EE 310 Electronic Circuit Design I

EE 310 Electronic Circuit Design I

Spring 2008

Experiment 6:

MOSFET Amplifier Design Using an Active Load

Chad Ostrowski

Abstract: In this lab a basic ac amplifier using complimentary metal-oxide-semiconductor (CMOS) devices, is constructed, using a constant-current source as an active load. First a basic common-emitter circuit is examined and the experimentally measured voltage transfer characteristic is compared to the expected results. Then a common-gate configuration is used and again compared to expected values. In this case the body effect is also taken in to account. Finally, the body effect current is characterized using a specially designed circuit.
Introduction:

Amplifiers are an important part of analog circuits. MOSFETs can be configured in such a way that a smallac input signal gives a significantly larger ac output signal. One common way of doing this is to use PMOS transistors to construct a current mirror to act as an active load for an amplifying NMOS transistor. Depending on the construction of the circuit, the body effect of the NMOS transistor can have a significant effect on the output characteristics. This effect is examined and characterized towards the end of this lab. The parameters measured for the devices used in this lab as measured in previous labs are given in table 1.

Procedure:

Task 1: Current-Mirror/Active Mode

First, a current-mirror had to be constructed in order to provide a more-or-less constant current to the amplifying device. For ease of calculation, a 50kΩ resistor is used in place of the amplifying circuit. This is because, theoretically, the amplifying circuit to be added later would have an input resistance of about 50kΩ. Unfortunately, this turns out to be not quite true, a problem corrected later in the analysis. The PMOS current mirror layout is shown schematically in Figure 1.

With the circuit set up as shown, for a nominal current of 100µA on both sides, it is found that:

100µA = 3 * 111µA V-1 * (Vsg – 1.466V)2  Vsg = 2.014V  Vg = 10-2.014 = 7.986

and thus R3(theoretical) = 7.986V / .1mA = 79.86kΩ and the actual measured value was 80.8kΩ.

When this R3 is connected in the circuit, a current of 117.64µA is measured through transistor Q2. To remedy this, R3 is increased to 99.8kΩ, for which a current of 100.63µA is measured, which is wonderfully close to the desired value.

Task 2: Common-Source (CS) Amplifier Circuit

Next, the full circuit needs to be set up so that the drain current will remain at the desired 100µA. Schematically, the circuit needs to look like Figure 2.

NMOS transistor Q1 needs to be biased via resistors R1 and R2 to have a gate-to-source voltage that allows the correct current. This is shown below:

VS = 100µA*50.7Ω = 5.07mV,

100µA = 1.5 * 199µA/V2 * (VG – 5.07mV – 1.111V)2  VG = 1.69V

When determining R1 and R2, it is preferable that the current through them be very small so as to minimize the power loss in them, since the current in them has nothing to do with the rest of the circuit. 10µA or less would be desirable, from which it is found

(R1 + R2) = 10V/10µA = 1 MΩ

and

VG = (R2 / (R2+ R1)) *10V  R1/ R2 = ( (1 – 1.69) / 10 ) / (1.69 / 10 ) = 4.92

so, in keeping with the magnitudes needed,

R1 = 4.91MΩR2 = 1MΩ

and actual values are

R1 = 5.13MΩR2 = .976MΩ

Task 3: Adjusting the Q-Point

Since the values for R1 and R2could not be the exact desired values, a 1MΩ resistor is used in series with R2 in order to adjust the Q-point to the desired value so that a current of 100µA can be obtained. Also desirable is for the value of Vo, as shown in Figure 2, to be 5V, or half of the total voltage drop.

The expected values in this circuit are the same as those measured in Task 1, namely

ID1 = ID2 = 100.63µA;VGS2 = -2.014

and, from this, the expected value of VGS1 can be found with

ID = kn’/2 * (W/L) * [ (VGS1 – VTN)2(1 + λNVDS) ]  VGS1= 1.68V

However, when values were measured in the circuit, the values were found to be

R1 = 5.13MΩ , R2 = 1.197MΩ , Vo = 5.01V , ID1 = ID2 = 99.5µA ,VGS1= 1.79V , VGS2 = -2.18V

In order to obtain such consistent results, it was necessary to re-tweak the value of R3 from that determined in Task 1, since Rin of the circuit was not quite 50kΩ as was assumed. The new value of R3 was 50.7kΩ.

Task 4: Measuring the Voltage Gain

Now that the circuit is biased properly, it is time to actually put an ac signal into node X of Figure 2, making an ac amplifier with a small-signal model as shown in Figure 3.

As seen in the figure, ro1 was calculated to be 1/(λnIDQ) = 1.37MΩ and ro2 was 201kΩ (which was not one of the resistor options in MultiSim, and so appears in the figure as 200kΩ).

The current I1 in the figure is gmVgs, where gm = 2(k’n*w/(2*L) *IDQ*(1+λnIDS) = 352µA/V.

