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I'm trying to analyze the Darlington transistor model, so I need to calculate the hybrid parameters for this circuit.

For example $$h_{ie} = \frac{v_{be}}{i_b}\bigg\rvert_{v_{ce=0}}$$

  • So which is the reasoning that leads to the solution? Do I need to consider the C.E. hybrid equations or just inspect the circuit?
  • Shouldn't \$i_{e1}\$ be \$i_{c1}\$?

This is my solution:

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I need to found \$v_{be}\$ in function of \$i_{b} = i_{b1}\$ where \$v_{ce} = 0\$

So the current \$i_{b1}\$ should pass through \$h_{ie1}\$,and then the sum of \$i_{b1}\$ with the current source, \$i_{b1} + h_{fe1}i_{b1} = i_1\$, should pass through the parallel of \$\frac{1}{h_{oe1}}\$ and \$ h_{ie2}\$. Therefore: $$\frac{1}{h_{oe1}} || h_{ie2} = \frac{h_{ie2}}{1+h_{oe1}h_{ie2}}$$ $$v_{be} = h_{ie1}i_{b1} + \frac{h_{ie2}}{1+h_{oe1}h_{ie2}}(1+h_{fe1})i_{b1}$$

So $$h_{ie} = h_{ie1} + \frac{h_{ie2}}{1+h_{oe1}h_{ie2}}(1+h_{fe1})$$

But the solutions that I found on my book is: $$ h_{ie} = h_{ie1} + h_{ie2}(1+h_{fe1}) $$

  • What's wrong?
  • How to calculate $$h_{fe} = \frac{i_c}{i_b}\bigg\rvert_{v_{ce=0}}$$

This is my solution for \$h_{fe}\$: $$i_{c} = h_{fe2}i_{b2} + i_{b2} = (1 + h_{fe2})i_{b2} = (1 + h_{fe2})\frac{1}{1+h_{oe1}h_{ie2}}i_1 = (1 + h_{fe2})\frac{1}{1+h_{oe1}h_{ie2}}(1+h_{fe1})i_b$$

So: $$h_{fe} = \frac{1+h_{fe1}+(1+h_{fe1})h_{fe2}}{1+h_{oe1}h_{ie2}}$$

  • It's ok?
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  • \$\begingroup\$ If we ignore \$h_{\text{oe}}= 0\$ we can see that \$I_{\text{e1}} = I_{\text{b2}}\$ Thus, \$ I_{\text{e1}} = I_{\text{b1}}(h_{\text{fe1}} + 1 ) = I_{\text{b2}}\$ Therefore \$V_{IN} = I_{\text{b1}}h_{\text{ie1}} + I_{\text{b1}}(h_{\text{fe1}} + 1 )h_{\text{ie2}}\$ and \$R_{IN} = \frac{V_{IN}}{I_{\text{b1}}} = h_{\text{ie1}} + (h_{\text{fe1}} + 1 )h_{{ie2}} \$ And in real life we can ignore this +1,therefore hie2 = hie1/hfe we have \$R_{IN} \approx 2 h_{\text{ie1}} \$ \$\endgroup\$
    – G36
    Nov 1, 2021 at 13:56

1 Answer 1

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Let's look at it another way without the confusing resistances and admittances,

consider $$I_{e1} = I_{b2}$$

$$I_{e1} = I_{b1} + I_{c1}$$

Since $$I_c = hI_b$$

Therefore $$I_{c1} = h_1I_{b1}$$

$$I_{e1} = I_{b1} + h_1I_{b1} = (1+h_1)I_{b1} = I_{b2}$$

$$I_{c2} = h_2I_{b2} = h_2(1 + h_1)I_{b1}$$

$$I_{c(darlington)} = I_{c1} + I_{c2}$$

$$I_{c(darlington)} = h_1I_{b1} + h_2(1 + h_1)I_{b1} = (h_1 + h_2(1+h_1))I_{b1}$$

$$I_{b(darlington)} = I_{b1}$$

Since $$I_c = hI_b$$

Therefore $$h_{(darlington)} = (h_1 + h_2(1+h_1))$$

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  • \$\begingroup\$ The site supports mathjax; it would make your answer much easier to read. \$\endgroup\$
    – Hearth
    Nov 1, 2021 at 3:09
  • \$\begingroup\$ @YoussefAly97 What is h? If you mean the \$\beta\$ current gain when \$I_{CBO}\$ is negligible, it is not what I answer. Did you want to say impedances instead of admittances? \$\endgroup\$
    – Simone
    Nov 1, 2021 at 10:57
  • \$\begingroup\$ @YoussefAly97 So the equation that I'm disappointed is \$I_c = hI_b\$ because from the second equation of C.E. hybrid parameters: $$i_{c1} = h_{fe1}i_{b1} + h_{oe1}v_{ce1}$$ but \$v_{ce1} \neq 0\$ because only the Dalington \$v_{ce} = 0\$. So in this case I can't write \$i_{c1} = h_{fe1}i_{b1}\$ \$\endgroup\$
    – Simone
    Nov 1, 2021 at 11:06
  • \$\begingroup\$ @simone h is \$\beta\$ but since you used \$h_{fe}\$ in your equations I didn’t want to confuse you. \$V_{ce}\$ may be small but not zero for darlington as well as a single BJT. \$I_{c1} = h_{fe1}i_{b1} + h_{oe1}v_{ce1}\$ is true but since \$h_{fe}\$ is typically between 10 and 400 or even more while \$h_{oe}\$ is typically on the order of 10e-6 then this term can be neglected. \$\endgroup\$ Nov 1, 2021 at 14:55
  • \$\begingroup\$ Oh, I know this but my problem is to find h-parameters without any approximation aside from \$h_{re}\$. \$\endgroup\$
    – Simone
    Nov 1, 2021 at 15:36

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