Dr. Krauss' answer summarized it nicely.

More generally, under equilibrium you can use the following two equations:

ND+ + p = NA- + n

n*p = ni2

Where ND+,NA-  are the concentrations of the ionized donors and acceptors, n,p are the concentrations of the electrons and holes, and ni is the intrinsic carriers concentration. For a known semiconductor with a given doping concentration(s), this results in two equations with two variables. The only pitfall is: can you assume full ionization of the dopants? For conventional dopants (As, B, P... in Si) this is a pretty good approximation, but this isn't the case for any dopant in any semiconductor. Having to consider the degree of impurity ionization makes the math more complicated. 

Dr. Krauss' answer summarized it nicely.

More generally, under equilibrium you can use the following two equations:

ND+ + p = NA- + n

n*p = ni2

Where ND+,NA-  are the concentrations of the ionized donors and acceptors, n,p are the concentrations of the electrons and holes, and ni is the intrinsic carriers concentration. For a known semiconductor with a given doping concentration(s), this results in two equations with two variables. The only pitfall is: can you assume full ionization of the dopants? For conventional dopants (As, B, P... in Si) this is a pretty good approximation, but this isn't the case for any dopant in any semiconductor. Having to consider the degree of impurity ionization makes the math more complicated. 

As nicely described in the two previous replies, you are talking about a process called counterdoping.

Let us assume that your semiconductor is p-doped; in this case, when you add an n-dopant, the electrons donated by this dopant will recombine with the previously existing mobile holes, meaning that the concentration of free holes will slowly decrease and approach that of an intrinsic semiconductor. In other terms, your semiconductor will lose its "p-type behaviour" and return an intrinsic semiconductor (at least in terms of free charge concentration). 

When the concentration of n-dopant is exactly the same as the concentration of the previously inserted p-dopant, as pointed out by Dr. Kornblum, the concentration of free electrons reaches that of free holes and the semiconductor (theoretically) returns intrinsic.

However, it should be noted that:

1) As observed by Dr. Krauss, the presence of dopant atoms inside the semiconductor crystal lattice alters the lattice periodicity thus decreasing the mobility of charges; in other words, your counterdoped semiconductor will have a lower "intrinsic" charge mobility than the pristine, undoped intrinsic semiconductor;

2) As observed by Dr. Kornblum, counterdoping is based on the assumption that all your dopants are perfectly ionised, meaning that once inside the semiconductor each donor (acceptor) atom is able to release a free, mobile electron (hole). This is an approximation of what really happens but, depending on the semiconductor and on the dopants, it does not always hold true;

3) It is experimentally impossible to control EXACTLY the quantity of dopants penetrating the semiconductor, so it is virtually impossible to dope the semiconductor with the precise quantity of a certain dopant necessary to counterdope the previous doping.

As a consequence of all these reasons, if the semiconductor is already doped you can try to counterdope it to bring it back to an "intrinsic-like" state but practically you will never be able to obtain exactly the same mobility you can have with the same pristine, undoped semiconductor. 

Simply...

There is a critical carrier concentration for the semiconductor material, above that level we can modify by certain types of doping.

Since, this possibility made researchers to work on that.

Dear Dr. Kornblum thanks for ur kind explanation, but i'm still in doubt that at low temperature when all the donors and acceptors are not completely ionized,  then what will happen? 

The two equations I wrote work at all temperatures (the second requires equilibrium). 

In low temperatures there will be changes is the fraction of dopants that is ionized. This is Arrhenius-dependent in the activation/ionization energy of each dopant. It is indeed likely that the p and n dopants will have different energetics and therefore each will freeze-out at a different temperature. This will mean that at a given temperature you may have a majority carrier type. It should further be noted that the intrinsic carrier density is temperature dependent as well, with half the bandgap / kT as the exponent. 

Dear Dr. Kornblum thanks for your answer..

It seems ok! 

However, i would like to that at low when the donors and acceptors both are being ionized, then will the conductivity be higher than the intrinsic case or not? 

Also, what will happen to Hall Voltage in this case? 

