# Darlington Transistor hybrid parameters from BJT h-parameters

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:

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?
• 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 $R_{IN} \approx 2 h_{\text{ie1}}$
– G36
Nov 1, 2021 at 13:56

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

• The site supports mathjax; it would make your answer much easier to read. Nov 1, 2021 at 3:09
• @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? Nov 1, 2021 at 10:57
• @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 $i_{c1} = h_{fe1}i_{b1}$ Nov 1, 2021 at 11:06
• @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. Nov 1, 2021 at 14:55
• Oh, I know this but my problem is to find h-parameters without any approximation aside from $h_{re}$. Nov 1, 2021 at 15:36