I'm designing an emitter follower circuit attached to the feedback node of a Buck/Boost SMPS in order to pull current out of the node and subsequently increase the output voltage by varying a DAC voltage, as described here: http://www.ti.com/lit/an/slva861/slva861.pdf. Here is the circuit I've been testing:
simulate this circuit – Schematic created using CircuitLab
Where NODE2 is the Feedback node on the Buck/Boost (VFB).
Originally, the resistor divider on the feedback node was R3 = 280k and R5 = 20k, in order to achieve a voltage of 800mV. However, I need an output voltage range of 1V to 14.4V and pulling current out of the feedback node only increases the output voltage. So I replaced the resistors to achieve a starting output voltage of 1V. I found my current resistor values using the following equations:
Where Req is:
I built the circuit on breadboard and replaced the Vout/VFB voltage divider with a 800mV power supply signal (since the Buck/Boost will keep the signal at 800mV). And tested the emitter voltage at 500mV increments of VDAC. Here are the results (note that I used a emitter resistor value of 1180 instead of 1194 in the actual test):
My question relates to the operation of the BJT in the context of this circuit. The emitter voltage varies fairly linearly with the DAC voltage, which is expected. However, I did not expect the voltage at the emitter to get any higher than 800mV (the voltage at the feedback node.) I was also trying to operate the BJT in its forward active mode (where VCE > VBE), but it looks like I'm operating outside of that range as soon as VDAC goes above 800mV. So what mode is the BJT operating in? Why is the voltage at the emitter greater than 800mV?
Apologies for the long report! Any feedback and thoughts are welcome!
