Some time ago, I reverse-engineered a low-cost inverter made by Vector Mfg. (now Black & Decker). It uses a classic 2-BJT astable multivibrator (AMV) but with a circuit design twist, as shown below. What is the purpose of R6?
The function of the AMV is to provide “sawtooth” waveforms for PWMing each half-cycle of the inverter H-bridge which outputs a bipolar square-wave. Each half-cycle of the bridge circuit requires its own sawtooth waveform.
The MMV operates mostly like the classic circuit. Suppose Q1 turns on, grounding the left terminal of C1. Then this 6.2 V decrease occurs on the other side of the capacitor, at the Q2 base, shutting it off. If the –b-e junction of Q2 has not broken down (which it will around 5 V to 7 V), then the Q2 base voltage waveform begins its exponential climb towards the target voltage at the top terminal of R3. When it crosses about 0.55 V, Q2 begins conducting appreciably, Q1 is similarly shut off, and the exponential waveform appears at its base. The BJTs alternate in conducting and being off and the circuit produces a “sawtooth” waveform at both collectors as shown below.
Because of the equal-value timing components, the time constants for each half-cycle are equal (R5·C3 = R3·C1) and the duty ratio is about 50 % which is needed for a symmetrical inverter bipolar square-wave output waveform. The collector voltages rise exponentially because as C1 or C3 charge, the timing-current path is through collector resistors R1, R2 which have a resistance that is not negligible relative to base resistors R5, R3. Each of the collector waveforms is input to a comparator with a reference voltage set by the inverter bridge voltage. If the bridge voltage increases, each half-cycle of the H-bridge turns on at a higher voltage on the exponential, which is later in the half-cycle. (This is called rising-edge PWM because instead of varying when turn-off occurs, the turn-on edge is varied instead.) The on-time thus decreases as does the duty ratio. This keeps the rms output voltage somewhat regulated.
The inverter rms output voltage is the peak voltage (which is the bridge voltage) times √D where D is the duty ratio. Ideally, D varies by the square of the bridge voltage, but for small changes around an operating-point, this scheme approximates a square by an exponential. It is not perfect because of the linear and cubic (and higher) terms in the exponential series expansion. Yet it is a somewhat elegant method for a low-cost inverter. Ideally, the PWM waveform would be parabolic instead.
In the classic 2-BJT AMV, R1, R2, R3, and R5 are returned to the supply, though in this case, it is 12 V, derived from the battery input to the inverter. For a 5 V supply instead, the BJT bases are driven off by a –5 V peak waveform and the b-e ¬ junctions do not break down. The addition of R4 reduces the voltage, to V = 6.2 V, as measured in a working unit. C2 is large enough to keep the voltage at the V node constant.
This leaves for us the opening question: why is R6 needed? Why not return vB to V ? R6 reduces the voltage further, to about 4.3 V, but there is an attenuated exponential waveform at this node. Each half-cycle exponential contributes to vB , resulting in a “full-wave” exponential waveform; the exponential part of the waveform repeats without any constant off interval between them. Because only one half of the circuit has an exponential base waveform at any one time, the opposite half-cycle timings do not interfere with each other. At the same time, R6 further reduces the collector supply voltage of V = 6.2 V to a lower VB = 4.2 V, a voltage low enough to not break down BJT b-e junctions. The higher collector voltage lets the collector resistors be larger and more like “long-tailed” current sources, causing the exponentials to be closer to a sawtooth waveform.
By having a higher target voltage of V for the timing circuits, PWM generation is less nonlinear. The PWM incremental (small-signal) transfer can be calculated from the following plot.
The PWM waveform target voltage is V > VPWM . VPWM is the voltage at which the half-cycle ends, at Ts , and is the highest PWM input voltage of the comparator, vC . At t1 , the comparator voltage is crossed and the inverter half-bridge output becomes high. Its on-time begins at t1 and ends at Ts , the end of a half-period of the inverter cycle. Thus, it is the off-time on the above plot that is the on-time of the bridge because of driver inversion, and the effective duty ratio is actually the plot complement, D ’. Applying the usual exponential formula to the above PWM waveform,
At the comparator voltage, vC ,
Then v (Ts ) = V and
Where VPWM ≈ 4.3 V. Then substituting,
Then to find the actual duty ratio for the inverter of D ’ = (1 – D ),
The change in D ’ with the scaled incremental bridge voltage, vC , is
Ideally, ln(1 – vC /V ) = 1, or vC << V . R6 makes it possible for V to be higher so that the ideal is more closely approached.