Transistor Voltage Regulator Advice

Thread Starter

staroflaw

Joined May 28, 2011
5
I am building a Voltage Regulator for my Laser driver boards.

This is what I have made.

LZ-PSU.png

The problem I am having is that it pulls a large amount of mA on start up.

(Load is -70mA / Pull from 15v supply is 400-500mA)
(Load is - 380mA / Pull from 15v supply is 1.2A)
Its very quick, but I can hear the PSU cracking and sparking.
If I run it on my bench top PSU it trips the fuse.

As its a laser I am powering. I don't want any current or voltage spikes going to the driver boards.

Am I missing something or is there a better design I should use?

-Info-

15V In
5V Out

3 Laser Driver boards each pulling a Max 500mA
I should also add that I am using analogue modulation on the lasers, So the Current will be fluctuating a lot.

Any help is welcome.


Thank you - Mick
 

steveb

Joined Jul 3, 2008
2,436
My guess is that Q1 is sourcing a lot of current on startup because the base is tied to a capacitor. This means that the base will be near ground potential initially (actually 0.7 V because of the protection diode on the bottom), and the emitter rises to 15 V faster than the base can keep up with it.

The protection diode itself might be adding to the trouble because it disconnects the ground on startup.

For how long does the high current last? Is it milliseconds, or many seconds, or even longer than than?
 
Last edited:

SgtWookie

Joined Jul 17, 2007
22,230
C1 should be to the left of R1.

There should be a resistor on the base of Q1; 4.7k or 5.1k would work.

The relay should have a diode & resistor in series connected in parallel with the coil, cathode towards positive.

Lasers require their current to be closely regulated, yet you are using voltage regulation. If the Vf of your laser changes over temperature (which is likely) you will probably subject your laser to more current than it is rated for, which will kill it.
 

Thread Starter

staroflaw

Joined May 28, 2011
5
For how long does the high current last? Is it milliseconds, or many seconds, or even longer than than?
Its only a few milliseconds.


SgtWookie

The current is regulated on the Driver (Offset & Gain). But only if the input voltage is locked at 5v (Red , Green)/ 9v (Blue)

If the input voltage differs. So will the Max current to the laser diode.
This is why its important that the Driver IN voltage dose not spike and remains at , or just under 5v/9v.

Here is a Block diagram of what I am trying to do.

LZ-Block.png


Also can you just sum up what the resistor on the base of Q1 will do, and what the diode & resistor on the relay is for.


Thank you all for your reply.
I appreciate your help.


Thank you - Mick
 
Last edited:

steveb

Joined Jul 3, 2008
2,436
Its only a few milliseconds.
So, it sounds like the capacitor on the base is the likely cause of the current spike. SgtWookie's advice would be a good start. The addition of a base current limiting resistor would be a good method to limit the current crudely. But even this is not enough, and his suggestion to move the capacitor from the base to the emitter side should allow the base voltage to keep up with the emitter voltage, thus preventing excessive forward biasing of the transistor on startup.

I'm not too comfortable with the protection diode at the bottom. It seems to take the load circuit off of a good DC ground. Is this really what you want to do? If you do want to do this, maybe you would prefer to go all the way and have an isolated load circuit. It's possible I just dont understand exactly what you are trying to do.
 

Thread Starter

staroflaw

Joined May 28, 2011
5
I'm not too comfortable with the protection diode at the bottom. It seems to take the load circuit off of a good DC ground. Is this really what you want to do?
No it was just for protection.

I just dont understand exactly what you are trying to do.
I want to have a regulated 5v that will never exceed 5v.
I dont need to worry about current regulation on the voltage regulator as current is regulated on the laser driver.

The input voltage to the laser driver is 5v.
I set the current limit on the laser driver to 400mA

As long as the input voltage to the laser driver does not exceed 5v, the current will stay at 400mA <.
If the voltage exceeds 5v the current will increase and destroy the laser diode.

I have moved C1 to the left of R1, That has stoped the large current pull on startup.


So having the diode on ground is causing a isolated load?


Thanks - Mick
 

steveb

Joined Jul 3, 2008
2,436
So having the diode on ground is causing a isolated load?


Thanks - Mick
No, I'm not saying that. Once power is on, the diode is forward biased and so it is not isolated. But, it might be making a poor grounding in your system, which may or may not be a concern. I can't say what you are doing is wrong, but it looks unusual. Normally, one would either make a fully isolated power supply, or try to tie the grounds together. Grounding through a diode creates a connection that is not so straightforward to understand. I just thought I'd mention it. If you think it does something useful and don't see any issues, that's fine. If you don't have a specific reason for putting the diode on the bottom side and you are just protecting against reverse polarity, you could put it in the top side.

I've seen grounding methods produce some unexpected results that don't always make themselves apparent right away. I'm not sure. Maybe SgtWookie could give his opinion about this? I trust his judgement more than my own on questions like this.
 

Thread Starter

staroflaw

Joined May 28, 2011
5
Yes I understand your way of thinking now.

Its been a steep learning curve with everything I have done in the past year, But I am having fun with it.
Going from Idea to Simulation to Breadboard to PCB. Its a great feeling.


Thank you for all your time and input.

If you would like to see more of what I am doing take a look here - > http://www.leeds49.co.uk


Thanks - Mick
 

Ron H

Joined Apr 14, 2005
7,063
The capacitor on the base was momentarily saturating the transistor, which applied 15V to the output of the 7805. You are lucky that you didn't destroy the 7805.
 

SgtWookie

Joined Jul 17, 2007
22,230
There are a number of folks that I'm trying to help at the moment, along with actually trying to get some things of my own accomplished (the nerve!) but you seem to need a much better regulation scheme than you have at the moment.

The 7805 regulators will vary by a significant amount (at least several percent) from regulator to regulator, and will also vary significantly over temperature.
 
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