I read that a JFET cascode amplifier has a reduced input capacitance thus lowering the Miller capacitance and will obtain wider bandwidth.

In order to obtain lower Miller cap, the voltage gain must be lower to reduce the multiplication of stray input capacitance. BUT, how does this circuit reduce the input capacitance?

I only notice that the signal is directly-coupled to the lower JFET while the upper JEFT has its gate in AC ground. But I cannot understand how was that suppose to lower voltage gain (Miller capacitance)enter image description here

  • \$\begingroup\$ Do you understand how a regular cascode works? You would probably find a lot more material searching up how a BJT or MOSFET cascode than a JFET cascode reduces the Miller capacitance. Or is this question specifically about the JFET cascode? They pretty much all work the same way: The top transistor clamps the voltage collector/drain voltage of the lower transistor (similar to how a the gate voltage of a source follower controls the voltage of the source resistor) which prevents Vin from charging up the drain-gate capacitance of the lower transistor. \$\endgroup\$ – DKNguyen May 17 '20 at 5:45
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    \$\begingroup\$ Iwatani, the basic idea is pretty simple. (Easier still, with a BJT in view.) If the source (emitter of BJT) of the cascoded device is a fixed (relatively) voltage value below the gate (base, in BJT case), then the drain of the lower device (in your schematic) remains at a relatively unvarying voltage (nearly so.) But the cascoded device must pass along the drain current of the lower device to the drain of the cascoded device. So the lower device "sees" very tiny variations of its drain and therefore almost no multiplication effect of the capacitance. \$\endgroup\$ – jonk May 17 '20 at 5:46
  • \$\begingroup\$ electronics.stackexchange.com/questions/416510/… \$\endgroup\$ – G36 May 17 '20 at 9:00
  • \$\begingroup\$ Well, what is the voltage gain of the input transistor? \$\endgroup\$ – user_1818839 May 17 '20 at 13:39
  • \$\begingroup\$ @jonk, I agree with you that "the cascoded device must pass along the drain current of the lower device to the drain of the cascoded device". But how? \$\endgroup\$ – Circuit fantasist May 22 '20 at 19:10

Let's say I disregarded the upper JFET and applied a fixed DC bias of 10 volts to the lower JFET's drain as per this diagram: -

enter image description here

Although it isn't a voltage amplifier any more, can you see that that the miller capacitance (shown as a red capacitor above) no longer has much of an effect? It still acts as a regular capacitor but instead of being connected to a (drain) node that is swinging about with the amplified input signal, it connects to a DC node?

Can you also see that the circuit still functions as a pretty phenomenal amplifier of current in that the input signal voltage still causes a significant AC signal current to be taken from the 10 volt source? Now consider the upper transistor circuit: -

enter image description here

I've disregarded the lower transistor and replaced its circuit with a red resistor called \$\color{red}{\text{LOAD}}\$. Can you see that if \$\color{red}{\text{LOAD}}\$ varies, despite this variation, the voltage at the source of the JFET will remain fairly constant at + 10 volts. I've estimated 10 volts based on what I know about JFETs i.e. the gate will be 2 or 3 volts below the source for a typical bias arrangement.

This means that when the two circuits are joined, the drain of the lower JFET is biased with a fixed (and steady) DC voltage. The upshot is that the lower JFET's drain produces very little AC signal voltage. This means that miller capacitance effects are very small and just like the top diagram.

However, now the upper JFET turns the AC signal current produced by the lower JFET back into an AC voltage at its drain hence, we get regular voltage amplification. And, it should be noted that the upper JFET doesn't suffer from its own miller capacitance effects because, its gate has a capacitor to ground that ensures that AC signal variations are decoupled to ground.

  • \$\begingroup\$ Very good teaching approaches - first, to split the circuit and remove the upper part... then to replace the sophisticated current source in the lower part with a humble resistor load... \$\endgroup\$ – Circuit fantasist May 22 '20 at 19:19

Suppose the 2 fets have a transconductance of 1/39 ohms. and a C_gate_drain of 10pF.

Apply 39 millivolt AC at Vin. The bottom fet will produce Iout = Vin * (1/39 ohm) == 1milliAmp small signal (AC) drain current.

This 1mA interacts with Rin_source of the upper fet; Rin(common_gate) also == 1/39 ohms, and you see 39mv on the node between the two fets.

The 1 mA small signal current continues on thru the upper fet.

At drain of upper fet, the 1mA and 3,900 ohms produces 3,900 millivolts Vout, which is gain of 100X.

What is the input capacity of the bottom fet? The C_g_d has 39mv on each of gate and drain, but these are out of phase; being out-of-phase, the voltage doubles, thus both the gate and the drain must provide DOUBLE the current; the Cin is doubled to 20pf.

On other hand, the capacity burden (the charging current) of Vout on Vgate of the upper fet is 1010pF, and a large bypass capacitor in needed.


The cascode helps to reduce the Miller Effect of the gate to drain capacitance.
Without Cascode FET
If there is no cascode FET then the gain from input gate to the drain of the FET is \$g_mR_D\$. Consequently, the input capacitance looking into the input gate is: $$C_{in} = (1+A_v)C_{gd} = (1+g_mR_D)C_{gd}$$ This reduces the bandwidth of the amplifier.
With Cascode FET
Now the impedance looking into the drain of the input FET is approximately \$\frac{1}{g_m}\$. Consequently, the input capacitance now is: $$C_{in} = (1+A_v)C_{gd} = (1+g_m.\frac{1}{g_m})C_{gd} = 2C_{gd}$$ This assumes both the cascode and the input FET have same transconductance.
Clearly, since the gain is high the input capacitance for the cascode case is smaller.


