IC power dissipation consists of standby and active leakage components. The active component is dominated by the switching or dynamic power component. In addition, the standby component can be made significantly smaller than the active component by changing the body bias condition or by power gating.
Reducing supply voltage is one way to help optimizing the power consumption. However, it will lead to slower current drive of the transistor.
Reducing supply voltage means reducing the VG or the gate voltage. Thus, due to the square component the current drive is substantially lowered too. However, we can also lower down the threshold voltage of the transistor. This is because the ON-state of the transistor is achieved when:
VGS > Vth, where Vth is the threshold voltage of the transistor.
The threshold voltage, in turn, can be controlled by body bias voltage as shown in below equation. VSB is the source-to-body substrate bias. Normally, it is a negative value, so the more negative the back bias voltage, the higher the threshold voltage will be.
That leads to harder criteria for transistor to turn-on. Another way of reducing the threshold voltage is by thinning the oxide (related to the gamma sign in the equation).
One penalty, though, in reducing the threshold voltage is the increase in sub-threshold current as shown in the figure below.
The ISUB is thus determined by an equation that looks like the IDS current equation itself, except for several exponential numbers.
Out of several possible reasons in the high IOFF. The sub-threshold current dominates as the technology node shrinks.
In addition, the sub-threshold current
increases with temperature. This is the reason why testing IDDQ at high
temperature can aggravate the soft-defects especially in catching
non-uniform oxide thickness issue.
Cheers,
Pungky
Reducing supply voltage is one way to help optimizing the power consumption. However, it will lead to slower current drive of the transistor.
Reducing supply voltage means reducing the VG or the gate voltage. Thus, due to the square component the current drive is substantially lowered too. However, we can also lower down the threshold voltage of the transistor. This is because the ON-state of the transistor is achieved when:VGS > Vth, where Vth is the threshold voltage of the transistor.
The threshold voltage, in turn, can be controlled by body bias voltage as shown in below equation. VSB is the source-to-body substrate bias. Normally, it is a negative value, so the more negative the back bias voltage, the higher the threshold voltage will be.
One penalty, though, in reducing the threshold voltage is the increase in sub-threshold current as shown in the figure below.
Out of several possible reasons in the high IOFF. The sub-threshold current dominates as the technology node shrinks.
Cheers,
Pungky
Awesome Pungky!! I like your ABC of electronic library.. hehe. :)don't stop posting..
ReplyDeleteBtw, I need some clarification here:
To reduce leakage current, we need to increase Vth right? Instead of lowering it down? I am assuming Vth is the threshold voltage here. :p
Because if Vth is low, transistor tends to be active all the time (transistor will be harder to turn off coz a little Vg will turn the transistor on already). If transistor always active, leads to more leakage.. That's why we want to have Vsb (by tie-ing bulk/body voltage to most negative supply) to get higher Vth as what you mention in the posting above. :)
Hi Jennifer,
DeleteThanks for the nice words. I will definitely continue posting more stuff.
Yes you are correct, Vth needs to be increased to avoid the subthreshold leakage. One way to is force more negative body bias. However, increasing Vth will also make the transistor harder to turn-on. If we are talking about process, we cannot ensure uniform Vth window.