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

  • \$\begingroup\$ A few: 1. Use a (ideal) diode. 2. Use a comparator. 3. Use an analog switch. 4. Use a MCU. To just get you started. \$\endgroup\$ – dannyf Jul 11 '17 at 21:00
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    \$\begingroup\$ You shouldn't accept a answer so quickly. Wait at least 24 hours to give everyone around the world one day/night cycle to see the question and possibly come up with new and different answers. \$\endgroup\$ – Olin Lathrop Jul 11 '17 at 21:45
  • \$\begingroup\$ @OlinLathrop You're right. But I accepted it because it was exactly what I was looking for: a simple solution that can be built using spare components. I un-accepted it in case something simpler/different comes out as you suggested. \$\endgroup\$ – mickkk Jul 11 '17 at 21:48
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    \$\begingroup\$ I get that out of my buck converters whether I want it or not... \$\endgroup\$ – Harper Jul 11 '17 at 23:46

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.


The AD8036 Voltage Feedback Clamp Amp from Analog Devices may suit. This device, when used in non-inverting mode can clamp the input voltage between an upper and lower limit.

enter image description here

Figure 2. Clamp-amp block diagram. (See page 16 of the datasheet.) \$ +V_{IN} \$ is applied to A1's non-inverting input until \$ +V_{IN} > V_H \$ at which point S1 applies \$ V_H \$ to A1. Similarly when \$ +V_{IN} < V_L \$ S1 applies \$ V_L \$ to A1.

There are a few limitations to the device - particularly the supply voltages (+/- 3 to +/- 5 V) and 6.3 V max \$ \Delta \$V between \$V_H\$ and \$V_L\$ - but you can work around those by, for example, working in the 0 to 1 V range on your inputs and putting an amplifier afterwards to restore the signal level.

See my answer to OpAmp Buffer configuration with max allowed output for further details.


In addition to the many good answers provided already. You can diode-or the feedback path of a set of buffer amplifiers. The concept is applied heavily in power electronics, allowing multiple feedback loops to overdrive the output (typically with priority to lowering output power).

A sample schematic is the following, enter image description here

Here there are two op-amp buffers with there outputs diode or'd. Neither opamp can push the output high, only pull the output low. So the minimum voltage overdrives the output and the result is the Min(A,B).

A sample simulation, enter image description here


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