more diode then diode.

[vlad777]

I measure square wave output with my voltmeter and I get middle value.(hopefully)
But when I add diode in series with voltmeter I get higher value?
Also transistor (base) connected to this diode wont open.
Diode is 1n4148. Or 1n4007.

Now if I ground diode output (cathode) with few kOhm resistor I get
normal reading.
This is like diode has inductance but I think it cant be that.

So why ungrounded diode output gives higher voltage?


Many thanks.

 

[Diver300]

The input capacitance of the voltmeter is storing the high voltage, but there is nothing to discharge the voltmeter when the square wave is low and the diode turns off.

 

[vlad777]

Sorry I wasn't clear. I measure with DC setting.
I don't actually know what is input circuitry of my voltmeter.
Also there's this transistor that won't go high, no cap involved just diode. (even at low 14kHz)
It works fine without the diode.

Sorry about the title, I should have named it: diode recovery.
Little later I remembered MOSFET datasheet that gives graph of diode recovery.
I found on the net that diode cant turn off immediately because electrons and holes
have to go back to their regions.

So if I have to use diode on my "clock" line, how do I get rid of recovery charge???

 

[RCinFLA]

There is R-C filtering on the DVM input. With a series diode it will be peak responding. Negative or lower voltage portion of square wave will not be discharging the capacitor on the DVM input filter due to the diode blocking any discharge.

Adding a resistor to ground will discharge the input filter cap during non-peak period of waveform.

 

[audioguru]

Transistors do not open and close. Doors do that.
Transistors turn on and turn off.

 

[crutschow]

So if I have to use diode on my "clock" line, how do I get rid of recovery charge???

Why do you need a diode in the "clock" line? What's its purpose?

 

[vlad777]

Why do you need a diode in the "clock" line? What's its purpose?

Kind of easyest way to do TTL to transistor is to put a diode at it's base, but I think I'll use schottky diode.

 

[Diver300]

If you are turning a transistor on and off with the square wave, you don't want a diode. The low level will help turn the transistor off.

If you leave the diode, you risk getting the same problem as you are getting with the DVM.

 

[MrAl]

Kind of easyest way to do TTL to transistor is to put a diode at it's base, but I think I'll use schottky diode.

Hi,

Where did you get that idea from?

 

[vlad777]

Hi,

Where did you get that idea from?

Well TTL low is up to 0.8V so to be sure I add a diode on base.
Why is it not needed?

 

[Diver300]

Well 0.8 V is the maximum. The minimum is probably negative, but it's normal to aim for as near to zero as possible.

Where does this square wave come from?

What TTL input are you trying to connect it to?

I suspect that you are over-complicating things. You may be able to connect the square wave directly to the TTL input. If you post a circuit and more description we may be able to help more.

 

[MrAl]

Well TTL low is up to 0.8V so to be sure I add a diode on base.
Why is it not needed?

Hello again,


I think you are confusing the input and output specifications for TTL.

The TTL *input* logical low requirement is 0.0 to 0.8v roughly, while the TTL *output* is 0.0 to 0.4v roughly. That means that for the specified fan out for the particular TTL chip you are using the output should stay under 0.4v when the output is in the logical low (0) state. Since the transistor input during this state will be fairly high impedance, you should not need a diode to drop the low level output because it will be low enough already and only have to drive a tiny leakage current from the transistor unless there is also some sort of pullup resistor, but if there is a pullup then it must meet the spec's of the chip output also. Also note that that 0.4v spec would be with maximum fanout, so a single transistor base would be much less than that (NPN).

You should however use a resistor to limit the logical high level current out of the TTL gate pin and into the transistor base. Depending on your speed and gain requirements this resistor can probably be on the order of 470 ohms to 5k, but you'll have to make sure that it provides enough gain and fast enough switching speed also as well as limits current out of the pin.

 

[audioguru]

A TTL low has a max voltage of 0.8V only when it is sinking a lot of current to ground (16mA) which doesn't happen when it is driving only the base of an NPN transistor.
If you connect a series diode between the output of a TTL gate and the base of an NPN transistor then the TTL output drives as much current as it can when high and nothing turns off the transistor when the TTL output is low.

Connect a series resistor of 220 ohms to 1k ohms instead of the diode so that the transistor is turned on when the TTL is high and the transistor is properly turned off when the TTL is low.

 

[vlad777]

Its LPT port. I do connect limiting resistor but without a diode NPN is always ON.

About the diode I think I solved the problem.
I just put 100nF in parallel with the diode and it works up to 92kHz and maybe more.

I think amount of diode charge is too low to build up voltage on cap. U=Q/C .
And that charge goes to cap instead to my NPN.

 

[MrAl]

Its LPT port. I do connect limiting resistor but without a diode NPN is always ON.

About the diode I think I solved the problem.
I just put 100nF in parallel with the diode and it works up to 92kHz and maybe more.

I think amount of diode charge is too low to build up voltage on cap. U=Q/C .
And that charge goes to cap instead to my NPN.

Hello again,

That's interesting because the TTL spec is 0.4v max for an output low and that would be with full fanout. You could simply use a resistor pulldown on the base to emitter then of much higher value than your series current limit resistor. That should work pretty good without bringing the complications of capacitor loading into the picture too.