circuit-designvoltage
Simple question, how do I turn the following block diagram to a real world working circuit?
I know I could use the following:
1) A comparator, but that will give me only either +Vcc or -Vcc.
2) I could use a rail-to-rail comparator and set V- at 0 Volts and the maximum allowed voltage at V+.
3) I could use a Zener diode… But I've zero experience with using them…
simulate this circuit – Schematic created using CircuitLab
That is a fairly husky Zener diode. The test current shown in the datasheet is 45 mA. You are only allowing 2.5 mA, which is apparently not enough to get over the diode's "knee".
Basically what you're asking is, do current controlled switching convertes exist. The answer is yes, as you can see here:
Yes, that looks complicated ! But it's not (for me at least). The MAX649 is a DCDC Buck converter controller IC, it does not have a build in switch, PMOS Q1 is the switch. If you would want a constant output VOLTAGE, you would need a voltage divider (2 resistors) and feed a divided value of the output voltage back to the FB (feedback) pin.
But in this circuit, has a constant output CURRENT, to charge a battery ! Instead of sensing the output voltage, now the output current is sensed by taking the voltage across R10, amplifying this voltage (the opamp does this, with a feedback network to set the gain) and where does ths feedback signal go ? Straight back in to the FB pin !
The simple buck converter in your schematic is too simple to be considered having a voltage or a current output. If you change the load then voltage and current change, it is not regulated in any way. Yes you can change the PWM of the switch signal but that PWM signal needs te be determined somehow. This is done with voltage or current feedback.
Best Answer
It depends on how accurate you need this to be. Here is a simple concept:
D1 and D2 provide the MIN function. The voltage at the top of the diodes is the minimum of Vin1 and Vin2, plus a diode drop. It should be obvious how to expand this to any number of inputs to take the minimum of.
D3 tries to compensate for the diode drop, so that Vout is the minimum of all the inputs with the diode drops cancelled out. If within a few 10s of mV is OK, then this might do.
Vhigh and Vlow are voltages you have to supply. Vhigh must be a bit higher than any input voltage of interest, and Vlow a bit lower.
The impedance of the input voltages need to low enough to overcome R1. The output impedance at Vout is higher due to R2 needing to be high to not interfere with the signal.
That's the basic concept.
The next step is to realize that BJTs can be thought of as diodes with gain. Here is the same concept carried out with BJTs using their gain to advantage:
Look carefully and you'll see its really the same thing, using the B-E junctions of the transistors in place of the diodes in the previous circuit. The advantage is that the input signals don't have to supply anywhere near as much current. The B-E diodes still do the MIN function, but most of the current comes from the negative supply.
The gain is used the other way around so that the B-E junction of Q3 loads the signal much less. Due to the gain of the transistors, this circuit has much lower output impedance while having higher input impedance.
One drawback of using the B-E diode of BJTs this way is that the max input range is narrower. This is because the reverse voltage characteristics of the B-E junction is usually fairly low. If you're only doing this over a 5 V range, then there should be no problem. Of course, always check the datasheet of whatever transistor you plan to use. Choose ones with high B-E reverse voltage capability if you want a wider input voltage range.