The cascode configuration (FET, BJT, tube…) is usually presented as a common-source amplifier (Q1) that drives by current a common-gate amplifier (Q2). At first glance, this connection does not make sense since the common-gate stage is designed to work as a voltage amplifier while here Q2 does not amplify... it only delivers the current. The other answers have explained why it is done this way. I will explain how this is done because it is not well explained in sources.

First of all, you have to have a clear idea what the problem is. Let's try to formulate it.

A. The problem

Virtual input capacitance. "Looking” at Q1 gate, the input voltage source "sees" two JFET capacitances in parallel - the gate-source Cgs and gate-drain Cgd. But in such a common-source configuration with a source resistor, they are not genuine capacitances since their other ends are not connected to the "stiff" ground but to "movable" points with varying voltage. As a result, the value of these capacitances are modified as follows:

Virtually decreased Cgs. The right end of Cgs is connected to the source voltage that follows the input voltage. The source voltage is subtracted from the input voltage so the voltage across and current through Cgs are decreased... as though this capacitance is decreased. This is another version of the classic Miller effect aka "bootstrapping". It is useful for us; so we will not beat it...

Virtually increased Cgd. The right end of Cgd is connected to the drain voltage that is an inverted copy of the input voltage. The drain voltage is added to the input voltage so the voltage across and current through Cgs are increased... as though this capacitance is increased. This is the classic Miller effect that, in its utmost form, is known as "virtual ground". In some cases (e.g., op-amp inverting integrator) this effect is useful but here it is harmful; so we have to beat it. How do we do it?

B. The solution

1. Revealing the contradiction. Q1 drain voltage varies because of the drain resistor Rd... so we should remove (zero) Rd, e.g. simply connecting Q1 drain to Vcc or other constant voltage. But we want Q1 drain voltage to vary since this is the output voltage of this common-source amplifying stage... so we should not remove Rd... we need it. How do we then solve this contradiction?

JFET cascode visualized

Fig. 1. JFET cascode visualized

2. The idea - neutralizing Rd. The clever solution in such seemingly hopeless situations is to neutralize (compensate, destroy...) the undesired voltage drop across the resistor (Rd) with equivalent voltage (Vds) "produced" by an active element (the upper transistor Q2) - Fig. 1. Thus, we have both a voltage drop on Rd (as an output voltage Vout) and a zero voltage drop on the "destroyed Rd" (fixed Q1 drain voltage). Op-amp inverting circuits and S-shaped negative resistance circuits exploit the same idea where the voltage drop across some positive resistance is neutralized by equivalent voltage drop across equivalent negative resistance. As a result, a virtual ground appears.

3. Implementation. This powerful idea is implemented in the ingenious cascode configuration by inserting an additional source follower (Q2) between Q1 and Rd. Q2 drain-source channel compensates the resistance Rd as follows.

4. Operation. When Q1 increases the common drain current, the voltage drop across Rd increases. However, Q2 decreases its channel resistance and accordingly, the voltage drop Vds2 across it. As a result, the total voltage drop VRd + Vds2 stays constant, Vd1 drain voltage too. However, the voltage drop across Rd varies, Vd2 (Vout) too. How does Q2 do this magic?

5. Explanation. Q2 can be thought as of a negative feedback system with constant input (gate) voltage that keeps constant its output (source) voltage; so it acts as a voltage source. It does not need the resistance Rd; so let’s first remove it. Q1 can be thought as of another negative feedback system with varying input (gate) voltage that keeps constant its output (drain) current; so it acts as a current source. This combination of two heterogeneous sources connected to each other is “favorite” for both of them since the current source “sees” the almost zero differential resistance of the voltage source and the voltage source “sees” the almost infinite differential resistance of the current source. When one of them tries to change its quantity by changing its static resistance, the other “helps” it by changing its static resistance in the according direction - Fig. 2. So this is a system of two interacting (mutually “helping”) sources. The considered CS-CG cascode circuit is presented in Fig. 2b.

V driving I Fig. 2a. Voltage source driving a current source (an example is the common mode of the differential pair where two source followers in parallel drive a common current source)

I driving V Fig. 2b. Current source driving a voltage source (the present CS-CG cascode circuit where a common-source stage drives a common-gate stage)

For example, if the input voltage increases, Q1 decreases its static drain-source resistance to increase its drain current. This "pulls down" Q2 source voltage... Q2 "senses" this input disturbance and begins increasing its drain current until it becomes equal to the desired Q1 current. Thus Q1 current is delivered to Q2 drain and we have a current output.

But we need a voltage output; so let's put Rd back in its place. As a result, the common current decreases... but Q2 "senses" this change and immediately reacts by decreasing its static drain-source resistance. The total resistance Rd + Rds2 + Rds1 + Rs and accordingly, the current is restored.

So Rd is a disturbance for Q2 negative feedback system (voltage stabilizer)... but it is compensated by the system.


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