[SOLVED] Unique Electronics Switching Design Problem (Please Help)

[hshah8970]

Hello everybody!

I'm working on a project and am facing a problem I simply cannot solve. I'm unable to design the electronic circuit that accomplishes the following task:

I want to couple 2 DC inputs in such a way that when Input 2 reaches certain voltage level (from zero), both inputs become connected in series. Before this threshold voltage level is reached by Input 2, we simply get Input 1 at the output. Check out the attached image to see what I mean.

For voltage values of DC Input 2 higher than the threshold voltage, the series combination is maintained.

I would highly appreciate it if somebody helped me out with this! Thanks! And an explanation to go along the circuit would be nice too. Again, thanks a lot to anybody who responds!

 

[vimalkhanna]

What is the transition time given for series addition of the second supply to the first one after the equal level is reached ?
I shall finalize the schematic on your answer .

 

[hshah8970]

Thanks for the reply Vimal!

The transition time hasn't been specified. You can use your best judgement. Obviously, the minimum the time, the better.

 

[luben111]

the simplest way is to use relays. If the voltage is over the threshold the relay is ON and connects the sources in serial

 

[hshah8970]

luben111, thanks for replying!

Unfortunately I don't know how I'll be able to use a solid state relay to accomplish this task. Keep in mind that the control circuit (DC Input 2) would itself become part of the overall circuit.

But I'm only a student. Perhaps there's something you're seeing and I'm not. Thus I'd greatly appreciate a diagram or a schematic with an explanation.

Thanks in advance!

 

[mtwieg]

Are the two DC inputs isolated from each other?

Also why do you think you'll need a solid state switch? Does the switching time need to be that fast? One issue with this is that if you have any bypass caps on the DC output, they are going to see tremendous spikes of current when the switch is made. If you need smooth, controlled transitions in the voltage, then you're looking at a much more complicated system.

 

[rjkrocks]

Ok!
Let us begin to solve this problem.
Please note this is my concept of solving this and you ma found some more suitable solution.

Now let us have an understanding of your problem first. Based on this we can solve the complete problem.

Matter given.
1. you have two inputs, let us name it input 1 & input 2. Both are DC
2. You have a threshold voltage to input 2, so when the input 2 reaches that threshold level, it has to be added to the output with input 1. ie it is connected in series.

Hope this is the problem description.

I am describing two methods (only concept, for actual implementation u have to work on the values of component. you can do, becoz u r an engineer)

Method 1

From the problem description input 2 has to be compared with a threshold value. So a comparator circuit is needed. then if it is higher than the threshold, it has to be added with the input 1 which is already available at the output. For this we can use a switching circuit followed by Summing amplifier.
have a look at this rough schematic

when the input 2 voltage level at INV terminal is less than NI terminal, th transistor will be conducting nd the voltage available at the input will be only saturation voltage. approximately zero. so out put of summing amplifier will be input 1+ 0 = input 1

but when the voltage at INV terminal is greater than NI, the transistor won't conduct. do the complete input 2 voltage will be available at the input of summing amplifier. So at the output of summing amplifier will be input 1 +input 2

thus your problem is solved. Note u can use a relay too for switching.....

Method 2

this method depends on what kind of threshold voltages that you are using. consider the figures below.

Here we uses the simple principle of unidirectional flow. using a battery, we put the diode in a reverse biased state and when the threshold of input 2 becomes larger, the effect of the battery is canceled and it will be forward biased. thus the conduction occurs. this can be fed into the input of summing amplifier along with input 1 as shown in figure.

the problem with this arrangement is that the voltage needed to overcome the reverse bias can't be added to summing amplifier. in order to rectify this, use this instead of comparator section in method 1 ( so that it will turn on the switching transistors). Hope that these answers are helpful.

 

[hshah8970]

First I want to thank everybody who is helping me solve this problem.

@mtwieg: Yes these inputs are isolated from one another. Also, smooth controlled transitions aren't required. I need to get the job done.

