# Is the dynamic resistance ignored in this output impedance formula?

Does the following tutorial neglect the dynamic resistance across the base and the emitter in derivation of the output resistance:? It assumes Vbe constant and ignores the dynamic resistance? But dynamic resistance still exists even there is no change in Vbe(?) Because it says the voltage drop Vin - Vout = (I/beta)* Rsource. Dynamic resistance seems to be ignored(?)

• What is dynamic resistance? I guess you mean small signal resistance? No it is not left out if you assume that in the bottom right picture the NPN with 10 V on the collector is an ideal current amplifier with current gain of $\beta$. The DC part of Vbe is indeed constant, only the AC (small signal) part varies which causes current I to vary, which is what you want. I do find the approach taken here in the example a bit unconventional, it takes some large steps here and there. You could draw the proper small signal schematic and do the calculations leading to the same result. – Bimpelrekkie Oct 30 '17 at 11:15
• Yes small signal resistance lets call it Rdyn. To me the equation should be: Vin - Vout = (I/beta)* (Rsource + Rdyn). But seems like Rdyn is ignored here. Where am I wrong in thinking? – atmnt Oct 30 '17 at 11:43
• It's the difference of AC vs DC thinking. They are, I think, talking about DC. Even then, it's actually a little more complicated. It should be that $r_{source}$ is divided by $\beta+1$ and not $\beta$. But since $\beta$ is so large, the difference here usually isn't important. At a more technical level, I believe the DC value is $Z_{OUT}=\frac{r_{source}}{\beta+1}\frac{V_{in}}{V_{in}-V_{BE}}+R_E\frac{V_{BE}}{V_{in}-V_{BE}}$. (But usually $V_{BE}$ is small compared to $V_{in}$, too.) But note that $r_e$ isn't present here. – jonk Oct 30 '17 at 11:46

Yes you are right the dynamic resistance of $V_{be}$ is neglected because it is usually very small in comparison with rest of the resistances in the circuit. If you write the exact equation (neglecting the output resistance of BJT), you will get: $$V_{in} - \frac{I}{\beta}r_{source} - \frac{\beta + 1}{\beta}I(\frac{\alpha}{g_m}) = V_{out}$$ where $\frac{\alpha}{g_m}$ is the dynamic resistance of $V_{be}$ Thus, $$V_{out} - V_{in} = -\frac{I(\beta + 1)}{\beta}(\frac{r_{source}}{\beta + 1} + \frac{\alpha}{g_m})$$ Since, $Z_{out} = (V_{out} - V_{in})/(\frac{-I(\beta + 1)}{\beta})$. Finally,

$$Z_{out} = \frac{r_{source}}{\beta + 1} + \frac{\alpha}{g_m}$$ To get the sense of numbers, usually $\beta$ ~ 100 so, $\alpha = \frac{\beta}{\beta + 1}\approx 1$.
$\frac{1}{g_m} = \frac{V_T}{I_c}$, at room temperature $V_T \approx 25mV$ and assuming $I_c$ ~ $1mA$, $\frac{1}{g_m}$ ~ $25\Omega$.
Thus last term in the expression for output impedance is of the order of few ohms.
The source resistance, on the other hand, is the coming from the output impedance of a previous amplifier stage and is usually high ~ $10^5\Omega -10^6\Omega$. So usually the first term dominates. So we can approximate the output impedance as:
$Z_{out} \approx \frac{r_{source}}{\beta + 1} \approx \frac{r_{source}}{\beta}$.

• I agree to the calculation of Zout. However, I do not agree to the conclusion that "usually, the first term dominates". In many cases we have a base bias network with an equivalent source resistance of some kohms only (to be divided by beta). Hence, we should NOT neglect the term 1/gm which very often is in the same order as the first term of Zout. – LvW Nov 5 '17 at 11:28
• @LvW for the case of op-amp, as an example, we use this emitter follower after a differential amplifier stage which has a high output impedance. This is done to reduce the output impedance of the overall amplifier and make it a better voltage amplifier. It is clear from the question as well where the user has said that the output impedance after emitter follower stage has reduced. In the context of the question, this seems to be the correct explanation. – sarthak Nov 5 '17 at 11:37
• Of course, there are examples, where the source resistor is relatively large. No doubt about this. But you wrote "usually" - and that was the point of my disagreement. There are a lot of examples where the source resistance usually is NOT in the range 1E5...1E6 (as mentioned by you). Why not stay at the Zout expression with the two terms and let the application decide if the second term can be neglected or not? Thats all! – LvW Nov 5 '17 at 14:30
• @LvW in most applications emitter follower is used to reduce the output impedance. That's why I said usually and that is why the reference that the user has provided in the question also says that the impedance is r_source/beta and neglected the alpha/gm term. Agreed it depends on application but it's true for most applications. – sarthak Nov 5 '17 at 14:41
• sarthak - no doubt about it; in most cases, the emitter follower is used to reduce the output impedance of an amplifier. HOWEVER, you spoke about values in the range of 100kOhms...1Megohm. That is pretty unrealistic! Please note that even in case of transconductance amplifiers (OTA), where a high ouput resistance is desired, it is hard to achieve values about 10....50 kohms. A realistice example is a value of 10k which - divided by a beta value 0f 200 - gives a resulting source resistance of 50 ohms only. – LvW Nov 5 '17 at 16:30

Anytime the source impedance is < 100 Ohms you should consider including $r_{\pi}=(\frac{\alpha}{g_m} )$ and adding this to the source. Given the wide tolerances on hFE and Hfe, this is just prudent but you may adjust this value as you choose.

For DC biasing considerations consider also the power rating AND temperature rise. PN junctions have a NTC voltage characteristic and a positive ESR or bulk (base-spreading=Rbb )resistance based on chip size or Pd rating of the device.

Rbb*Ib values affect Vbe above 0.6V when the junction is saturated. ( approximation)