Dual Bridge Rectifiers - Same Circuit

Thread Starter

blwngazkit

Joined Oct 13, 2011
11
I'm having an issue here.

In the attached schematic, I only want LED2 to light when the switch S1 is pressed.

As the circuit is currently drawn, LED1 is always lit since it's wiring is backfed and I obviously have a ground issue...

It's been 6 years since school and I haven't touched electronics since; I've obviously forgotten something.

Any help?



ETA: For simplicity, I've left out the pull-down resistors on the inputs of the AND gate.
 
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Thread Starter

blwngazkit

Joined Oct 13, 2011
11
Sorry, I didn't realize I'd left out the details on the components...

IC1 & 2 are 7805 regulators

C1 & 2 are 0.33uF caps

B1 & 2 are bridge rectifiers obviously

IC3 is a simple AND gate Vss & Vdd sourced from IC2

T1 is a 24VAC 30VA transformer
 

crutschow

Joined Mar 14, 2008
24,919
You cannot connect two bridge rectifiers like that without a common ground.

If you want to switch one of the voltages then use one bridge to power both regulators and just switch the input DC to one of the regulators.
 

Thread Starter

blwngazkit

Joined Oct 13, 2011
11
You cannot connect two bridge rectifiers like that without a common ground.

If you want to switch one of the voltages then use one bridge to power both regulators and just switch the input DC to one of the regulators.
I was trying to do it this way since the switch is actually not part of MY circuit, but a switch in another component.

The power is coming from a control device that has (3) outputs; 1)24VAC, 2) Common, 3) 24VAC switched.

Both 24VAC sources are from the same transformer and share the common.
 

elec_mech

Joined Nov 12, 2008
1,500
A couple of comments:

1) Tie your grounds together, specifically the ground from IC1, IC2, and IC3. Otherwise you could have a floating voltage or voltage difference between grounds.

2) SW1, when open, allows power to go into IC2 but not IC1. Is this intentional?

If yes, I assume you're using SW1 to turn input to pin 1 on IC3 high and low?

If yes again, your problem should be solved by tying the grounds together.

A better method would be to use a single bridge rectifier and voltage regulator and feed the 5V output directly to pin 2 of IC3 and put SW1 between the 5V output and pin 1.

However, I presume you're using two bridge rectifiers and two regulators for a reason, though I'm not sure why. Could you elaborate?
 

Thread Starter

blwngazkit

Joined Oct 13, 2011
11
A couple of comments:

1) Tie your grounds together, specifically the ground from IC1, IC2, and IC3. Otherwise you could have a floating voltage or voltage difference between grounds.

2) SW1, when open, allows power to go into IC2 but not IC1. Is this intentional?

If yes, I assume you're using SW1 to turn input to pin 1 on IC3 high and low?

If yes again, your problem should be solved by tying the grounds together.

A better method would be to use a single bridge rectifier and voltage regulator and feed the 5V output directly to pin 2 of IC3 and put SW1 between the 5V output and pin 1.

However, I presume you're using two bridge rectifiers and two regulators for a reason, though I'm not sure why. Could you elaborate?
I may have answered your question above. If not, let me know.
 

elec_mech

Joined Nov 12, 2008
1,500
Yes, I saw that 2 seconds after I posted.

Do tie the DC grounds together though - AC sources having the same common does not mean DC power generated from them are also shared.
 

Thread Starter

blwngazkit

Joined Oct 13, 2011
11
A couple of comments:

1) Tie your grounds together, specifically the ground from IC1, IC2, and IC3. Otherwise you could have a floating voltage or voltage difference between grounds.

2) SW1, when open, allows power to go into IC2 but not IC1. Is this intentional?

If yes, I assume you're using SW1 to turn input to pin 1 on IC3 high and low?

If yes again, your problem should be solved by tying the grounds together.

A better method would be to use a single bridge rectifier and voltage regulator and feed the 5V output directly to pin 2 of IC3 and put SW1 between the 5V output and pin 1.

However, I presume you're using two bridge rectifiers and two regulators for a reason, though I'm not sure why. Could you elaborate?

Further info:

- BR1 & IC1 are switched from a separate component which outputs 24VAC.
- BR2 & IC2 are 'hot' constant fed from the same transformer that outputs the switched 24VAC.
- Effectively IC3 pin2 is always '1' and pin1 switches from '0' to '1' based upon the other device outputting 24VAC.

I realize it would be simpler to only have the single bridge and regulator to control the LED. My intention is to expand this circuit and the expansion would not allow this.
 
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Thread Starter

blwngazkit

Joined Oct 13, 2011
11
Yes, I saw that 2 seconds after I posted.

Do tie the DC grounds together though - AC sources having the same common does not mean DC power generated from them are also shared.

When I tied the grounds together, the second rectifier was showing voltage at the outputs with AC switched OFF.

B2: 27.45VAC in = 23.20VDC out

B1: (ON) 27.45VAC in = 23.20 out | (OFF) 0VAC in = 15.3VDC out

If I add a diode between the grounds, the voltage at B1 when OFF is 1.4VDC.
 

elec_mech

Joined Nov 12, 2008
1,500
So, with SW1 off, 0VAC is measured between the B1 inputs, but 15.4VDC is measured at the outputs?

What is the output voltage of IC1 when SW1 is off?
 

Thread Starter

blwngazkit

Joined Oct 13, 2011
11
As it's drawn (no diode like I was experimenting with):

SW1 = Open

B1: 0.452VAC on the input side | 15.?VDC on the output

IC1 Vout = 2.53VDC
 

Thread Starter

blwngazkit

Joined Oct 13, 2011
11
FYI, this problem is only occurring when the ground is shared.

With the gate removed from the circuit for testing, isolated grounds resolves the issue.

Obviously I can't use isolated grounds in my actual circuit since there wouldn't be a reference point and I could over-voltage the AND gate.
 

elec_mech

Joined Nov 12, 2008
1,500
Hmm, I think the problem stems from the fact that SW1 is only disconnecting one of the two AC inputs to B1. I can't explain why connecting the grounds is having this effect, but somehow a partial power path is completed by having one AC input high and the grounds shared to B1.

So, for experimenting's sake, with the transformer de-energized, can you remove the other AC connection to B1 (the one not connected to SW1), turn SW1 off, reapply power and check the output of both B1 and IC1?

If the output is zero, then you'll need to replace SW1 with a DPST switch. If this is not an option, we need to find a different way to accomplish this task or find someone smarter than I (shouldn't be too hard). :rolleyes: What purpose is the AND gate serving - to determine if both transformers are outputting 24VAC? Is the second transformer, feeding B2, always on? Or do you need to verify this just in case it loses power?
 

Thread Starter

blwngazkit

Joined Oct 13, 2011
11
Yes, the AND gate is essentially to verify that there is power from BOTH sources.

B2 SHOULD always be ON.

SW1 is physically a SPST relay on a separate device and can't be changed. I suppose I could have an AC relay in my circuit controlled by SW1. (see below)

This isn't really ideal since it adds a relay and takes a large amount of space on a small PCB... As such, I'd like to avoid it!!!

 

elec_mech

Joined Nov 12, 2008
1,500
Hmm, that is a good idea. I could see where a relay would take up some real estate.

I'll have to give this some thought, but a quick and dirty way to do this is add a small AC-powered neon lamp and put a photoresistor or similar right next to it, ideally enclosing them together with some heat shrink. This way you've got an opto-isolator and you should be able to use it in your latest circuit and replace the relay.
 
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