Rin is simply R1||R2 = 970kΩ and Rout = ro1||ro2 = 175kΩ.

The gain of the circuit is calculated by first finding vo divided by vi:

vi = vgs;vo = -(gmVgs)(ro1||ro2) Av = vo/vi = gm*(ro1||ro2) = -61.6V/V

When an oscilloscope was hooked up to the above circuit, however, the graph in figure 4 was obtained, which gives a voltage gain of -36.2V/V. The calculated value is 1.7x larger than this. The loading effect when measuring vo was partially taken care of with a 10X probe, but there is still a surprisingly large discrepancy.

Task 5: Common Gate (CG) Amplifier Circuit

To make a common-gate amplifier circuit, the same setup can be used as shown in Figure 2, Vi simply needs to be moved to node Y and node X be plugged into ground (hence the name). This configuration will give the ac equivalent circuit shown in Figure 5.

When measuring the output voltage as compared to the input voltage, which is the same in this case to make reading the graphs easier, the graph in figure 6 is obtained, which gives a voltage gain of Av = 51.7V/V.

gm is the same as that measured before because the bias current has not changed. ro1 and ro2, likewise, do not change with the new setup. Namely, all of them are gm=352µA/V, ro1=1.37MΩ, and ro2=201kΩ.

Rout is simply the resistance across the three parallel branches in parallel with ro2, but since Vgs is set to zero when determining Rout, the first resistance simply reduces to ro1. So rout becomes ro1||ro2 = 1.75kΩ.

Rin is simply the resistance across all three parallel branches in parallel with ro2 in parallel with 51Ω, but this is simply Rout || 51Ω = 51Ω.

Av is more complicated. First, it helps to note that vgs= -vi. Then it can be found that

vo = -(vgsgm(1+η) +(vo-vi)/ro1)ro2 = -ro2Vgsgm(1+η) - ro2*vo/ro1 + ro2*vi/ro1

so

vo = vi*(ro2/ro1 + ro2*gm*(1+η) ) / (1 + ro2/ro1)

and hence

vo/vi = Av = (ro2 + gm(1+η)ro1ro2) / (ro1 + ro2) = (assuming η=0.1) 51.7V/V

The measured value is 1.3x larger than the calculated value, which is not quite as much of a difference as measured for the CS amplifier.

Av for the CG amp was 1.1x(calculated)/1.4x(measured) larger than that for the CS, which is due to the body effect (as shown in the next task).

For CS, Rin was 970kΩ as compared to 51Ω for CG, which is immensely smaller than that for CS. Rout for CS was 175kΩ as compared to 1.75kΩ for CG, 100x larger.

Task 6: Deducing values for the body effect

For CS, |Av| = gm(ro1||ro2) = gm*ro1*ro2/(ro1 + ro2) = 36.2V/V

For CG, Av = ro2+gm*ro1*ro2*(1 + η) / (ro1 + ro2) = 51.7V/V

So

51.7- (ro2 + gmb*ro1*ro2) / (ro1 + ro2) = gm*ro1*ro2 / (ro1 + ro2) = 36.2

from which

gmb = ( -(36.2 – 51.7)*(ro1 + ro2) – ro2 ) / (ro1 + ro2) = 87.7µA/V

and so

η = gmb/gm = 87.7/352 = 0.249

Task 7: Directly Measuring the Body Effect

In order to directly measure the body effect, rather than using the two previous configurations and doing all of the algebra presented in task 6, the same basic setup can be used as shown in Figure 2 and node Z can be connected to vi while both X and Y are grounded. This makes vgs = 0, since both the gate (node X) and the source (node Y) are at zero, and so the only amplification is a consequence of the body effect, as shown in the ac equivalent schematic of figure 7.

From the schematic, vo = -gmb*vbs*(ro1||ro2) and vi = vbs, hence

Av = vo/vi = -gmb*(ro1||ro2)

When the above circuit is hooked up and the input (same as before) and output plotted, the graph in Figure 8 is obtained, which gives a gain of -36.3V/V. This gain gives

gmb = Av/(ro1||ro2) = 207µA/V

and hence

η = gmb/gm = 0.588.

These values are 2.3x larger than the values obtained the other way. Something somewhere is incongruent, it seems, though exactly where is unclear. Checking the results with fellow researchers gave only more confusion, and worse, confirmation that confusion was apparently the correct place to be about all of it.

Due to the large number of steps saved by using the process outlined in task 7 rather than that of tasks 4-6, task 7 is likely more accurate and is certainly easier to perform than the earlier method.

Discussion and Conclusion:

MOSFETs are an effective and fairly simple way of amplifying an ac signal, as long as the amplitude is small enough. In this lab, a basic common-source amplifier was constructed and characterized and then a common-gate, which allowed the body effect to also be examined. The body effect is a consequence of the body and source terminals being at different voltages, and can add to or take away from the total amplification. In the case of the common-gate amplifier, it was seen to add to the total amplification. By clever circuit setup, the body effect can be characterized very directly.