Transistor to switch on negative pull down voltage - Page 1
Forum section: electronics / beginners

author: Alex30
author_rank: Regular Contributor
timestamp: December 20, 2014, 02:23:38 pm
content: So say I have a signal normally at +5V and I have a negative voltage rail of -5V and I wanted to send a positive voltage signal in order to pull the +5V down to -5V, how do I do this? I am trying to use an NPN or PNP but I'm just confused as to how to set up the rails in order to get this to work? Thanks.

author: Simon
author_rank: Unknown
timestamp: December 20, 2014, 02:26:33 pm
content: You need to provide much more information. It sounds like you are trying to signal a transistor connected to an input that has a pull up resistor but your description is very confusing. Perhaps a circuit diagram would help.

author: Alex30
author_rank: Regular Contributor
timestamp: December 20, 2014, 02:48:39 pm
content: Sorry. I have drawn a rough equivalent in LTSpice and it appears to work but I am pretty confused by what I measure and am unsure if this should work in real life. I am using the opamp as a comparator and just pulsing it on and off. I measure +15V -> -5V at the output of the opamp as expected, but as shown on the graph I measure -4.98V-> -4.23V at the base of the NPN (first picture) but then it appears to actually pull down the load at the emitter to +5V -> -5V (second picture). What is going on here? Should this actually be working? Why am I reading weird voltages at the base?

author: Simon
author_rank: Unknown
timestamp: December 20, 2014, 02:59:51 pm
content: Can you explain what your trying to do. I think maybe your negative supplies need the + connected to GND

author: madires
author_rank: Super Contributor
timestamp: December 20, 2014, 03:06:51 pm
content: The voltage at the collector (-5V) is lower than at the emitter (+5V).

author: Alex30
author_rank: Regular Contributor
timestamp: December 20, 2014, 03:07:22 pm
content: Sorry I'm just not that good at explaining this one. The circuit I have drawn actually works in LTSpice. The voltage at the emitter is going between -5V and 5V as the comparator is turned on and off with the pulse wave shown at the inverting input of the opamp (which is generating a square wave of 0V -> 5V). At the output of the opamp the resulting voltage is going between +15V to -5V as expected. However, at the base of the NPN I don't measure a positive voltage, instead I measure a negative voltage that goes between -4.98V-> -4.23V as shown in the picture. I don't understand why this is so and am wondering if this circuit should be working at all, even though in LTSpice it is doing what I want.Edit:Essentially in my real circuit this is what I want to do. I have a comparator that will go high when I want the signal shown at the emitter of the NPN to go to -5V.

author: Alex30
author_rank: Regular Contributor
timestamp: December 20, 2014, 04:18:01 pm
content: Actually I think I have this figured out. When the voltage at the output of the comparator is -5V this causes the voltage at the base pin to be about -4.9V due to some leakage current. But the npn is turned off as the base sees -5V and the emitter is higher (+5V). But then when the comparator goes to +15V we have to consider this as a two step process. First the base is higher than the emitter (15V compared to 5V) so the transistor turns on. But then now that it is on the emitter sees -5V and there must be a ~0.8V drop over the PN junction which causes the base to be at about -4.2V. Thats where the numbers come from. Im satisfied this should work in real life nowthanks anyway hope I didnt confuse you all too much by my bad explanations

author: IanB
author_rank: Super Contributor
timestamp: December 20, 2014, 07:01:20 pm
content: I think you have confused yourself and us by not really explaining what it is you are trying to accomplish. OK, you have some input voltages and a comparator. The comparator will switch on the input voltages, so far so good.But now you have a transistor. What is the transistor for? Is it a power switch or a level switch? What is it switching?You have an asymmetrical +15 and -5 V supply on the comparator. Why do you need these voltages, and why asymmetrical? Nothing about the circuit shown suggests that a 0 to +20 V wouldn't work just as well.Why does the transistor have to be on the same supply as the comparator? If the transistor is switching a power device it might be more logical for it to be on a separate power rail.Your LTSpice diagram is all jumbled up and confusingly laid out. It makes it very hard to deduce the intended function. Clarity of mind and clarity of diagram often go hand in hand. It is no wonder you have been getting so confused.If you would try to explain what you are trying to achieve it would be much easier to help you.

