Physics and component models

Mini project:Realization of an Amplifier

Abbezzot CdricNguyen Thanh H Giang

AEI 4th year01-2007

Introduction

To illustrate the fundaments of MOS devices, the MOS modeling, we have a choice to realize a mini project by using of nMOS and pMOS.

Amplifier is a basic cell of all electronic circuit which consists in multiplying the amplitude of a sinusoidal Vin by a significant factor. We would like to begin with a simple and basic thing, by consequence, we decided to learn more about the models of an amplifier and build it by 3 different ways: amplifier MOS device, differential amplifier and push-pull amplifier.

I. The simplest amplifier: MOS device:

This single amplifier consists a Mos device with a load. In the first time, we realized and tested an amplifier: n- channel MOS with a p-channel Mos load. In the second time, we did the same thing with n- channel MOS and R load.

I.1 MOS device with Pmos load:

Principe: In interesting zone, the transfer function between Vin and Vout has a linear shape. By consequence, if we have a small variation ids is added to Ids, we have a variation vout added to Vout.

ids= gm*vgs

From the general figure, we design on Microwind the amplifier.

We can see here the tensions of Vin (green) and Vout (red)

At the beginning, our AO didnt work because we were out of zone of function. From the figure Vout in function of Vin, to stay in this interesting zone, Vin must be in interval [0.3 0.5], we chose Vin a offset 0.42 with variation sinusoidal 0.1.

The GAIN optimal which is tangent with the shape, is equal to 2.3 in interesting zone.

We notice that when we increase the length of Nmos or decrease the width of polysilicon, we could increase the Gain because it would favour the induct courant of electrons in channel. The relation of current is: I = K.W / L, with L width of grid and W length of doping.

I.2 MOS device with R load:

With R load, we found the same thing as with Pmos load. But the gain is bigger; its equal to 2.9.

II/ Simple differential amplifier:

The goal of this differential amplifier is to compare 2 analog signals and to amplify their difference.

Vout = K ( Vp- Vm)

Principe: Without effect of Vbias, if Vp = Vm, the currents crossing the 2 pMOS are equal (due to the mirror effect). This same current I also crosses the 2 nMOS. Consequently, no current crosses the load connected to exit. Thus, we have Vout = Vref (potential of grid of the pMOS). If there is now Vp = Vm + , a current I' superior to the

preceding current crosses the first branch. This current I' also crosses the pMOS of the second branch, while the nMOS is always traversed by the same current I (V-N' does not change). The difference between the current I' and the current I is then evacuated by the exit, and Vout increases until there is a balance of currents I and I'. There is then a great variation of Vout for a small variation of Vp, which is the goal of the differential amplifier. Vout being limited by the supply, one attends a rapid saturation in tension.

The improvement of the Differential Amplifier consists to insert a cell nMOS between the differential pair and the mass. The grid voltage Vbias is to control the increase of current circulating in the two branches. Vbias decreases the effects of saturation.

We put the numbers for each part and assembly the linked parts to draw a differential amplifier on Microwind:

Vp takes 0.45 V, 0.35 V like value.Vm is constant equal 0.4 V

With Vbias= 0.5, we have the result with a gain of 7.4

In the first time, we were mistaken to think that the gain is measured like in the amplifier Mos devise. In fact, the figure Vout( Vp) gives us only the interesting zone. In this case, look at the figure, the interesting zone is around 0.35 V.

To calculate the gain, we made several measures. for Vbias= 0.417 Vif Vp=Vm= 0.6 V, Vout=0.44 Vif Vp=1V, Vm= 0.6 V => Vout= 1V

Gain= V out / 0.4 = (1- 0.44 )/ 0.4 =1.4 V

Realize with different Vbias, we have these results:Vp Vm Vout Vp Vm Vout GAIN

Vbias=0.471V 0.6 0.6 0.44 1 0.6 1 1.4

Vbias=0.6 V 0.6 0.6 0.29 1 0.6 0.71 1.05

Vbias=0.8 V 0.6 0.6 0.24 1 0.6 0.49 0.625

We notice that a high Vbias reduces the gain. A low Vbias features a larger voltage range, specially at high voltage values. A high Vbias leads to a slightly faster response, but reduces the input range and consumes more power as the associated nMos transitor drives an important current.

III/ Amplifier Push pull:

This amplifier consists to amplify the difference between Vp and Vm:

This amplifier uses also the current mirrors to copy the currents. Function in general is the same with the differential amplifier.

By using the method of put the number in common zone, we realize the figure on microwind

To facility the observation, we chose Vminus constant equal to 0.5V, Vplus varied between 0.5 V and 0.6 V. We find the gain of 9.9

. Compare with precious method, the push-pull amplifier gives us the highest Gain

We dont have enough time to improve in this amplifier.

Conclusion

During this project, we learn how to realize an amplifier in particular and a Mos cell in general. Before doing this project, we hadnt thought that there are as many waysto realize an amplifier. Its an area not easy to understand. Each technique has its limit. We try to optimize each method to increase the length of conduit channel or decrease its width to maximize the current, so that have a high Gain. However, we must always stay in the linear zone, and the value of Gain is limited by the current of saturation. Moretechnique is complicated, more NMOS, PMOS are used, higher the Gain is. Measuring the gain was a difficulty we met.