Could this circuit work?

[MstrKurt]

Hey guys i'm just wondering if the circuit attached could be changed a little to receive an infrared signal and use the signal for DC Motor Control?.

Any input is appreciated thanks.

 

[BobK]

Yes. Simply delete the circuit and make a new one. In all seriousness, that circuit is pretty much useless. You could eliminate most of the components and it would do the same thing, turn on an LED with a switch.

Bob

 

[Harald Kapp]

What's this circuit supposed to do anyway?

Here is a simple IR receiver.

 

[MstrKurt]

Haha, cheers guys was just a circuit I was trying out that I found online. The LED doesn't switch on unless there's a signal coming in from the Diode with a current supply acting as a IR Receiver. (All things i've read online)

 

[KrisBlueNZ]

The circuit shows the diode as a 1N4001, which is just a standard silicon diode. It doesn't produce any kind of signal. You would need to use a photodiode if you want the circuit to do anything useful.

 

[Harald Kapp]

I thinks that's what the 20µA current source is for. By not telling us the whole truth he fooled us here.

Harald

 

[gorgon]

You will never get 20uA through there, since the 1M resistor will limit it to 3uA(@3V). (The symbol is also wrong, indicating a voltage generator.)
The hFE is typical 270 for this transistor and I can't see that the LED will more than glow if turned on at all, in this design. With a Vf of the LED in the 1.7V range, the headroom for the base voltege is only 0.6V. The base current will then be 0.6uA, maybe slightly enough to turn the transistor and the LED on.

 

[Harald Kapp]

You will never get 20uA through there, since the 1M resistor will limit it to 3uA(@3V

The currrent is supposed to flow into the base of the transistor. A base current of 20µA at hfe=270 will result in 5mA LED current.
Putting the LED and current limiting resistor into the collector circuit could improve the circuit since the current source (photodiode) then sees only Vbe.

Harald

 

[gorgon]

The currrent is supposed to flow into the base of the transistor. A base current of 20µA at hfe=270 will result in 5mA LED current.
Putting the LED and current limiting resistor into the collector circuit could improve the circuit since the current source (photodiode) then sees only Vbe.

Harald

So in effect this virtual 'current generator' is in fact a voltage generator of around 3V in addition to the current in question. A 5mA current through the 100 ohm resistor and the LED will remove 0.5V of the minimal headroom available.

I wonder how you manage to get 20uA through a 1M resistor at 3V, much less 0V? Are we talking solar cells here?

This is all from a practical point of view, we can all dream up circuits that 'should' work as we like it, but seldom do?

 

[Harald Kapp]

No, it is not a voltage generator. A photodiode generates a photocurrent. The voltage is only a consequence of the current flowing through a resistor (or in this case the LED and the base-emitter junction of the transistor).

As a thought experiment:
Remove R1 from the circuit. The current from the currrent source will flow into the base of Q1 and out of the emitter through the resistors and the LED back to the current source. This is a closed loop and the relevant laws are all satisfied.
This base current is amplified by the transistor's hfe and generates an additional emitter current hfe*20µA which is around 5mA and therefore sufficient to drive the LED.
The photodiode is operated in photoconductive mode (reverse biased). That's what the 1Meg resistor is for.

 

[gorgon]

No, it is not a voltage generator. A photodiode generates a photocurrent. The voltage is only a consequence of the current flowing through a resistor (or in this case the LED and the base-emitter junction of the transistor).

I just wonder where you get the 20uA from in this circuit. What you say is like pulling yourself up with the hair, speaking in a metafor.

If I understand you correctly, you say that the photodiode is biased from the 1M over 3V? The bias current can then only be max 3uA, if you drop all the 3V over it. To activate the LED(current through it) via the transistor, you need to have the base at a voltage potential above 0V, of at least the Vf of the LED plus the Vbe, or about 2.2 - 2.5V. This leaves a 0.5 to 0.8V left over the 1M, or 0.5-0.8 uA through it.

I can't see where the 20uA is flowing in this picture?

 

[(*steve*)]

Think of the diode as a very small solar panel.

 

[MstrKurt]

So do you think this circuit could be connected in someway to a H bridge circuit?

Where the H Bridge operates from the signals received from the IR Receiver (or in this case the Diode in parallel with the current source).

 

[Harald Kapp]

I just wonder where you get the 20uA from in this circuit. What you say is like pulling yourself up with the hair, speaking in a metafor.

No hair to pull at ...


The 20µA current source sits smack in the middle of the circuit. It is named I1-

 

[Harald Kapp]

or in this case the Diode in parallel with the current source).

To model the photodiode you don't need the diode in the circuit. The current source alone is sufficient. The diode seems to irritate more than to help understanding the circuit.

Where the H Bridge operates from the signals received from the IR Receiver

You can use the output of the transistor circuit as a control signal for an H-bridge. You cannot power the bridge from a photodiode.
Also I advise to add some safety measures to prevent the H-bridge from destruction.
It would be best to use the output from the IR-receiver as an input to a suitable H-bridge controller.

If that is your application, you'd be better of using a complete IR-receiver. See this thread.

Harald

 

[MstrKurt]

Thanks for the help guys.