# Working with 2 voltages across circuits

Discussion in 'The Projects Forum' started by burger2227, Feb 10, 2014.

1. ### burger2227 Thread Starter Member

Feb 3, 2014
192
24
I am working with a one battery Joule thief chip that can be controlled by the voltage on Pin 1. Voltage above .5 volts turns it off and lower voltage turns it on. Since a positive voltage would not turn it on, I added a PNP transistor to send positive voltage to the pin until a signal from a device turns the transistor off. Then the CDS can determine if it should light as shown in the circuit below:

I wanted to add motion detection to turn on the LED driver, but the PIR sensor module needs more than 3 volts to run properly. Although it sends about 2.4 volts to base of the PNP transistor, it is just turning it off. That part of the circuit has worked well.

Using the same logic for a low input signal, I figured that the reverse would work with a NPN transistor. Should I set the base to ground or the positive battery? I am afraid that a higher voltage through the 4.7 K resistor will end up on the battery bus as both circuits have the same common.

Thanks!

2. ### Alec_t AAC Fanatic!

Sep 17, 2013
5,980
1,138
Can you use a diode in series with your 'Logic low' input (diode anode connected to the 4.7k/1k junction)?
BTW, shouldn't there be a pull-down resistor on pin 1 of the IC?

3. ### burger2227 Thread Starter Member

Feb 3, 2014
192
24
Well I assume that the voltage from the other device would have to be positive if it is not low so I could just pull the voltage low and let the device run the transistor when it is high too. Then the transistor would block the voltage wouldn't it? That's what I'm doing with the PNP to shut it off. I want as few components as possible, but when I drew the NPN base connections it looked dangerous if the input voltage would take it up to 5 volts on the chip. That might put double the peak voltage on the LED's.

Pin 1 apparently has a pull down resistor built in as the chip is enabled when no connection is made to it. Normally Pin 1 monitors a solar panel's voltage to enable the LED's at night. It also sends charging current to the battery through an internal Shottkey diode to Pin 2. A CDS can also be used to switch the driver chip on.

The 5252F chip also does not allow the battery voltage discharge to go below .9 volts to preserve its life. It is rated up to 5 volts max but not the LED's. If there is no battery voltage or the battery is completely dead, the voltage on Pin 1 can run the LED's so I also want to avoid that by not sending outside voltage to the pin directly.

Thanks for your input, any other thoughts?

4. ### TANDBERGEREN Member

Jan 20, 2014
74
4
This cirquit will always give a voltage higher than .5V on pin 1 on the 5252F.
This because both transistors is biased to conduct.

Reverse the bias, and you would be good to go.

5. ### burger2227 Thread Starter Member

Feb 3, 2014
192
24
Yes, it would only work correctly as shown if there were a logic low voltage on the NPN and a logic high voltage on the PNP. I only use it as a reference for either case, normally not both.

If the CDS is not used, you could control Pin 1 directly if you want. I just didn't want an outside voltage overriding the photocell in this case. Also a higher voltage might be used by the chip to charge the battery and that may not be a great idea either.

That brings up another thing. If I put a diode with anode on Pin 1, will a high logic on the cathode put Pin 1 high even with an internal pull down resistor? I can test that pretty easily, but what do ya think?

Last edited: Feb 10, 2014
6. ### Alec_t AAC Fanatic!

Sep 17, 2013
5,980
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With or without the transistors present?

7. ### TANDBERGEREN Member

Jan 20, 2014
74
4
Why a diode with anode on pin 1?

8. ### burger2227 Thread Starter Member

Feb 3, 2014
192
24
I was thinking about a set up without transistors. The diode could block higher voltages from coming on to Pin 1 and putting it into charging mode. A low signal could pull Pin 1 down if the battery voltage is sent to Pin 1 to keep it off until there is a low signal. Only problem there might be that a low signal would pull power from the chip battery possibly.

Actually I like the transistor set up best as it relies on outside circuits to run the switching while not using the LED battery at all. The chip draws little current as long as Pin 1 is set high. Using the battery voltage to keep it high drains nothing until the logic circuit turns off the transistor.