@rjkrocks: I don't have any other power source available other than these 2 DC inputs. I am very weak in electronics but I have an idea how the summing amplifier could work. Please see the attached image and let me know if it's the right or wrong way to go.

In the diagram, let's assume the threshold voltage is 1.4V and that each diode will have a drop of 0.7V. But the problems are obvious: I don't know how to give the amp its Vcc because no other power source. As for adding the overcoming diode voltage to the summer, we can use this instead of the comparator, like you said.

Plus I've got a completely amateurish question: shall I simply join the negative terminals of both supplies to ground?

 

[rjkrocks]

I don't have any other power source available other than these 2 DC inputs. I am very weak in electronics but I have an idea how the summing amplifier could work. Please see the attached image and let me know if it's the right or wrong way to go.

In the diagram, let's assume the threshold voltage is 1.4V and that each diode will have a drop of 0.7V. But the problems are obvious: I don't know how to give the amp its Vcc because no other power source. As for adding the overcoming diode voltage to the summer, we can use this instead of the comparator, like you said.

Well if you gave the threshold voltage as 1.4v then you can use the series combination of the diode. So when the total voltage is above 1.4v only the diode will conduct. Note that, you are using the transistor as a switch ( by biasing it to the saturation point).so when there is 700mV at the base of the transistor only, it will conduct. (ie 0.7V) so. only one diode is needed. Because transistor junction will provide other 0.7v drop.

Please check the Method 1 diagram. here the input 2 dc voltage is given as its Vcc. since the transistor is a switch, when it becomes off, the input 2 dc (Vcc) will be available at the input of the summing amplifier.

I don't know how to give the amp its Vcc because no other power source.........

Note if you are talking about the power supply of summing amplifier op-amp, you should provide it. Note:LM 324 is a low power single power supply op-amp. In case of single power supply you can to have 3v to 32V. This can be obtained from a voltage regulator ( using zener diode, or 7803,LM 317 etc) from input 1, provided input 1 is having a minimum of 4 V. otherwise you have to design a power supply (may be from line, or from external circuit if you are integrating this as a part of other circuits.)

Plus I've got a completely amateurish question: shall I simply join the negative terminals of both supplies to ground?

This question is a good one. if you are using a single power supply, you can give the common point as ground (also the -ve). But if you are using a dual power supply, one having +ve, -ve , and a common point, then you can give +ve to +V, -ve to -V, and common point to the ground. ( common point is the point that is having zero potential.) if you are using 78xx for +ve supply, it has its own ground. Similarly 79xx (-ve regulator) has its own ground. these ground can be joined together. as you see, there is no need to join the -ve terminal to ground. because it is needed for our ckt and doing so may result in short circuit. You can have your common point by adding equal +ve and -ve voltage. and that can be connected to ground without risk. If you are a risk taking person, go ahead and check it by connecting both. I would advise the use of pot and a multimeter to check current and voltages regularly. if that are with in limits, then can connect it safely.

 

[hshah8970]

@rjkrocks: Thanks a lot for taking out the time to respond in detail! I appreciate it!

Please clarify the usage of DC Input 2 as Vcc. Shall I simply plug it into the +Vcc of the comparator? More specifically, this is the portion I wasn't able to understand "Please check the Method 1 diagram. here the input 2 dc voltage is given as its Vcc. since the transistor is a switch, when it becomes off, the input 2 dc (Vcc) will be available at the input of the summing amplifier."

@vimal: If you're still working on the schematic, I appreciate the time you're putting into this. Thanks!

 

[rjkrocks]

Please check the Method 1 diagram. here the input 2 dc voltage is given as its Vcc. since the transistor is a switch, when it becomes off, the input 2 dc (Vcc) will be available at the input of the summing amplifier.

when I read your previous post, I thought that you were worried about the Vcc (the collector supply voltage) of the transistor. That is why I told like that. But to guve the V+ and V- supply voltages to comparator, I have explained it in a section starting from "Note if you are talking about the power supply of summing amplifier.....

Hope that this clarified your confusion...