IIR coefficients that simply boost the signal
I will assume that your IIR filter is an order-2 biquad. But the reasoning remains the same if the implementation is different. 

Original filter : $H(z) = \frac{b_0+b_1\cdot z^{-1}+b\cdot z^{-2}}{1 + a_1\cdot z^{-1} + a_2\cdot z^{-2}}$

If you want to boost your signal by G

$H_G(z) = G\frac{b_0+b_1\cdot z^{-1}+b_2\cdot z^{-2}}{1 + a_1\cdot z^{-1} + a_2\cdot z^{-2}}$

simply distribute G to the numerator

$H_G(z) = \frac{G\cdot b_0+G\cdot b_1\cdot z^{-1}+G\cdot b_2\cdot z^{-2}}{1 + a_1\cdot z^{-1} + a_2\cdot z^{-2}}$

Simply multiply your b0-b2 coefficients by G.

In the case where you simply want to boost the signal without filering it at all. Set $b_0$ to $G$, $b_1$, $b_2$ to $0$, as well as  $a_1$ and $a_2$ to 0.