1 and Ib2 << Io
3) Vbe = Vcc-R1*Ir1
4) Vbe = Io*R2 => Io = Vbe/R2
6) Io = Vcc/R2 -R1/R2*Ir1 Vbe muste be between .6 and .7 for a good performance.

[Anonymous No.5560083]

I2=(Ut/R2)*ln(I1/I2) where Ut=k*T/q

ugly iteration

[neckbeard_the_pirate No.5560099]

>>5559876
Imagine if Q1 were removed. Due to the gain of Q2, a voltage drop will occur across R2. If Vcc or the temperature is changed, the gain of Q2 will change and the Vdrop on R2 will change. Now imagine you add Q1 back in. Q1 is matched to Q2. Q1 takes half of the current from the base of Q2. As the Vcc or temperature changes, the gain on Q1 will counter act the changes in Q2.

[Anonymous No.5560116]

>>5560099
>>5560083
>>5560028
So I'm thinking you'd probably need to do a KVL down the left side of the circuit if you wanted to find, say, the current <span class="math">I_R1[/spoiler], and since Q1 is diode-connected, it would mean <span class="math">V_CE[/spoiler] is equal to <span class="math">V_BE[/spoiler], so you'd get something like <span class="math">V_CC=R_1*I_1+V_BE[/spoiler] right?

[Anonymous No.5560120]

>>5560116
Why didn't the <span class="math"> tags work here guys?[/spoiler]

[Anonymous No.5560126]

>>5560116
Why didn't the math tags work on this post guys?

[Anonymous No.5560130]

>>5560099
>>5560083
>>5560028
So I'm thinking you'd probably need to do a KVL down the left side of the circuit if you wanted to find, say, the current <span class="math">I_{R1}[/spoiler], and since Q1 is diode-connected, it would mean <span class="math">V_{CE} [/spoiler] is equal to <span class="math">V_{BE}[/spoiler], so you'd get something like <span class="math">V_{CC}=R_1*I_1+V_{BE}[/spoiler] right?

[neckbeard_the_pirate No.5560137]

>>5560116
Yes.
The Vbe of Q1 will be the same as Vbe of Q2. Any current trying to go into the base of Q2 will split in half and go into the base of Q1. If heat causes the Vbe of Q1 to get smaller, the same heat will cause the Vbe of Q2 to do the same. If something causes the gain of the transistors to change, one transistor will null the changes in the other transistor.

[Anonymous No.5560154]

>>5560137
Okay, so, for <span class="math">R_2[/spoiler], doing a KVL down the right would yield <span class="math">V_A=V_{CE2}+I_O*R_2[/spoiler], but in this case the transistor isn't diode-connected, so how do you deal with the <span class="math">V_{CE2}[/spoiler]?

[neckbeard_the_pirate No.5560186]

>>5560154
Ic of Q2 is the same as Ic of Q1. Remember, Q1 and Q2 are matched pairs, and the base current for both transistors are split equally giving the same current gain for both transistors. To calculate the V drop of R2, use the known collector current of Q1. I*R=V

[Anonymous No.5560200]

>>5560186
So what you're saying is that <span class="math">I_O = I_{C2} = I_{R1}[/spoiler]?

[neckbeard_the_pirate No.5560203]

>>5560200
Yes. Due to the nature of the circuit, both matched transistors are fed the exact same current, and have the exact same current gain, and have the same current flowing through each collector-emitter.

[neckbeard_the_pirate No.5560208]

>>5560200
Oops. Wanted to add:
The current in the collector is the same as the current in the emitter minus the base current. Minor oversight on my part. Sorry.

[Anonymous No.5560260]

>>5560208
<span class="math">so I_{E2} = I_O+I_B? That will help me get the voltage drop across R_2?[/spoiler]

[Anonymous No.5560262]

>>5560208
so<span class="math">I_{E2} = I_O + I_B[/spoiler]? That will help me get the voltage drop across <span class="math">R_2[/spoiler]?