At low temperatures, in the so called freezeout region, each carrier density has an Arrhenius temperature dependence but with a different activation energy, that is a specific magnitude for a given dopant in a given semiconductor. For conventional dopants this energy is much smaller than Eg/2 (which determines the intrinsic carrier density) and thus the ionized dopants will surpass the intrinsic ones rapidly upon heating.

Please take a look at chapter 1 in the seminal book of S.M. Sze  Physics of semiconductor devices . Please look at Figures 12-13 and the discussion around them.

Let me know if you have further questions and I'll try my best

Dear Dr. Kornblum,

thanks for your answer. I agree to your suggestion. Based on it, I think that :-

1)- Conductivity in the so called freez-out region will increase as compared to intrinsic one but Hall Voltage will be negligible.

2)- There will be two Fermi levels, one for electrons and other for Holes in this region.

Do you agree?

I actually have to disagree on both items;

1. If I'm not mistaken, at a given current, a higher conductivity yields lower Hall voltages (lower carrier density will yield a larger Hall coefficient and Hall voltage). 

2. The Fermi level (which is more accurately the electron's chemical potential) is defined for electrons. Keep in mind that a hole is a missing electron in the valence band, so basically you can "count" them together.

The only reference that I know of for using two Fermi levels is the so called 'quasi Fermi level' description, of "imreF" according to some textbooks. Basically they are just another way of describing non-equilibrium carriers within equilibrium band diagrams and they are indeed defined separately for electrons and holes. However, I haven't seen them popularly used outside of textbooks (although I'm not a device person and there may be fields where this is more standard)

In case of intrinsic semiconductors, Hall voltage is very small as electrons and holes produce Hall voltage of opposite sign. Some small voltage is observed due to different mobilities of electrons and holes. Same effect should be observed in the unfreezes regions where electrons and holes both are being generated by doner and accepter levels.

 I think, it may behave like an intrinsic semiconductor, if Na = Nd

Because, the acceptor states lower than conduction band, the electrons in donner site will be going to the acceptor states. therefore, for moderate temperature which the electron in acceptor states cannot ionized, the semiconductor would behave like an intrinsic semiconductor

I agree with the interpretation of the question given by Lior Kornblum and Hongjeon Kang.

The question is not well set, however I would like to give my opinion as below-

Case- I consider  the donor impurities with carrier concentration ND cm-3 have been added to semiconductor crystal (intrinsic), then retaining electrical neutrality, the total negative charge ( electron and ionized acceptors) equals to the total positive charge (hole and ionized donors. Therefore  n = N+D + p

where  n: the electron density in the conduction band, p : the hole density in the valence band and N+D : the number charge donor.

Case- II consider the acceptor impurities of carrier concentration NA cm-3 have been added to said above crystal. Same equation may be written for the electrical neutrality.

Therefore p = N-A + n

where  p : the hole density in the valence band, n: the electron density in the conduction band, and N+A : the number charge acceptors.

The intrinsic carrier concentration np = ni2 (mass-action law), the product of carriers is independent of the added impurities. Hence Fermi level of intrinsic semiconductor normally lies very close to the middle of the band gap.

At elevated temperature (moderate and high temperature) donors and acceptors would be ionized, therefore neutrality condition becomes

n + NA = p + ND.

Therefore in my opinion, the doped intrinsic semiconductor crystal could show the intrinsic properties. For this theory number of text book are cited In literature, however you may refer our text paper (on ResearchGate)

Dear Suresh,

If an intrinsic semicondcutor is doped by an equal amount of donors and receptors such that Nd=Na, then it will be said that it is compensated semicondcutor and behaves as if it is intrinsic. Please look at this proof,

According to the neutrality,

Na- +n0= Nd+ +po,

Assuming all maturities are ionized, which may be the case in room temperature

Then n0= po,

and p0n0=ni^2

Then p0=n0=ni

So, the material will be inrtinisic

Completely compensated semicondcutor behaves as intrinsic.

Best wishes

Agree with Dr Abdelhalim Zekry explanation at room temperature. This is additional information for the low temperature (RT), as earlier ours answer was at elevated temperature. Therefore, the completely compensated semiconductor behaves as intrinsic from the RT to elevated temperature.