author: Alex30
author_rank: Regular Contributor
timestamp: December 21, 2014, 02:24:23 am
content: Quote from: IanB on December 20, 2014, 07:01:20 pmQuote from: Alex30 on December 20, 2014, 03:07:22 pmIf you just ignore my example diagram, this is what I am trying to do. Not sure how to make it simpler than this explanation, admittedly my previous explanations were a bit confused, it was late when I started this thread and I was pretty tired.Anyway don't worry about it, I am satisfied this works.

author: IanB
author_rank: Super Contributor
timestamp: December 21, 2014, 03:07:39 am
content: Quote from: Alex30 on December 21, 2014, 02:24:23 amBut this is what I'm getting at. Todosomething implies some action, some result, something changed. Setting some voltage at some point in a circuit to some arbitrary value achieves nothing, does nothing. Without context, it is pointless. If that is all you want, it is most easily achieved by removing the transistor altogether and swapping the connections to the (+) and (-) inputs of the comparator. Presumably you haven't done that because you are trying todosomething with the transistor. But you haven't told us what, so everyone (including yourself) will remain in a fog.

author: IanB
author_rank: Super Contributor
timestamp: December 21, 2014, 03:36:00 am
content: To help illustrate the confusion, I have redrawn your circuit below in a way that makes it easier to follow. The first question would be, why is your transistor upside down? You have the emitter at a more positive voltage than the collector.

author: Alex30
author_rank: Regular Contributor
timestamp: December 21, 2014, 06:11:37 am
content: Thanks for taking the time to help. Even though I am fairly confident this should work perhaps you could still clarify it for me. The reason I don't want to provide my actual circuit is because it is very rough and hard to follow and would take a long time to redraw. The schematic I have presented is similar enough to my actual circuit that the problem presented here will translate to what I am doing, so it would be much easier for me and anyone who wants to help if you just assume that the line I have labelled as load in my new schematic is fixed and that it must use the comparator to change the voltage on that load line from 5V to -1V (I have since changed this requirement from -5V to -1V). Anything else can be changed provided these two elements stay in and the comparator inputs remain the same too.As shown in my schematic in the first picture I have redrawn to be similar to yours (which is much clearer) I have changed a couple of things. I have reversed the NPN to the right way around, I have added D1 at the output of the opamp to prevent reverse current, I have made the negative rail on the opamp go to ground and I have changed the voltage as mentioned above.My first question though, is why does it still work if the transistor is the opposite way around and the diode is removed? (Picture 2)My second question is how do you account for the gate voltage going between -220mV to -420mV? (Picture 3)Cheers

author: Zero999
author_rank: Super Contributor
timestamp: December 21, 2014, 02:10:59 pm
content: 1) Transistors do work when connected up backwards, just not very well. It may be different if you build it in real life.2) The comparator's output is at 0V and the emitter of the transistor is at -1V so the voltage on the base (not the gate, I think you're confusing a BJT with a FET) will be somewhere in-between the two.The diode isn't needed to prevent reverse current flow but to ensure the transistor turns off. Without it, the comparator's output will be at 0V and the emitter of the transistor at -1V which exceeds the necessary 0.6V required to forward bias the base-emitter junction.

author: Alex30
author_rank: Regular Contributor
timestamp: December 21, 2014, 03:45:37 pm
content: Quote from: Hero999 on December 21, 2014, 02:10:59 pmThanks and yeah I was getting mixed up between BJT and FET. I sort of had an inkling to what you described here, that's why I was fairly sure it would work but after you explained this it has confirmed it for me and connected the dots.Also what you said about the diode sort of makes sense. But if it needs to be at -0.4V or less to forward bias the junction, does that mean that in the current configuration I have the gate at a floating voltage? Do I need to do something here or is it okay to leave it in the current configuration?

author: IanB
author_rank: Super Contributor
timestamp: December 21, 2014, 04:15:44 pm
content: Your circuit would be easier to understand and analyzer if you arranged it like this: