I came across this circuit for a high-side current sense using BJT current mirror, and I am trying to derive a formula for how I could calculate the sense voltage output (going to the FB pin) in terms of Isense and the resistors used. I am having a hard time understanding how they derived their formulas. Firstly where does the 1.25 come from to calculate their RFB1, and when it says suggested bias current of 1mA, is this talking about base current, and why 1mA?

Resistor RB sets a bias current through the right-hand transistor. The suggested bias current for the PNP transistors is 1 mA. RB is selected by dividing the typical output voltage minus one diode drop by 1 mA. RB = (VO – 0.6) / 0.001 = 32.6 kΩ (5) RB = 32.4 kΩ 1% (6) RFB1 is set to bias the left-hand PNP at 1 mA, using the following expression. RFB1 = 1.25 / 0.001 = 1.25 kΩ (7) RFB1 = 1.24 kΩ 1% (8) RFB2 is set to amplify the current sense signal to equal the feedback voltage: RFB2 = (IF x RSNS x RFB1) / 1.25 (9) RFB2 = (1.0 x 0.2 x 1240) / 1.25 = 198Ω (10) RFB2 is 200Ω 1% (11)

Texas Instruments Application Note: AN-1696

using "1" for left transistor and "2" for right transistor:

From my understanding the base of the two transistors will be at VO - Vbe2 (forward drop of right-hand transistor). Doing KVL around the top loop, you Have (RsenseIsense) + Vbe2 = Vbe1 + (RFB2Ie1). For a matched pair, where Vbe1 ~ Vbe2, this gives: (RsenseIsense) = (RFB2Ie1), so RFB2 has the same potential across it as Rsense, and for a given sense resistor and load current, Ie1 is set by RFB2 as: Ie1 = (Rsense*Isense) / RFB2.

Since Ie1 >> Ib1, then Ic1 ~ Ie1, so Vsense ~ Ie1RFB1. so: Vsense = (RsenseIsense)*(RFB1 / RFB2).

So is this a correct formula to use for the Vsense output?

I think I'm not understanding something correctly, because I don't get what effect RB actually has on this formula? The formula I derived above does not include RB, but I think the base current of both transistors is set by RB, and the collector/emitter current should be related to base current as an approximation: Ib = Ie / Beta, Where Beta is the current gain - but if I try to derive a formula that way I can't seem to get the same result.

So where should base current and Beta come into these calculations?

Could you not just replace the transistor pair (emitter diodes) with two regular diodes OR'd together and connected to RFB1, and arrive at the same formula, and functionality?

I came across this circuit for a high-side current sense using BJT current mirror, and I am trying to derive a formula for how I could calculate the sense voltage output (going to the FB pin) in terms of Isense and the resistors used. I am having a hard time understanding how they derived their formulas. Firstly where does the 1.25 come from to calculate their \$R_\mathrm{FB1}\$, and when it says suggested bias current of 1mA, is this talking about base current, and why 1mA?

Resistor \$R_\mathrm{B}\$ sets a bias current through the right-hand transistor. The suggested bias current for the PNP transistors is 1 mA. \$R_\mathrm{B}\$ is selected by dividing the typical output voltage minus one diode drop by 1 mA. $$ R_\mathrm{B} = \frac{V_\mathrm{O} – 0.6}{0.001} = 32.6\mathrm{k}\Omega\label{1}\tag{5} $$ $$ R_\mathrm{B} = 32.4\mathrm{k}\Omega\pm 1\% \label{2}\tag{6} $$ \$R_\mathrm{FB1}\$ is set to bias the left-hand PNP at 1 mA, using the following expression $$ R_\mathrm{FB1} = {1.25 \over 0.001} = 1.25\mathrm{k}\Omega \label{3}\tag{7} $$ $$ R_\mathrm{FB1} = 1.24 \mathrm{k}\Omega\pm 1\% \label{4}\tag{8} $$ \$R_\mathrm{FB2}\$ is set to amplify the current sense signal to equal the feedback voltage: $$ R_\mathrm{FB2} = \frac{I_\mathrm{F} \times R_\mathrm{SNS} \times R_\mathrm{FB1}}{1.25} \label{5}\tag{9} $$ $$ R_\mathrm{FB2} = \frac{1.0 \times 0.2 \times 1240}{1.25} = 198\Omega \label{6}\tag{10} $$ $$ R_\mathrm{FB2}\text{ is }200\Omega\pm 1\% \label{7}\tag{11} $$ Texas Instruments Application Note: AN-1696

Now using "1" for left transistor and "2" for right transistor:

From my understanding the base of the two transistors will be at \$V_\mathrm{O} - V_\mathrm{BE2}\$ (forward drop of right-hand transistor). Doing KVL around the top loop, you have $$ R_\mathrm{SNS}\cdot I_\mathrm{SNS} + V_{BE2} = V_{BE1} + R_\mathrm{FB2}\cdot I_{E1}. $$ For a matched pair, where \$V_\mathrm{BE1} \sim V_\mathrm{BE2}\$, this gives $$ R_\mathrm{SNS}\cdot I_\mathrm{SNS} = R_\mathrm{FB2}\cdot I_{E1}, $$ so \$R_\mathrm{FB2}\$ has the same potential across it as \$R_\mathrm{SNS}\$, and for a given sense resistor and load current, \$I_\mathrm{E1}\$ is set by \$R_\mathrm{FB2}\$ as: $$ I_\mathrm{E1} = \frac{R_\mathrm{SNS}\cdot I_\mathrm{SNS}}{R_\mathrm{FB2}}. $$ Since \$I_\mathrm{E1} \gg I_\mathrm{B1}\$, then \$I_\mathrm{C1} \sim I_\mathrm{E1}\$, so \$V_\mathrm{SNS}\sim I_\mathrm{E1}\cdot R_\mathrm{FB2}\$. So: $$ V_\mathrm{SNS} = R_\mathrm{SNS}\cdot I_\mathrm{SNS}\cdot\frac{R_\mathrm{FB1}}{R_\mathrm{FB2}}. $$

So is this a correct formula to use for the \$V_\mathrm{SNS}\$ output?

I think I'm not understanding something correctly, because I don't get what effect \$R_\mathrm{B}\$ actually has on this formula? The formula I derived above does not include \$R_\mathrm{B}\$, but I think the base current of both transistors is set by \$R_\mathrm{B}\$, and the collector/emitter current should be related to base current as an approximation: $$ I_\mathrm{B} = \frac{I_\mathrm{E}}{\beta}, $$ where \$\beta\$ is the current gain - but if I try to derive a formula that way I can't seem to get the same result.

So where should base current and \$\beta\$ come into these calculations?

Could you not just replace the transistor pair (emitter diodes) with two regular diodes OR'd together and connected to \$R_\mathrm{FB1}\$, and arrive at the same formula, and functionality?